Dimensions beyond Our Imagination
Lim Ju Boo
Synopsis:
Most scientists as well as ordinary people-in-the-street are only familiar with small numbers, common scales and measurements which they encountered in their daily lives. Most of us are only used to small measurements like a few meters, kilometers, kilograms, watts, joules and calories, cubic meters, and square meters.
In this talk, we shall deal with distances, mass, energy, speed and other dimensions well outside the tethers of our imaginations. This talk centres on the applications of simple mathematics in astronomy to give us some beautiful conclusions that is amazingly outside our understanding. We will take a ride outside this infinitesimal world of ours into the realms of the unimaginable.
The talk shall be well illustrated by lots of coloured slides. However, because of the large amount of information collected, and the limited amount of time available, it would not be possible to include all the information in the talk. Hence these short notes below shall complement the rest of the slides in the talk.
The Horrendous Scales
Speed of light = 299 792 458 ms -1
1 light year is approximately = 9.46 x 10 12 km = 9.46 x 10 15 meters = 9.46 x 10 18 mm (based on average year). This is the distance light takes to travel in a year
1 Astronomical Unit (AU) = 149,597,870.691 kilometers = 1.495978706 x 10 11 meters = 1.495978706 x 10 14 mm. This is 7.59 times the distance from Earth to the Sun
It takes light only 499 seconds (8.3167 minutes) to cross this distance
One light year = 63,239.7 astronomical units
The nearest known quasar is 2S 0241 + 622 discovered in 1978 is 800 million light years away
The number of sheets of papers needed to represent the distance to the nearest quasar = 800 million sheets, each sheet is 1 mm thick. Each 1 mm is equivalent to 1 light year
This means the nearest quasar 800 million light years away has to be represented by a column of papers stretching 800 km high, each sheet 1 mm thick representing a light year of 9.46 x 10 12 km.
The most distant quasar is OQ 172, discovered in 1973 by Lick Observatory. It lies 18000 million light years (1.7 x 10 23 km or 1.136 x 10 15 AU) away. If a light year of nearly 10 million, million km is drastically scaled down to just a single sheet of paper 1 mm thick, then the columns of papers has to pile up till it stretches out to 18,000 km into outer space. This is 7.59 times the distance from Earth to the Sun. Remember each 1 mm sheet is 1 light year across.
But if a single sheet represents 1 AU, then the column to quasar OQ 172 shall be more than 1136, 000,000 km long. If a plane could cross all that distance to OQ 172 at a constant speed of 1,000 kph, it will take the plane 1.94 x 10 16 years (nearly 20,000 million, million years) to arrive to the furthest quasar
The average Sun-Pluto distance is 39.4 AU = 5,894,156,102 km
But we scale down the Sun-Pluto distance as 1, then Quasar OQ 172 will stretch to 2.89 x 10 13 (28.9 million, million) times the Sun-Pluto distance
Total quantity of matter in the Universe
The amount of matter in the Universe is equal to the mass of Sun x total no. of suns in the Universe
= (2 x 10 33) x 10 x 22 = 2 x10 55 gm = 20 septendecillion gm
Estimated no. of particles (nucleons & electrons) in the Universe = 2.2 x 10 79 = 22 quinvigintillion
Size of atomic particles
Effective radius of an atom 10 -10 metres (10 -7 mm)
Effective radius of nucleus = 10 -14 metres (1 x 10 -11 mm)
Effective radius of an electron = 10 -15 metres (10 -12 mm)
Thus if an atomic nucleus is enlarged to the size of a Malaysian 10 cents coin 2 cm in diameter, the effective radius of the electron will be
2 cm ÷ (10 -14) x (10 -15) = 0.2 cm
The infinitesimal to the infinite:
Since the speed of light = 299 792 458 ms -1
Light will take only 6.67128 x 10 -18 of a second to cross the diameter of an atom
‘Resonance particles’ have a life span of 10 -23 of a second (ten trillionth of a trillionth of a second)
Resonance particle exists only sufficiently long enough for light to cross 2.99 x 10 -15 metres
This means it can cross only 1.495 x 10 -3 = 0.0015 % the diameter of an average atom in the life span of a resonance particle
But for a hydrogen atom, the time light takes to cross its diameter is 4.503 x 10 -19 second = 0.45 of an attosecond = 0.45 quintillionth of a second (1 attosecond = 10 -18 second)
Light takes 8.34 x 10 -24 seconds to cross the diameter of a proton or neutron, each roughly 2.5 x 10 -15 meters in diameter, or 2.5 quadrillionth of a meter.
The largest naturally occurring nucleus is that of uranium 238 which has 92 protons and 146 neutrons. Its diameter is 1.55 x 10 -14 metres so that it is 6.2 times as wide as a proton.
In any atom, the nucleus has about 1 / 100,000 the diameter of the atom of which it forms a part
If the atom were a hollow sphere, one could fill it with about 10 15 (1 quadrillion) nuclei.
The diameter of a hydrogen atom is 1.35 x 10 -10 meters or a little over a ten-billionth of a metre. More massive atoms are a little bit larger, but even the most massive atoms known is probably not wider than 8 x 10 -10 meters in diameter
This means that 405 hydrogen atoms can be squeezed into a cubic nanometer, and that the spindle tuber disease virus, the smallest known fragment of life can contain perhaps 75,000 atoms The spindle tuber disease in potatoes virus has a mass of 8 x10 -22 kg (8 x 10 -19 gram)
Energy Values:
The energy value of petrol = 35 048 096.18 Joules (8 374 853.25 calories) per litre.
1 Imperial gallon = 4.54609188 litres. One calorie is approx. equal to 4.18 Joules
Human daily food requirements is about 2800 kilocalories (1 kilocalorie = 1,000 calorie)
An average car consumes about 1 Imperial gallon of petrol every 35 miles (12.39 km / litre). This is 38,072,852 calories (159,144,523 Joules)
This is the energy equivalent to 13.59 times the daily food requirements of an adult
But the Sun pours out 3.83 x 10 26 watts of energy through hydrogen fusion into helium. This is equivalent to 3.83 x 10 26 J sec -1 or 3.83 x 10 33 ergs / sec (nearly 4000 nonillion ergs sec -1). 1 Joule = 107 ergs
This is the energy value of 1.092 x 10 19 litres of petrol. Hence if we could put in all the energy the Sun releases in just one second into our petrol tank instead of using petrol, then it could push a car round and round, and round this Earth up to (1.092 x 10 19) ÷ 12.39 = 8.8 x 17 (880 quardrillion) km. This is 2.2 million, million times round the Earth since the equatorial circumference round the Earth is 2πr = 2 π x 6378.137 km = 40,075 km
Mass & Speed:
If light could bend round the curvature of Earth, it could whip round the equatorial circumference 7.48 times in one second
The mass of the Sun is 1.9891 x 10 30 kg (nearly 2 nonillion kg)
Mass of Earth is 5.974 x 10 24 kg (nearly 6000 sextillion kg)
Hence the mass of the Sun is 332 960 times as massive as Earth
The diameter of a human hair ranges from 1.27 x 10 -4 m to 2.54 x 10 -5 m (127 -25.4 microns)
A car speeding at 100 kph (27.78 m sec-1) crosses the diameter of a human hair 75 microns (0.075 mm) in just 2.7 x 10 -6 second.
This is the time sufficient for even light to travel only 809.439 metres. Light is 10 792 528.49 times faster than a 100 kph car.
But light takes only 2.5 x 10 -13 second (0.25 of a picosecond) to cross the breath of a human hair 75 micron in diameter
Hydrogen atoms permeate the entire diameter of the Universe. The diameter across the width of the hydrogen atom is taken as the smallest unit reference of length in the entire universe since the hydrogen atom is the symbol of the commonest element in the entire Universe.
Size of atomic particles:
Effective radius of a hydrogen atom is 10 -10 metres (10 -7 mm)
Effective radius of nucleus = 10 -14 metres (1 x 10 -11 mm)
Effective radius of an electron = 2.81794 x 10 -15 metres (2.81794 x 10 -12 mm)
Effective volume of an electron = 9.373 x 10 – 44 m-3
Effective volume of an atomic nucleus = 4.189 x 10 -42 cubic metres
Hence the volume of an average atomic nucleus is 44.7 times that of the electron
Radius of Sun: 695 500 000 m
Volume of Sun = 4/3 π r 3 = 1.409 x 10 27 cubic metre
If the volume of an average atomic nucleus were to be enlarged to the size of the Sun, then the volume of its electron shall be 3.152675 x 10 25 cubic metres. (3.15 x 10 16 cubic km).
The radius of Jupiter, the largest planet in the Solar System, is 71,492,000 meters (71, 492 kilometers). Its volume is 1.53 x 10 24 cubic metres (1.53 x 10 15 cubic km). It is placed over a mean distance of 778,330,000 kilometers away. The electron shall be 20.6 times bigger than Jupiter
Earth mean distance from the Sun is 149,600,000 km. This means an electron is 5.2 as far away its nucleus as Earth is from the Sun. The volume of Sun is 921 times that of Jupiter.
But in the case of an average atom, if the nucleus were to be magnified to the volume of the Sun, its nucleus would be about 45 times larger than its electron, and it would be placed 6.96 x 10 12 metres (6,960,000,000 km) away from its nucleus at ground state. Pluto’s average distance from the Sun is 5, 913,520, 000,000 metres. This is 7.59 times farther away than Jupiter is. But our magnified electron would slightly further than the Sun-Pluto’s distance by 1.18 times.
Thus if an atomic nucleus is enlarged to the size of a Malaysian 10 cents (2 cm) coin, the distance of an orbiting electron at ground state will be = 2 /100 metres ÷ (10 -14) x (10 -15) = 200 cm (2 metres) away
The infinitesimal to the infinite
Since the speed of light = 299 792 458 ms -1
Light takes only 6.67128 x 10 -18 of a second to cross the diameter of an atom
‘Resonance particles’ have a life span of 10 -23 of a second (ten trillionth of a trillionth of a second)
Resonance particle exists only sufficiently long enough for light to cross 2.99 x 10 -15 metres
This means it can cross only 1.495 x 10 -3 = 0.0015 % the diameter of an average atom in the life span of a resonance particle
But for a hydrogen atom, the time light takes to cross its diameter is 4.503 x 10 -19 second = 0.45 of an attosecond = 0.45 quintillionth of a second (1 attosecond = 10 -18 second)
Light takes 8.34 x 10 -24 seconds to cross the diameter of a proton or neutron, each roughly 2.5 x 10 -15 meters in diameter, or 2.5 quadrillionth of a meter.
The largest naturally occurring nucleus is that of uranium 238 which has 92 protons and 146 neutrons. Its diameter is 1.55 x 10 -14 metres so that it is 6.2 times as wide as a proton.
In any atom, the nucleus has about 1 / 100,000 the diameter of the atom of which it forms a part
If the atom were a hollow sphere, one could fill it with about 10 15 (1 quadrillion) nuclei.
The diameter of a hydrogen atom is 1.35 x 10 -10 meters or a little over a ten-billionth of a metre. More massive atoms are a little bit larger, but even the most massive atoms known is probably not wider than 8 x 10 -10 meters in diameter
This means that 405 hydrogen atoms can be squeezed into a cubic nanometer, and that the spindle tuber disease virus, the smallest known fragment of life can contain perhaps 75,000 atoms The spindle tuber disease in potatoes virus has a mass of 8 x10 -22 kg (8 x 10 -19 gram)
The Atom and the Solar System
But in most cases, the nucleus of an average atom has a diameter of about 10 -15 meter, whereas the atomic diameter is about 10-11 meter. This can vary according the quantum of energy supplied to exit the atom. The electron can jump obit to a higher or lower level unlike the planets round the Sun according to how excited it is by the energy (heat or light) pumped into the atoms. Generally, the nucleus has a diameter 10,000 times smaller than the atom. The great amount of empty space in an atom can be illustrated by the following analogy.
An Analogy
Imagine the nucleus to be the size of a golf ball. Then on this scale the first electron shell would be about one kilometer from the golf ball, the second shell about four kilometers, the third nine kilometers and so on. If you find that hard to visualize let’s try another method. The full stop at the end of this sentence probably is about 1/2 a millimeter in diameter. If that dot represents the nucleus, then the electrons in the first shell would be orbiting with a diameter anything from a meter to 50 meters around the nucleus.
This distance varies from shell to shell, meaning the radius of the electronic orbit varies according to the amount of energy it has. Their orbits are similar to all the 9 planets of our Solar System except that the planets of our Solar System do not jump from orbit to orbit. The planets’ distances from the Sun change at different times of the year only because of their eccentricity as they execute an elliptical path round the Sun. Their orbital pathways are always the same.
In fact, the actual diameter of an atom is very small and it would require some two hundred million of them side by side to form a line a centimeter long.
Let’s look again at some data:
• The diameter of a single proton is 10-15 metres
• The diameter of a hydrogen atom is 10-10 metres
• The diameter of the universe by the latest estimates is at least 30 or 40 billion light years in diameter.
This means that light traveling at a velocity of 299 792 458 metres per sec that will take only:
• 3.34 x 10-18 seconds (3.34 quintillionth of a second) to cross the diameter of a hydrogen atom at ground state.
• 3.34 x 10-23 seconds (33.4 septillionth of one second) to cross the diameter of a single proton which is one of the smallest atomic particles.
• But will take 4 x 1010 (40 billion) years to cross from one end of the universe to the other
• Thus the Universe is 3.78 x 1023 km x 1000 ÷ 10-10 metres = 3.78 x 10 36
(3.78 undecillion or 3780 decillion) times the width of the hydrogen atom
The Ultimate of Time:
Probably the shortest time possible, is to use the diameter across the width of a hydrogen atom as the smallest distance, and the speed of light in a vacuum as the fastest of all speeds as reference yardsticks to measure space & time. From the above light will take 3.34 x 10-18 seconds (3.34 quintillionth of a second) to cross the diameter of a single hydrogen atom.
The Ultimate of Temperature:
In similar light, the ultimate of all theoretical possible temperatures can only be:
V = 0.158 √ (T ÷ m).
By using a series of equations, it can be shown that the above equation may also be expressed as:
T = 40 mv2
The factor 40 only holds if we use units of temperature in (degrees) Kelvin, and velocity in km / sec.
where, V = velocity of any atomic particles in km / sec (molecular / atomic velocity)
T = temperature in degrees Kelvin
M = mass of particles at rest
Let us set the value of ‘v’ (velocity of the molecules as a gas gets heated up) as the maximum possible speed of 299, 792 km / sec. – the speed of light. Applying this value into the equations, we will get what seems the maximum possible temperature.
When the temperature reaches higher and higher, the atomic particles move faster and faster until it reaches the velocity of light. Since nothing can travel faster than light, then the highest possible theoretical temperature can only be:
T max = 3.59 x 1012 = 3,600,000,000,000 (degrees) Kelvin (3.6 million, million K)
At last! It is not as simple as that! At searing temperatures of millions of degrees all molecules and atoms break down into mere particles. Fusion reactions between simple nuclei are possible so that complicated nuclei can be created. At even still higher temperatures (if this is theoretical possible), then the reverse is true, and all nuclei must break down into even simpler particles. This gives rise to further complications between creation / destruction of matter, temperature, pressure, dimension and time.
Mass Increase with Velocity
But by then, as the velocity increase to near that of light, the mass of the particles will also increase according to Fitzgerald-Lorentz contraction derived from Einstein Theory
M = m 0 / m ÷ [√ (1- v2 / c2)]
Where M= final mass, m 0 is the initial rest mass, v2 = velocity of the particle at rest, c2 = the velocity of light.
As the particles go faster and faster their mass get more and more massive, until their mass reach infinity at the speed of light. So when v = c, the equation tells us the final mass (M) becomes infinite - exceeds that of the mass of the Universe itself. That is not possible. This means that even the temperature of 3.6 million, million Kelvin can never be reached.
A Journey to the Edge of the Universe:
Let us imagine we have a plane than can drift through the immense intergalactic valley, end to end, an immense abyss of space spanning 40,000 million light years, or 3.78 x 1023 km (378 sextillion km) across.
For accomplish that the plane theoretically has to fly non-stop at a constant speed of 1000 kph almost 48000 million, million Earth years. Fancy that! I salute the super-pilot who can live and endure that kind of journey. To solve that, he may have to marry abroad, bear children over 1.6 x 1015 (1600 million, million) generations to take over the piloting once each child attains the age of 30 years. The span of 30 years is taken as one generation.
Better Idea:
This is not possible. So, I have a better idea. It is possible for the pilot’s sperm and his wife’s eggs be frozen in liquid nitrogen as they are left to drift into the frigid coldness and darkness of space where the temperature is almost 0 Kelvin. The pilot may remain back on Earth, or he may come along, but kept in suspended animation to follow his genes aboard. A robot is programmed to take over which will only be activated towards the end of the journey.
The awaken robot will then take out the sperm-ovum from deep freeze, fertilize them in oxygen-rich nutrient broth. The robot will then nurse the foetus-baby and bring him (them) up. The robot-nurse will teach the child the world and galaxy from where he came. It will then t each the human child his language, culture and civilization, and shown images of the world of his root.
It will tell and teach him or her, the purpose of his mission, their destination, his fate and destiny. The child’s tear-filled eyes will be shown pictures and images of his biological parents, their world, humans civilizations on Earth, its plants, trees, forest, animals and all other living things from the world they came. All the images of his world will be beamed towards the plane from the day he plane left Earth and the Solar System into the frigid reaches of extra-galactic dimensions
All scenes of Earth will be continuous in time frame such that the entire length of 40,000 million years of history from the beginning to the edge of this Universe shall be continuously shown.
The TV transmissions will not be broadcast the usual way. The present technology will ‘dilute’ the energy of the transmission over a wider and wider volume into the emptiness of space. The entire energy of the signals will have to be concentrated into just a very narrow beam towards the direction of the plane in order to focus the pictures clearly on arrival without being spread, diluted and weaken out
The transmission should be continuous so that there is no gap in time in receiving the images. Even then, towards the end of the journey, all the images would have been at least 40,000 millions years old for the child. He will perhaps never be able to learn the ‘current’ scenario of this Earth, except his origin, in a distant galactic world that has long faded into oblivion over the immensity of time and space.
Even for light this is a horrendously long journey. It will take even light traveling at 299792458 metres per second ÷ 1000 (to change into km / sec) x 60 sec (to change to km per min) x 60 min (to change to km per hour) x 24 hr in a day x 365.25 days in a year x 40,000, 000,000 solar years = 3.78 x 1023 km (378 sextillion km) to span this chasm.
In order to cross this chasm, light will have to take at least 4 x 1010 (40 billion) years to transverse from one end of the Universe to the other end. This is a horrendously bizarre distance.
We can only wish our very distant generation the very best of luck for their survival.
Saturday, April 24, 2010
How Far Can We See from A Cruising Height?
Dear Captain KH Lim,
Before I ask my question, I have done some home-work to come out with some data first to determine visibility of a place on ground from a height, and also the distances between two locations along the curvature of the Earth. My explanation first, and my question comes last below.
The Explanation:
The distance d in miles to the true horizon on earth seen from a plane or from any height is approximately:
First Equation: d = ?? (1.5 h), where h = height in feet of the eye (?? = square root)
This is a very simple formula which is applicable from most heights. Thus from an aircraft flying at 33,000 feet, then the distance to the true horizon seen by a pilot or a passenger is 222.485 miles (358 km) away.
Standing on a hill or tower of 100 feet high or even from an aircraft, the height (h) is much smaller than the equatorial radius (R) of the Earth of 3963.189 statute miles (6,378.135 km).
The exact formula for distance from the viewpoint to the horizon, applicable even for satellites, is:
Second Equation: d = ?? (2Rh + h2), where, (?? = square root)
d = distance is also the true distance to the horizon from a height
R = Equatorial radius of the Earth (6,378,135 metres)
Applying the second equation, the horizon seen from an aircraft flying at 10 km high would also be 357.299 km (222.0154 miles) away.
The calculations using the above two formulae give the straight-line distance from the plane cockpit, passenger pothole, or a hill to the horizon, and NOT the distance to the horizon along the ground which would be longer for low heights such as seen from a hill.
In the second equation, both the radius (R) of the Earth, and the height (h) of the observer must be given in the same units (e.g. kilometers), but any consistent units will work.
The above two formula for d is only for the straight line of sight distance to the object of view, say the horizon.
A different relationship involves the arc length distance s along the curved surface of the Earth to the bottom of object. In this case, we apply:
The Third Equation:
l = cos-1 [r ?? (r+h)] x [2?? r ?? 360]
(?= square root)
where,
l = the distance from the observer to the horizon along the curve (Great Circle) of the planet (Earth), along the ground
r (equatorial radius of Earth) = 6,378.135 km
h (height of plane) = 10,000 metres (10 km)
(r+h) = 6388.135 km
r ?? (r+h) = 0.998 434 597
Circumference of Earth (2?? x 6378.135 km) = 40075 km
2?? r ?? 360 = 111.3194559 km
Cos-1 [r ?? (r+h)] is the angle in degree between the observer and the horizon, measured from the centre of Earth = 3.206324 degrees
Thus at an altitude of 10 km (33.000 ft) or 6.25 miles, the typical ceiling altitude of an jet airliner, the actual arc length (l) along the curved surface of the Earth to the bottom of an object (say a town) is 356.926 km (221.78 miles) away.
The distances along the curvature of Earth for low heights less than that of a jet plane at 10,000 meters, it would be greater than that of a direct line-of-vision. The ceiling altitude of 10 km of a jet liner is about the limit where the straight line-of-vision of a pilot to the horizon slightly exceeds the actual arc distance measured on the ground. However, several calculations using varying measurements for the radius of the Earth showed that the differences were less than a kilometer provided the observer did not fly above that of a commercial jet plane.
The Limit of Comparative Distances:
At much further distances, say 150 ?C 300 km away as seen from a satellite, or even further out as seen by a space traveler from outer space, say from the Moon, and the extra distance due to the curvature of the Earth will become less and less significant compared to the greater and greater distances or ??height?? away from the Earth. As an observer recedes into outer space, the distance to the horizon continue to increase until it far exceeds even that of the entire circumference of the Earth, let alone just a small extra arc of the curvature. The observer could be light-seconds, light-minutes, or light hours away, while the entire circumference of Earth is just only 0.133 light second round, if light could whip around the Earth. .
But for low heights, such as from a small hill, a tall building, a low flying plane, or from a watch tower, the extra arc distance along the ground to the horizon would always be greater than that of a straight line vision from above.
Actually the visual horizon is slightly further away than the calculated visual horizon, due to the slight refraction of light rays due to the atmospheric density. There may be also the effect of mirage playing tricks on the eyes, lifting an invisible distant city above the horizon.
1 statute mile = 5,280 feet = 1,609 meters
Geographical Locations of Cities:
Location Longitude (E) Latitude (N)
Penang 100 0 15 50 25
Phuket 98 0 22 80 0
Kuala Lumpur 1010 41 30 9
Singapore 1030 51 10 17
Distances between Neighboring Cities
Using the above geographical coordinates, and applying spherical trigonometry or the Haversine formula, the shortest distances along the curvature of the Earth between Malaysia and neighboring countries are:
Kuala Lumpur - Penang: 297.9 km = 185.1 statute miles = 160.8 nautical miles
Kuala Lumpur ?C Singapore: 318 km = 198 miles = 172 nautical miles
Penang - Phuket (South Thailand): 354.64 km = 220.36 miles = 191.49 nautical miles
The Question Now:
Having calculated out the distances along the Great Circle of some of the towns and cities between Malaysia, Thailand and Singapore, and we have shown that are within the theoretical distances that can be seen from an aircraft flying 10,000 metres above the ground, can we actually see Penang and Singapore from Kuala Lumpur, or Phuket above Penang from a plane flying at 10,000 metres over these areas? We presume there is no blanket of cloud cover, and that the weather is perfect with visibility up to infinity?
Frankly, I have never been able to locate where the horizon is whenever I fly. All I saw were just sheets, and sheets of thick white clouds below, and they stretched as far away as I could see. I seek your valuable experience and your expert comments.
Thank you Captain.
JB Lim
Malaysia
Hi Dr JB Lim,
Thank you for the explanation on how to calculate the visibility of an object on the ground from a particular height. Yes, pilots have been using the same formula (First Equation) to work out the the line-of-sight distance, not so much to look at any particular object, but to find out the range when they could start to receive a VHF radio transmission.
You see, VHF (Very High Frequency) radio transmissions travel in a straight line and could not bend or follow the curvature of the Earth like the HF (High Frequency) transmission could.
The latest weather of a particular airport is provided by the ATIS (Aerodrome Terminal Information Service) transmistted on VHF. When I am flying at 35,000 feet, I know that at around 220 nautical miles, I would be able to receive the latest weather from the ATIS of an aerodrome.
In this sense, the formula you used to calculate the line of sight distance is useful. As to actually see a city from 220 nautical miles at 35,000 feet, it is not always possible to do so even if the visibility is up to infinity (due to traces of haze or the effect of oblique visibility).
Perhaps, at night, one may be able to see the lights of the cities on a good day when the weather is perfect, but otherwise, just like you, I would see sheets and sheets of thick white clouds below me most of the times! However, if the cities were in the range of around 100 nautical miles, I probably could see and recognise them in the horizon (crosschecking with the radar on board if needed!)
Before I ask my question, I have done some home-work to come out with some data first to determine visibility of a place on ground from a height, and also the distances between two locations along the curvature of the Earth. My explanation first, and my question comes last below.
The Explanation:
The distance d in miles to the true horizon on earth seen from a plane or from any height is approximately:
First Equation: d = ?? (1.5 h), where h = height in feet of the eye (?? = square root)
This is a very simple formula which is applicable from most heights. Thus from an aircraft flying at 33,000 feet, then the distance to the true horizon seen by a pilot or a passenger is 222.485 miles (358 km) away.
Standing on a hill or tower of 100 feet high or even from an aircraft, the height (h) is much smaller than the equatorial radius (R) of the Earth of 3963.189 statute miles (6,378.135 km).
The exact formula for distance from the viewpoint to the horizon, applicable even for satellites, is:
Second Equation: d = ?? (2Rh + h2), where, (?? = square root)
d = distance is also the true distance to the horizon from a height
R = Equatorial radius of the Earth (6,378,135 metres)
Applying the second equation, the horizon seen from an aircraft flying at 10 km high would also be 357.299 km (222.0154 miles) away.
The calculations using the above two formulae give the straight-line distance from the plane cockpit, passenger pothole, or a hill to the horizon, and NOT the distance to the horizon along the ground which would be longer for low heights such as seen from a hill.
In the second equation, both the radius (R) of the Earth, and the height (h) of the observer must be given in the same units (e.g. kilometers), but any consistent units will work.
The above two formula for d is only for the straight line of sight distance to the object of view, say the horizon.
A different relationship involves the arc length distance s along the curved surface of the Earth to the bottom of object. In this case, we apply:
The Third Equation:
l = cos-1 [r ?? (r+h)] x [2?? r ?? 360]
(?= square root)
where,
l = the distance from the observer to the horizon along the curve (Great Circle) of the planet (Earth), along the ground
r (equatorial radius of Earth) = 6,378.135 km
h (height of plane) = 10,000 metres (10 km)
(r+h) = 6388.135 km
r ?? (r+h) = 0.998 434 597
Circumference of Earth (2?? x 6378.135 km) = 40075 km
2?? r ?? 360 = 111.3194559 km
Cos-1 [r ?? (r+h)] is the angle in degree between the observer and the horizon, measured from the centre of Earth = 3.206324 degrees
Thus at an altitude of 10 km (33.000 ft) or 6.25 miles, the typical ceiling altitude of an jet airliner, the actual arc length (l) along the curved surface of the Earth to the bottom of an object (say a town) is 356.926 km (221.78 miles) away.
The distances along the curvature of Earth for low heights less than that of a jet plane at 10,000 meters, it would be greater than that of a direct line-of-vision. The ceiling altitude of 10 km of a jet liner is about the limit where the straight line-of-vision of a pilot to the horizon slightly exceeds the actual arc distance measured on the ground. However, several calculations using varying measurements for the radius of the Earth showed that the differences were less than a kilometer provided the observer did not fly above that of a commercial jet plane.
The Limit of Comparative Distances:
At much further distances, say 150 ?C 300 km away as seen from a satellite, or even further out as seen by a space traveler from outer space, say from the Moon, and the extra distance due to the curvature of the Earth will become less and less significant compared to the greater and greater distances or ??height?? away from the Earth. As an observer recedes into outer space, the distance to the horizon continue to increase until it far exceeds even that of the entire circumference of the Earth, let alone just a small extra arc of the curvature. The observer could be light-seconds, light-minutes, or light hours away, while the entire circumference of Earth is just only 0.133 light second round, if light could whip around the Earth. .
But for low heights, such as from a small hill, a tall building, a low flying plane, or from a watch tower, the extra arc distance along the ground to the horizon would always be greater than that of a straight line vision from above.
Actually the visual horizon is slightly further away than the calculated visual horizon, due to the slight refraction of light rays due to the atmospheric density. There may be also the effect of mirage playing tricks on the eyes, lifting an invisible distant city above the horizon.
1 statute mile = 5,280 feet = 1,609 meters
Geographical Locations of Cities:
Location Longitude (E) Latitude (N)
Penang 100 0 15 50 25
Phuket 98 0 22 80 0
Kuala Lumpur 1010 41 30 9
Singapore 1030 51 10 17
Distances between Neighboring Cities
Using the above geographical coordinates, and applying spherical trigonometry or the Haversine formula, the shortest distances along the curvature of the Earth between Malaysia and neighboring countries are:
Kuala Lumpur - Penang: 297.9 km = 185.1 statute miles = 160.8 nautical miles
Kuala Lumpur ?C Singapore: 318 km = 198 miles = 172 nautical miles
Penang - Phuket (South Thailand): 354.64 km = 220.36 miles = 191.49 nautical miles
The Question Now:
Having calculated out the distances along the Great Circle of some of the towns and cities between Malaysia, Thailand and Singapore, and we have shown that are within the theoretical distances that can be seen from an aircraft flying 10,000 metres above the ground, can we actually see Penang and Singapore from Kuala Lumpur, or Phuket above Penang from a plane flying at 10,000 metres over these areas? We presume there is no blanket of cloud cover, and that the weather is perfect with visibility up to infinity?
Frankly, I have never been able to locate where the horizon is whenever I fly. All I saw were just sheets, and sheets of thick white clouds below, and they stretched as far away as I could see. I seek your valuable experience and your expert comments.
Thank you Captain.
JB Lim
Malaysia
Hi Dr JB Lim,
Thank you for the explanation on how to calculate the visibility of an object on the ground from a particular height. Yes, pilots have been using the same formula (First Equation) to work out the the line-of-sight distance, not so much to look at any particular object, but to find out the range when they could start to receive a VHF radio transmission.
You see, VHF (Very High Frequency) radio transmissions travel in a straight line and could not bend or follow the curvature of the Earth like the HF (High Frequency) transmission could.
The latest weather of a particular airport is provided by the ATIS (Aerodrome Terminal Information Service) transmistted on VHF. When I am flying at 35,000 feet, I know that at around 220 nautical miles, I would be able to receive the latest weather from the ATIS of an aerodrome.
In this sense, the formula you used to calculate the line of sight distance is useful. As to actually see a city from 220 nautical miles at 35,000 feet, it is not always possible to do so even if the visibility is up to infinity (due to traces of haze or the effect of oblique visibility).
Perhaps, at night, one may be able to see the lights of the cities on a good day when the weather is perfect, but otherwise, just like you, I would see sheets and sheets of thick white clouds below me most of the times! However, if the cities were in the range of around 100 nautical miles, I probably could see and recognise them in the horizon (crosschecking with the radar on board if needed!)
How Much Heat is Needed to Vaporize Away all the Oceans on Planet Earth
Some Traumatic Thoughts
Running from Emergency Medical Care to Astronomy
How Much Heat Energy is needed to vaporize away all the oceans on Earth?
Facts, Calculations, Results & Figures
An Accident I Saw
Just this morning when I was driving along Jalan Batu Caves for a meeting, I saw a motorcyclist skidded on a sandy patch just as he was negotiating a bend at the traffic lights to join Karak Highway.
I felt ethically and morally duty bound to stop to help. I have neither gloves nor water in my car, and I am one who is very obsessed with washing my hands afterwards touching any drop of blood smeared on my hands after attending a bleeding patient. It is not just my fear of HIV or hepatitis B infections when handling infected blood, but my fear of even common skin infections such as Staphylococcal to Streptococcal infections. Of course it was silly of me as the risk of contacting these infections by touching a patient and not having gloves or water to wash my hands afterwards is very remote.
Infections caused by Streptococcus bacteria for instance can lead to streptococcal sore throat, also called strep throat. Then we also think of the risk of tetanus infection causing lockjaw, not to ourselves as doctors and health-care providers, but to the accident victim from the manure-rich soil particularly if the wounds are deep and penetrating. Of course this can easily be managed with an anti-tetanus serum (ATS) injection afterwards.
A Strep throat acquired in childhood for instance may lead to rheumatic fever and rheumatic heart disease later in life. It begins with a history of strep throat from streptococcal infection. The infection leads to bacterial endocarditis, a dangerous infection of the heart's lining or valves. The heart valves, especially the mitral valves are affected. It causes the blood to regurgitate from the ventricles back to the atrium at each ventricular contraction as the heart valves can neither fully open nor close. But that’s another story. Moreover, it is highly unlikely that I will ever get rheumatic mitral valve incompetence just attending to a bleeding patient. There is no connection at all I admit, or even there is, the probability statistically speaking is very remote indeed, perhaps a chance in a million. It is just my obsessive thoughts about getting infections from touching blood that separates my mind from reality.
But risk of all types of infections is always there. The pathogens are always in the air, water, soil, and even in your hands, food and clothes. Any health care provider working in a hospital is at risk. This includes even the patient, not just the doctor, the nurse or the paramedics. Infections that are acquired while a patient is in a hospital are referred to as nosocomial infections; a term derived from 'nosos' the Greek word for 'disease'. Nosocomial infections are diseases that we, as physicians and heath care professionals, give to our clients. Hospitals and clinics are places where sick people go with the expectation that they will get better. Unfortunately, there is a risk that clients may become infected because of their visits to these places.
After examining the motorcyclist, I found he had some lacerations on his arms and legs. There was no fracture as far as I could assess clinically. He is aware of the surroundings, respond to my questions. In short his LOC (Level of Consciousness) is a full 15 score on the GCS (Glasgow Coma Scale), and there was no evidence of neurological deficit, which would have been indicative of neuro-spinal injuries. His airways were clear, breathing normally, and his haemo (circulatory) dynamics were not compromised. He was not in shock After examining and assessing him through a primary and secondary survey, I was of the opinion it was not much of a life-threatening situation for him.
He had just some moderate lacerations and tear, moderate bleeding, which could well managed with pressure dressings, and later the lacerations closed with 6-8 sutures, and a prophylactic ATS injections. These could be done at any small private clinic, and there was no necessity of calling an ambulance to spare the ambulance from more urgent and emergency needs elsewhere. I could have easily driven him in my car to any nearby clinic (not necessary to an appropriate state-of-art hospital) to get some decent dressings for his wounds, if not for his bike, which he had to leave behind. So I proposed he pick up his bike and ride along, and that I shall follow him behind in my car just in case …?.
But to me, it was only a very minor event.. As I drove behind him, a far, far, greater fear ran through my mind. It was a thought how precious water is to our lives on this planet. Even little drops of water to wash my hands from blood and infected biological fluids suddenly became so obsessively urgent to me at that moment. What, I asked, as it ran through my mind, if the entire Earth is depleted of water. Could the oceans boil off, or could all the water in the oceans seep beneath the ocean floors if the floors opened up into a vast chasm generated by a colossal tectonic drift? What happens if the Greenhouse Effect comes to pass, and the ocean waters boiled off as superheated steam enveloping this entire Earth with a mantle I thought?
My mind transfixed from a very minor medical emergency in the street (a minor duty) to a much more fearful thoughts of grotesque dimension in astronomy which is my major interest of all the sciences, except in human physiology, pharmacology, human nutrition, and medical research.
Here what I fantasize. Read on, and follow my (science fiction) logic below:
Introduction to an Astronomical Nightmare
(Some basic data and information needed
for this Nightmare):
Radius of the Sun = 696 000 km
Volume of Sun = 4/3 r 3 = 1.4123 x 10 18 km 3
Equatorial Radius of the Earth = 6 378.140 km
Volume of Earth = 4/3 r 3 = 1.0868 x 10 12 km 3
Sun / Earth Volume Ratio = 1.3 million: 1
Hence the Sun is 1.3 million times bigger than the Earth in volume
Mass of Sun is 1.9891 x 1033 grams (1.9891 x 10 30) kg. This makes it 330,000 times more massive than the Earth. The Sun destroys itself at a rate of 5 x 10 9 kg sec –1 to 3.827 x 10 26 watts (3.827 x 10 26 joules of energy per second)
The Sun can maintain this current output of energy for about 5,000 million years more
The source of this solar energy is the proton-proton cycle in which hydrogen nuclei are converted to helium nuclei. Today, after more than 45000 million years of fusion in the core the concentration by mass of H 2 has been reduced from 75 % to about 35 %. Fusion is accompanied by a mass loss, which is converted, into energy
In the process of generating this vast amount of energy (4 x 10 26 watts), the heat and light is spread out into space. A tiny part of this heat is intercepted by Earth from an average distance (semi major axis) of 149.6 million km
In so doing, Earth receives 135.3 2.0 milliwatts / per cm (1.94 0.03 calories / cm 2 / minute. To put it in another way in the SI System, this is equivalent to 1353 watts per sq. metre. To express it another way, it is 1353 joules per second per square metre (1353 J s –1 m -1). This is called the Solar Constant.
Since the Radius of Earth is 6 378.140 km (6378 140 metres), and since the surface area of a sphere is 4r2, the surface area is 5.112 x 10 14 m –2 (approximately).
This means that, for every square metre of the Earth’s surface, 1353 watts x 5.112 x 10 14 m –2 = 6.917 x 10 17 watts (joules per second) will fall on it. This is spread out evenly day and night as the Earth rotates. The poles may be much colder than the equator for 6 months a year, but for the remaining 6 months of the year, it will receive the Sun’s energy continuously even at “night” Thus nearly about the same amount of energy is distributed over the Earth’s surface evenly over a long time frame.
The specific heat (also called specific heat capacity) is the amount of heat required to change a unit mass (or unit quantity, such as mole) of a substance by one degree in temperature
The specific heat capacity (abbreviated C, also called specific heat) of a substance is defined as the amount of heat energy (measured in Joules) required to raise the temperature of one kilogram of the substance by one Kelvin (K). The SI unit for specific heat capacity is the joule per kilogram Kelvin. Specific heat capacity is therefore heat capacity per unit mass.
The Kelvin (K) scale is a thermodynamic temperature scale, in which the lower fixed point is absolute zero, and the higher fixed point is the triple point of water at exactly 273.15 K. The melting point of ice based on the triple point is 273.15 K. Temperatures (t) on the Celsius scale can be converted to temperature T on the Kelvin scale: T/k = t/0C + 273.15. However, for practical purpose to simplify our calculations we shall still use the Celsius scale which most of us can understand better. Morever, we are not dealing with nano-physics of the world of atoms here.
The Specific Latent Heat of Vaporization is the amount of heat required to convert unit mass of a liquid into the vapour without a change in temperature
For water at its normal boiling point of 100 ºC, the latent specific latent heat of vaporization is 2260 kJ.kg-1. This means that to convert 1 kg of water at 100 ºC to 1 kg of steam at 100 º C, the water must absorb 2260 kJ of heat. Conversely, when 1 kg of steam at 100 º C condenses, it gives out 2260-kilo joules
Specific Latent Heat of ice = 3.4 x 10 5 J kg -1
Specific Heat Capacity of water (Cw) = 4.2 x 10 3 J kg –1 0 C -1
Latent Heat of Evaporation of water = 2.26 x 10 6 J kg -1
Heat needed to increase temperature of melted ice from 0 0 C to 0 C
= mil + miCw ( - 0)
There are approximately 400 glaciers and icebergs with a combined weight of (2.2067 x 10 19) kg
Density of ice = 917 kg/m3
Density of Liquid water = 1000 kg m-3
1 cubic km = 1000 m x 1000 m x 1000 m = 1 x 10 9 cubic metres
Since Density = Mass / Volume,
Therefore total mass of ice on planet Earth = density of ice x volume of ice = 917 kg x 24,064,000 cubic km x 10 9 (2.4064 x 1016 cubic metres) x 917 kg = 2.2067 x 1019 kg
(The above are the basic knowledge needed for my scientific nightmare. Now let me argue).
Average Temperature of All the Oceans:
The average temperature for all ocean waters is 3.51°C and its average salinity is 34.72 parts per thousand. For the ocean surrounding Antarctica (south of 55°), the average temperature is 0.71° and the average salinity is 34.65 parts per thousand. Of the major ocean regions, the North Atlantic is the warmest and saltiest (averages: 5.08°, 35.09 parts per thousand)
Source: Penguin
The Cold Dark Ocean Floors:
Sunlight cannot penetrate below a depth of about 660 feet, around the start of what's known as the bathyal zone (it ends where the water temperature drops to 4 degrees Celsius -- at about 6600 feet). Some fish and crustaceans at these depths are blind; other animals -- as many as half of the creatures in the deep oceans -- have become bioluminescent, producing their own light in specialized organs called photophores.
Without sunlight, there is no photosynthesis, and without phytoplankton to kick start the food web, animal life is sparse. Because of the scarcity of food in the deep sea, many fish have evolved bizarre adaptations to help them get what they can.
The greatest ocean depth has been sounded in the Challenger Deep of the Marianas, a distance of 10,294 m (35,798 ft) below sea level in the Pacific Ocean. It is located 338 km (210 miles) SW of Guam. It is the deepest at 10,294 metres (35,798 ft) known depression on the earth's surface. God only knows what lurks inside there.
Even the height of Mount Everest is only 8850 metres (29035 feet) high, which means the entire Mt Everest would be submerged into the Mariana Trench if it was placed there. We are unsure what are the temperatures of waters conceal in some of these awesomely deep trenches. Some of the ocean floors have vents and abyss where hot water may sprout out from underground volcanic activities. The hot water may dilute the remaining relatively cold masses of surrounding water. Then the hot and cold water may circulate around deep in the ocean floors. So the picture is very complicated. This makes calculations exceedingly complicated. Even if we do know all the information available, probably we may need a supercomputer to execute the equations containing all those ever-changing variables. But I think this is not necessary as we are looking at this planet as a whole, and not as a intricate changing fabric of physical variables. So we will use some oceanographic data that tells us that the average temperature of all the ocean waters on Earth is 3.51°C. We will adopt this figure for our calculations, and this estimate is not going to deviate very much from truth
Amount of Water on Earth
There is a lot of water on our planet. There are about 1 354 728 800 cubic kilometer (km3) of it. This water is spread over different kinds of places as given in table below:
Total amount of water in cubic kilometers (km3) (%)
Seas and Oceans 1 321 920 000 (97.57)
Ice-caps and glaciers 24 064 000 (1.77)
Deep groundwater 4 216 000 (0.31)
Shallow groundwater 4 216 000 (0.31)
Inclination and capillary water 68 000 (0.005)
Rivers 1 360 (0.0001)
Salt or brackish lakes 108 800 (0.008)
Fresh water lakes 122 400 (0.009)
Water-vapour in the air 12 240 (0.0009)
Total amount of solid water (ice) + liquid water = 1 354 716 560 km 3 This excludes all the water vapor in the air including all the clouds as they cannot be included in the calculation since they have already been “vaporized”
1 cubic km (km 3) = 1000 m x 1000m x 1000 m = 1 x 10 9 cubic metres (10 9 m 3)
We also need to realize that water fills ¾ of the Earth surface. So we can expect that ¾ of the heat delivered to Earth falls and warm up the oceans, the remaining heats up the dry land. So we may think we should not include this part of heat into the calculations? Are we right? Wrong! All the heat is now trapped, and the entire Earth is a heat trap. It is now like an oven in which a cake is being baked. It is now the total heat distributed and available everywhere, and not just some parts going into the oceans. The scenario then we will be entirely different from what is now where Earth receives the balance between what it receives, and what escapes back into space. This heat balance just equates so that Earth is not overheated at the moment. This heat balance also allows us to enjoy the cooler sea breeze in the day, and the warmer land air being discharged into the sea at night due to convection air currents. But this wind-heat dynamics over land and sea will no longer work when everything on Earth-land and oceans are all heated up share the same temperature. There may not be a difference in heat gradients by then. So the heat from the land through this heat-transfer mechanism is also heating up the oceans and seas.
The heat dynamics would be different unlike now. If it would still be the same as it is now, I would have taken that into consideration, and divided that up into the calculations. But I have not, because I cannot foresee there would be heat exchange between land and sea by then. The oceans will be actively boiling be then, and superheated steam will be everywhere – land sea, atmosphere, and some diffusing even into interplanetary space. The whole Earth is just heated up uniformly like an oven. There is just no temperature difference at all.
Stage One of Calculation (Ice on Earth):
Density = Mass x Volume
Since density of ice is 917 kg per cubic metre (917 kg m 3)
Therefore the total mass of ice on planet Earth = Density x Volume = 917 kg x (2.4064 x 10 16) m 3 = 2.207 x 10 19 kg
Hence, to melt 2.207 x 1019 kg of ice into water from 0 0 C – 0 0 C (same temperature), requires (3.4 x 105) x (2.207 x 1019) = 7.5 x 10 24 Joules
Stage Two (Changing All the Ice into Liquid Water):
To heat up 2.207 x 1019 kg of melted ice from 0 0 C to 100 0 C, requires;
Specific heat capacity of water (Cw) x weight of water in kg
= 4.2 x 103 J kg-1 x 2.207 x 1019 = 9.3 x 10 22 Joules
Stage Three (Changing All the Liquid Water into Gaseous Water):
To boil off 2.207 x 10 19 kg of water at 100 0 C completely into steam, requires:
Specific Latent Heat of Vaporization (of water) x Weight of water = 2.26 x 10 6 J kg-1 x 2.207 x 10 19 = 4.98 x 10 25 Joules
Therefore, to boil off 2.207 x 10 19 kg of ice-caps + glaciers + permanent snow completely into steam requires: (7.5 x 10 24) + (9.3 x 10 22) + (4.98 x 10 25)
= 5.74 x 10 25 Joules
Liquid Water (The Oceans, Seas, Rivers and Lakes):
The Average Temperature of all the oceans in the world is 3.51 0 C (see information earlier). Hence to raise this temperature to just boiling point (100 0 C) requires a temperature difference of 100 – 3.51 = 96.49 0 C
There are all in 1 330 652 560 cubic km (km 3) of liquid water on this planet. This is an estimated value. So we will round this up to 1330652560 x 10 9 = 1.33 x 10 18 cubic metres. (m 3). It includes all the oceans, seas, lakes, underground water, rivers, etc. Since the seas and the oceans make up the most waters (97.57 %) and they are at an average temperature of 3.51 0 C, the minor sources are assumed to be at the same temperature. This excludes all the ice. The calculation for ice has to be done separately, but the water vapour cannot be included in the calculation as it is already in the “gaseous” form and they are not going to be condensed again as we are going to heat up the entire planet after this.
One cubic metre of water = 1000 kg at 0 0 C, but we assume this is also true at 3.51 0 C. The difference is so insignificant that we can ignore it as we are dealing with astronomical figures.
Hence to bring the temperature of all the liquid waters (1.33 x 10 18 cubic metres) from 3.51 to 100 degrees Celsius requires:
(.1.33 x 10 18) x (4.2 x 10 3) J kg –1 x 96.49 0 C x 1000 kg =
5.39 x 10 26 Joules
Now to Change All the Liquid Water into Steam:
Now we need to change all that water (1.33 x 10 21 kg) at 100 0 C completely into steam
(1.33 x 10 21) kg x (2.26 x 10 6) J kg –1 =
3 x 10 27 Jolues
Finally, to vaporized all the masses of ice caps, glaciers, icebergs, permanent snow, plus all the liquid water on Earth completely into steam requires:
(5.74 x 10 25) + (5.39 x 10 26) + (3 x 10 27) =
3.6 x 10 27 Joules
This means it is 3100, 000 000, 000 000, 000 000, 000 000 (31000 million, million, million, million joules of heat energy from the Sun). In the American English language this Nightmare Figure is called
3.6 Octillion Joules
Just How Much is Lost?
The Sun destroys itself at a rate of (5 x 10 9) kg per second. In the process, it generates 3.827 x 10 26 joules of heat and light per second. We have seen it requires 3.6 x 10 27 joules of energy to vaporize all the oceans, seas, ice, lakes, river, and all the waters from Planet Earth (see calculation above). In order to supply this amount of energy to boil off all the waters, including all the glaciers and all the permanent ice on Earth, the Sun will have to destroy (5 x 10 9) (3.827 x 10 26) x (3.1 x 10 27) =
4.05 x 10 10 kg of matter
This will yield a stupendous 3.6 octillion joules.
The Second Nightmare!
Imagine 3.6 octillion joules of heat are needed to vaporize all the water on Earth until it is bone-dry to the ocean floors. And how long will that take?
(3.1 x 10 27) (3.827 x 10 26) =
8 seconds
Eight seconds, that’s all it takes if Earth were to be thrown into the Sun. And 1.3 Million Earths can jolly well be thrown into the Sun – A Colossal Lake of Fire that Will Burn with A Searing Temperature of
15 million °C
This nuclear fire will burn if not throughout
All Eternity, but for at least another
5,000 Million Years More
Save Your Souls if you ever get inside!
The Sun’s heat falls off inversely as the square of its distance. Hence, if Earth were to stay put in her orbit at a respectable mean distance of 149.6 million km away around the Sun, then that will take:
(3.1 x 10 27) joules (6.917 x 10 17) joules per second (watts) = 4 479 768 786 seconds
But one Solar Year = 365.25 = 60 x 60 x 24 x 365.25 = 31 557 600 seconds. Hence it will take 4 479 768 786 31 557 600 =
141.95 years to empty all the oceans
The Third Nightmare
This means, not a drop to drink, let alone a drop to wash my hands
Save my OCD (Obsessive Compulsive Disorders)
This obsessive No Water thought is astronomically far, far too much for me to bear.
Now anyone cares for a drink? Get it now before 150 years
Beginning From
The Era of
Total Green House Effect
An Era When No Heat Can Escape
From Planet Earth
This is a Major TRAUMA psychologically for me with my OCD and washing hands obsession at the touch of every drop of blood!
But that chap who skidded off his motorbike this morning was relatively was just a minor drop of blood in the ocean compare to this psychological trauma. It would be easier to handle a much bigger traumas and MVA (motor vehicle accidents) in our streets. But not this one when all the oceans all boiled off.
A Warning Comment
Not So Easy As That!
Having said all that, the scenario of what will actually happen, and the time it will for all the water to disappear from the surface of Earth is not a picture as easy to paint from some simple calculations as that. The picture is going to be far more complex. The time taken, and the amount of heat needed may differ as much as 10 fold or more (or less) ? Many things can happen which no one can predict. The only way to know is to wait until it happens, and then take measurements. That is the only actual practical way to monitor the progress. Using updated measurements every now and then is the only approach to evaluate and reevaluate the status again and again, and only after much observation can we forecast and project the trend theoretically using statistical analysis. There is no easy way out. That’s the only way to tell. We can’t assume that 150 years is all that is needed to dry up all the oceans after a total green house effect has taken place? It is not as simple as that.
However, the laws of nature and physics have to be obeyed no matter what the scenario. They will not change, and they will never change even if Earth came to an end. The amount of heat requires to melt ice, the amount of heat needed to raise a certain amount of pure water from a certain temperature to another temperature, and the amount of heat needed to boil off boiling water completely into steam under a certain a pressure are all governed by physical laws. These values are all fixed, and they can never change. It is on these physical laws that we are basing our calculations. That is provided other confronting factors such as changing pressures; water-vapour equilibrium, concentration of the solute, etc don’t interfere, and may alter the initial physical constants. So our estimates are based on simple straight forward physic that other factors within the system don’t change or interfere. If they don’t, then those results holds true, accurate, and predictable. But they probably will not be as far as we can see.
Samples of many of the things that can happen to greatly alter the calculations are given below:
The Current Status of the Atmosphere:
An ocean of air envelops the surface of the Earth. It is estimated that this mantle of air that surrounds the Earth weighs some 5000 million, million (5 x 10 15) tons which exerts an atmospheric pressure (force / surface area) of 1013 mill bars or 760 mm of mercury Since air can be compressed, it is not surprising to find that most of the mass of air is concentrated at the lower altitudes. A quarter of the mass of the atmosphere lies below 2,000 metres, half of the mass below 5,000 metres, and around 75 % or three quarter at an altitude of around the height of Mt Everest, which is 8850 metres (29035 feet).
Despite the mass of the atmosphere, it is actually just only 0.3 % of the mass of the water in the ocean, so we can imagine what happens if all the water and ice on the surface of Earth were to evaporate and mixed up with the air above.
We can of course easily calculate out the additional volume being added One kilogram of water at 00 C occupies a space of just one litre (density of water is 1 kg per litre), but if it was be brought to boil, and converted into steam and maintained at 100 0 C, then for every kilogram of it, it will occupy a space of 1.673 cubic metres, and exert a pressure of 101.3 kPa (1.013 bars). But that’s under ideal lab conditions.
Even using the present parameters of just 0.0009 % of all the water available on Earth floating in the air, the pressure, density, composition and temperature varies differently at varying attitudes. What if all the waters and ice were to evaporate and mixed up with the air, plus the vastly increased in carbon dioxide levels from total greenhouse effect? What will these parameters be, and how would they affect and change the atmospheric scenarios? Even then, we are only assuming that the temperature will be maintain at 100 0 C and no higher. The Sun’s energy being trapped into the altered atmosphere is not going to stop there. The continuous penetration of solar energy into the altered atmosphere will continue to expand the super-heat steam-air-carbon dioxide mantle to a state where no scientist can predict.
What then will be the nett effect on the balance of temperatures, pressures, densities, layers of atmosphere, composition among others? Even right now it is impossible to give a definite upper limit of our atmosphere. For instance, the “lapse-rate” meaning the rate of which temperature decreases with increases altitude above sea level is 1.6 0 C for every 300 metres. There maybe temporary “inversion” or increase in temperature at low level. But generally the lapse rate does not vary very much. At about an altitude of about 11 km the lapse rate dropped to an almost constant value of – 55 0 C. We see these changing values even for our normal and “stabilized” atmosphere currently. What if the whole scenario is upset by entirely new dynamics, with much more heat plus all the oceans waters added in.
At the present moment we have the lowest layer (troposphere) where all the normal clouds and weather occurs, and where commercial jets fly. After this, comes the stratosphere where the ozone layer is, and where supersonic, hypersonic and probably surveillance, spy and military jets cruise. This stratosphere extends up to some 30 km up before coming to the ionosphere laying some 150 –250 km above the Earth’s surface. This is the layer of ionic discharge where radio signals are reflected back to Earth.
Above the ionosphere lies the exosphere, which lies with the border of interplanetary space. It lies between 720 upwards to some 2,500 km where it merges with the interplanetary medium. Hence it is not possible to give a definite answer to the upper boundary to the atmosphere. There is just no sharp border. The mass and density decrease until it merges with the density of interplanetary space.
So what would the picture like if all the waters on Earth just evaporate away? How far will the atmosphere be? As said earlier, one kilogram of water at 00 C occupies a space of just one litre (density of water is 1 kg per litre), but if it was be brought to boil, and changed into steam, then an additional space of 1.673 cubic metres is needed, and this will exert a pressure of 101.3 kPa (1.013 bars). This is provided it is maintained at 100 0 C. But it is not ?
Then again we are assuming that it is clean fresh water from the oceans all the time as it is being boiled off. We forget as the oceans are being evaporated off, and no rain can come down to replace the loss, the salt get gets more and more concentrated, and as it does so, boiling point also rises. This means more and more energy is needed as the ocean waters get more and more concentrated. This result is longer and longer time for the oceans to dry up since the Sun’s energy is constant per unit area. This affects the results of the simplistic calculations above. Of course in the initial stages, the melting glaciers will dilute the concentration of the salts. But this cannot last too long as the ice is only 1.77 % of all the waters on Earth. Soon that too will have to evaporate off.
Even if all the water is now boiled off, initially we may think it would be easy to use just straight logic on how much addition volume will be added to the air using the water-steam conversion data. We can even determine the addition thickness the atmosphere would be generated by the additional steam and even the new surface area over the steam-air mantle. Calculations like that are very simple and straightforward, but in reality, the problem is far from being as easy as that.
First of all the atmosphere is greatly going to continue to expand due to the continuous injection of heat from the Sun. The surface area of the new water-air mantle will continue to change, and so are all the parameters within. This cannot be worked out using straightforward mathematics. The problem is going to be very, very complex indeed, as it has to take into considerations other variables that cannot be handled by simplistic mathematical methods. Second, we do not even know how would all these superheated mixtures of gasses, water, including the heavier carbon dioxide have on the thermodynamics of all these gasses. Carbon dioxide is added only from just from the burning of fossil fuels, but much large qualities of them are also released from the oceans due decreased solubility by increasing ocean temperatures. We know the solubility of gasses in water decreases with increasing temperature, but what happens when additional vast quantities are released into the atmosphere, and also when the volume of water decreases as dissolved gasses in the oceans are also discharged into the atmosphere.
The vicious cycle will even be more disastrous and complex then we can think, let alone attempt to calculate. My imagination tells me nothing will be left over this hard and hot planet. Nothing volatile like superheated steam and highly energetic gases can remain. They will all escape the clutches of the gravity of Earth. The velocity of escape from Earth is currently about 10,000 meters per second. Any highly heated particle, whether air or water molecules, will exceed that speed and they will be thrown out into space. The Earth will probably be a dead world, devoid of any, not just more water, but air as well. It will be just like we see them in the rest of the other planets in our Solar System.
We may be wondering if we could then treat this Earth as a whole as if it was some kind of a black body emitting some electromagnetic radiation such as infrared. We might then apply the laws of thermodynamics of “black body radiation” to determine just exactly how much of this radiation heat managed to escape from the planet to be in thermal equilibrium with the surroundings Heat radiation enclosed in an isolated cavity such as within the walls of the atmospheric mantle with a temperature T may be used. We are thinking of a hypothetical Earth that emits such radiation in conformity with the classical thermodynamic system. But this works in our case where the atmosphere may undergo adiabatic expansion of the radiation. where T V –1/3. From there we might even be tempted to work out Wein’s law, and Planck’s law of radiation. But can it apply? I am unsure, and I am not even going to try.
JB Lim
An Amateur Astronomer in Thought
Running from Emergency Medical Care to Astronomy
How Much Heat Energy is needed to vaporize away all the oceans on Earth?
Facts, Calculations, Results & Figures
An Accident I Saw
Just this morning when I was driving along Jalan Batu Caves for a meeting, I saw a motorcyclist skidded on a sandy patch just as he was negotiating a bend at the traffic lights to join Karak Highway.
I felt ethically and morally duty bound to stop to help. I have neither gloves nor water in my car, and I am one who is very obsessed with washing my hands afterwards touching any drop of blood smeared on my hands after attending a bleeding patient. It is not just my fear of HIV or hepatitis B infections when handling infected blood, but my fear of even common skin infections such as Staphylococcal to Streptococcal infections. Of course it was silly of me as the risk of contacting these infections by touching a patient and not having gloves or water to wash my hands afterwards is very remote.
Infections caused by Streptococcus bacteria for instance can lead to streptococcal sore throat, also called strep throat. Then we also think of the risk of tetanus infection causing lockjaw, not to ourselves as doctors and health-care providers, but to the accident victim from the manure-rich soil particularly if the wounds are deep and penetrating. Of course this can easily be managed with an anti-tetanus serum (ATS) injection afterwards.
A Strep throat acquired in childhood for instance may lead to rheumatic fever and rheumatic heart disease later in life. It begins with a history of strep throat from streptococcal infection. The infection leads to bacterial endocarditis, a dangerous infection of the heart's lining or valves. The heart valves, especially the mitral valves are affected. It causes the blood to regurgitate from the ventricles back to the atrium at each ventricular contraction as the heart valves can neither fully open nor close. But that’s another story. Moreover, it is highly unlikely that I will ever get rheumatic mitral valve incompetence just attending to a bleeding patient. There is no connection at all I admit, or even there is, the probability statistically speaking is very remote indeed, perhaps a chance in a million. It is just my obsessive thoughts about getting infections from touching blood that separates my mind from reality.
But risk of all types of infections is always there. The pathogens are always in the air, water, soil, and even in your hands, food and clothes. Any health care provider working in a hospital is at risk. This includes even the patient, not just the doctor, the nurse or the paramedics. Infections that are acquired while a patient is in a hospital are referred to as nosocomial infections; a term derived from 'nosos' the Greek word for 'disease'. Nosocomial infections are diseases that we, as physicians and heath care professionals, give to our clients. Hospitals and clinics are places where sick people go with the expectation that they will get better. Unfortunately, there is a risk that clients may become infected because of their visits to these places.
After examining the motorcyclist, I found he had some lacerations on his arms and legs. There was no fracture as far as I could assess clinically. He is aware of the surroundings, respond to my questions. In short his LOC (Level of Consciousness) is a full 15 score on the GCS (Glasgow Coma Scale), and there was no evidence of neurological deficit, which would have been indicative of neuro-spinal injuries. His airways were clear, breathing normally, and his haemo (circulatory) dynamics were not compromised. He was not in shock After examining and assessing him through a primary and secondary survey, I was of the opinion it was not much of a life-threatening situation for him.
He had just some moderate lacerations and tear, moderate bleeding, which could well managed with pressure dressings, and later the lacerations closed with 6-8 sutures, and a prophylactic ATS injections. These could be done at any small private clinic, and there was no necessity of calling an ambulance to spare the ambulance from more urgent and emergency needs elsewhere. I could have easily driven him in my car to any nearby clinic (not necessary to an appropriate state-of-art hospital) to get some decent dressings for his wounds, if not for his bike, which he had to leave behind. So I proposed he pick up his bike and ride along, and that I shall follow him behind in my car just in case …?.
But to me, it was only a very minor event.. As I drove behind him, a far, far, greater fear ran through my mind. It was a thought how precious water is to our lives on this planet. Even little drops of water to wash my hands from blood and infected biological fluids suddenly became so obsessively urgent to me at that moment. What, I asked, as it ran through my mind, if the entire Earth is depleted of water. Could the oceans boil off, or could all the water in the oceans seep beneath the ocean floors if the floors opened up into a vast chasm generated by a colossal tectonic drift? What happens if the Greenhouse Effect comes to pass, and the ocean waters boiled off as superheated steam enveloping this entire Earth with a mantle I thought?
My mind transfixed from a very minor medical emergency in the street (a minor duty) to a much more fearful thoughts of grotesque dimension in astronomy which is my major interest of all the sciences, except in human physiology, pharmacology, human nutrition, and medical research.
Here what I fantasize. Read on, and follow my (science fiction) logic below:
Introduction to an Astronomical Nightmare
(Some basic data and information needed
for this Nightmare):
Radius of the Sun = 696 000 km
Volume of Sun = 4/3 r 3 = 1.4123 x 10 18 km 3
Equatorial Radius of the Earth = 6 378.140 km
Volume of Earth = 4/3 r 3 = 1.0868 x 10 12 km 3
Sun / Earth Volume Ratio = 1.3 million: 1
Hence the Sun is 1.3 million times bigger than the Earth in volume
Mass of Sun is 1.9891 x 1033 grams (1.9891 x 10 30) kg. This makes it 330,000 times more massive than the Earth. The Sun destroys itself at a rate of 5 x 10 9 kg sec –1 to 3.827 x 10 26 watts (3.827 x 10 26 joules of energy per second)
The Sun can maintain this current output of energy for about 5,000 million years more
The source of this solar energy is the proton-proton cycle in which hydrogen nuclei are converted to helium nuclei. Today, after more than 45000 million years of fusion in the core the concentration by mass of H 2 has been reduced from 75 % to about 35 %. Fusion is accompanied by a mass loss, which is converted, into energy
In the process of generating this vast amount of energy (4 x 10 26 watts), the heat and light is spread out into space. A tiny part of this heat is intercepted by Earth from an average distance (semi major axis) of 149.6 million km
In so doing, Earth receives 135.3 2.0 milliwatts / per cm (1.94 0.03 calories / cm 2 / minute. To put it in another way in the SI System, this is equivalent to 1353 watts per sq. metre. To express it another way, it is 1353 joules per second per square metre (1353 J s –1 m -1). This is called the Solar Constant.
Since the Radius of Earth is 6 378.140 km (6378 140 metres), and since the surface area of a sphere is 4r2, the surface area is 5.112 x 10 14 m –2 (approximately).
This means that, for every square metre of the Earth’s surface, 1353 watts x 5.112 x 10 14 m –2 = 6.917 x 10 17 watts (joules per second) will fall on it. This is spread out evenly day and night as the Earth rotates. The poles may be much colder than the equator for 6 months a year, but for the remaining 6 months of the year, it will receive the Sun’s energy continuously even at “night” Thus nearly about the same amount of energy is distributed over the Earth’s surface evenly over a long time frame.
The specific heat (also called specific heat capacity) is the amount of heat required to change a unit mass (or unit quantity, such as mole) of a substance by one degree in temperature
The specific heat capacity (abbreviated C, also called specific heat) of a substance is defined as the amount of heat energy (measured in Joules) required to raise the temperature of one kilogram of the substance by one Kelvin (K). The SI unit for specific heat capacity is the joule per kilogram Kelvin. Specific heat capacity is therefore heat capacity per unit mass.
The Kelvin (K) scale is a thermodynamic temperature scale, in which the lower fixed point is absolute zero, and the higher fixed point is the triple point of water at exactly 273.15 K. The melting point of ice based on the triple point is 273.15 K. Temperatures (t) on the Celsius scale can be converted to temperature T on the Kelvin scale: T/k = t/0C + 273.15. However, for practical purpose to simplify our calculations we shall still use the Celsius scale which most of us can understand better. Morever, we are not dealing with nano-physics of the world of atoms here.
The Specific Latent Heat of Vaporization is the amount of heat required to convert unit mass of a liquid into the vapour without a change in temperature
For water at its normal boiling point of 100 ºC, the latent specific latent heat of vaporization is 2260 kJ.kg-1. This means that to convert 1 kg of water at 100 ºC to 1 kg of steam at 100 º C, the water must absorb 2260 kJ of heat. Conversely, when 1 kg of steam at 100 º C condenses, it gives out 2260-kilo joules
Specific Latent Heat of ice = 3.4 x 10 5 J kg -1
Specific Heat Capacity of water (Cw) = 4.2 x 10 3 J kg –1 0 C -1
Latent Heat of Evaporation of water = 2.26 x 10 6 J kg -1
Heat needed to increase temperature of melted ice from 0 0 C to 0 C
= mil + miCw ( - 0)
There are approximately 400 glaciers and icebergs with a combined weight of (2.2067 x 10 19) kg
Density of ice = 917 kg/m3
Density of Liquid water = 1000 kg m-3
1 cubic km = 1000 m x 1000 m x 1000 m = 1 x 10 9 cubic metres
Since Density = Mass / Volume,
Therefore total mass of ice on planet Earth = density of ice x volume of ice = 917 kg x 24,064,000 cubic km x 10 9 (2.4064 x 1016 cubic metres) x 917 kg = 2.2067 x 1019 kg
(The above are the basic knowledge needed for my scientific nightmare. Now let me argue).
Average Temperature of All the Oceans:
The average temperature for all ocean waters is 3.51°C and its average salinity is 34.72 parts per thousand. For the ocean surrounding Antarctica (south of 55°), the average temperature is 0.71° and the average salinity is 34.65 parts per thousand. Of the major ocean regions, the North Atlantic is the warmest and saltiest (averages: 5.08°, 35.09 parts per thousand)
Source: Penguin
The Cold Dark Ocean Floors:
Sunlight cannot penetrate below a depth of about 660 feet, around the start of what's known as the bathyal zone (it ends where the water temperature drops to 4 degrees Celsius -- at about 6600 feet). Some fish and crustaceans at these depths are blind; other animals -- as many as half of the creatures in the deep oceans -- have become bioluminescent, producing their own light in specialized organs called photophores.
Without sunlight, there is no photosynthesis, and without phytoplankton to kick start the food web, animal life is sparse. Because of the scarcity of food in the deep sea, many fish have evolved bizarre adaptations to help them get what they can.
The greatest ocean depth has been sounded in the Challenger Deep of the Marianas, a distance of 10,294 m (35,798 ft) below sea level in the Pacific Ocean. It is located 338 km (210 miles) SW of Guam. It is the deepest at 10,294 metres (35,798 ft) known depression on the earth's surface. God only knows what lurks inside there.
Even the height of Mount Everest is only 8850 metres (29035 feet) high, which means the entire Mt Everest would be submerged into the Mariana Trench if it was placed there. We are unsure what are the temperatures of waters conceal in some of these awesomely deep trenches. Some of the ocean floors have vents and abyss where hot water may sprout out from underground volcanic activities. The hot water may dilute the remaining relatively cold masses of surrounding water. Then the hot and cold water may circulate around deep in the ocean floors. So the picture is very complicated. This makes calculations exceedingly complicated. Even if we do know all the information available, probably we may need a supercomputer to execute the equations containing all those ever-changing variables. But I think this is not necessary as we are looking at this planet as a whole, and not as a intricate changing fabric of physical variables. So we will use some oceanographic data that tells us that the average temperature of all the ocean waters on Earth is 3.51°C. We will adopt this figure for our calculations, and this estimate is not going to deviate very much from truth
Amount of Water on Earth
There is a lot of water on our planet. There are about 1 354 728 800 cubic kilometer (km3) of it. This water is spread over different kinds of places as given in table below:
Total amount of water in cubic kilometers (km3) (%)
Seas and Oceans 1 321 920 000 (97.57)
Ice-caps and glaciers 24 064 000 (1.77)
Deep groundwater 4 216 000 (0.31)
Shallow groundwater 4 216 000 (0.31)
Inclination and capillary water 68 000 (0.005)
Rivers 1 360 (0.0001)
Salt or brackish lakes 108 800 (0.008)
Fresh water lakes 122 400 (0.009)
Water-vapour in the air 12 240 (0.0009)
Total amount of solid water (ice) + liquid water = 1 354 716 560 km 3 This excludes all the water vapor in the air including all the clouds as they cannot be included in the calculation since they have already been “vaporized”
1 cubic km (km 3) = 1000 m x 1000m x 1000 m = 1 x 10 9 cubic metres (10 9 m 3)
We also need to realize that water fills ¾ of the Earth surface. So we can expect that ¾ of the heat delivered to Earth falls and warm up the oceans, the remaining heats up the dry land. So we may think we should not include this part of heat into the calculations? Are we right? Wrong! All the heat is now trapped, and the entire Earth is a heat trap. It is now like an oven in which a cake is being baked. It is now the total heat distributed and available everywhere, and not just some parts going into the oceans. The scenario then we will be entirely different from what is now where Earth receives the balance between what it receives, and what escapes back into space. This heat balance just equates so that Earth is not overheated at the moment. This heat balance also allows us to enjoy the cooler sea breeze in the day, and the warmer land air being discharged into the sea at night due to convection air currents. But this wind-heat dynamics over land and sea will no longer work when everything on Earth-land and oceans are all heated up share the same temperature. There may not be a difference in heat gradients by then. So the heat from the land through this heat-transfer mechanism is also heating up the oceans and seas.
The heat dynamics would be different unlike now. If it would still be the same as it is now, I would have taken that into consideration, and divided that up into the calculations. But I have not, because I cannot foresee there would be heat exchange between land and sea by then. The oceans will be actively boiling be then, and superheated steam will be everywhere – land sea, atmosphere, and some diffusing even into interplanetary space. The whole Earth is just heated up uniformly like an oven. There is just no temperature difference at all.
Stage One of Calculation (Ice on Earth):
Density = Mass x Volume
Since density of ice is 917 kg per cubic metre (917 kg m 3)
Therefore the total mass of ice on planet Earth = Density x Volume = 917 kg x (2.4064 x 10 16) m 3 = 2.207 x 10 19 kg
Hence, to melt 2.207 x 1019 kg of ice into water from 0 0 C – 0 0 C (same temperature), requires (3.4 x 105) x (2.207 x 1019) = 7.5 x 10 24 Joules
Stage Two (Changing All the Ice into Liquid Water):
To heat up 2.207 x 1019 kg of melted ice from 0 0 C to 100 0 C, requires;
Specific heat capacity of water (Cw) x weight of water in kg
= 4.2 x 103 J kg-1 x 2.207 x 1019 = 9.3 x 10 22 Joules
Stage Three (Changing All the Liquid Water into Gaseous Water):
To boil off 2.207 x 10 19 kg of water at 100 0 C completely into steam, requires:
Specific Latent Heat of Vaporization (of water) x Weight of water = 2.26 x 10 6 J kg-1 x 2.207 x 10 19 = 4.98 x 10 25 Joules
Therefore, to boil off 2.207 x 10 19 kg of ice-caps + glaciers + permanent snow completely into steam requires: (7.5 x 10 24) + (9.3 x 10 22) + (4.98 x 10 25)
= 5.74 x 10 25 Joules
Liquid Water (The Oceans, Seas, Rivers and Lakes):
The Average Temperature of all the oceans in the world is 3.51 0 C (see information earlier). Hence to raise this temperature to just boiling point (100 0 C) requires a temperature difference of 100 – 3.51 = 96.49 0 C
There are all in 1 330 652 560 cubic km (km 3) of liquid water on this planet. This is an estimated value. So we will round this up to 1330652560 x 10 9 = 1.33 x 10 18 cubic metres. (m 3). It includes all the oceans, seas, lakes, underground water, rivers, etc. Since the seas and the oceans make up the most waters (97.57 %) and they are at an average temperature of 3.51 0 C, the minor sources are assumed to be at the same temperature. This excludes all the ice. The calculation for ice has to be done separately, but the water vapour cannot be included in the calculation as it is already in the “gaseous” form and they are not going to be condensed again as we are going to heat up the entire planet after this.
One cubic metre of water = 1000 kg at 0 0 C, but we assume this is also true at 3.51 0 C. The difference is so insignificant that we can ignore it as we are dealing with astronomical figures.
Hence to bring the temperature of all the liquid waters (1.33 x 10 18 cubic metres) from 3.51 to 100 degrees Celsius requires:
(.1.33 x 10 18) x (4.2 x 10 3) J kg –1 x 96.49 0 C x 1000 kg =
5.39 x 10 26 Joules
Now to Change All the Liquid Water into Steam:
Now we need to change all that water (1.33 x 10 21 kg) at 100 0 C completely into steam
(1.33 x 10 21) kg x (2.26 x 10 6) J kg –1 =
3 x 10 27 Jolues
Finally, to vaporized all the masses of ice caps, glaciers, icebergs, permanent snow, plus all the liquid water on Earth completely into steam requires:
(5.74 x 10 25) + (5.39 x 10 26) + (3 x 10 27) =
3.6 x 10 27 Joules
This means it is 3100, 000 000, 000 000, 000 000, 000 000 (31000 million, million, million, million joules of heat energy from the Sun). In the American English language this Nightmare Figure is called
3.6 Octillion Joules
Just How Much is Lost?
The Sun destroys itself at a rate of (5 x 10 9) kg per second. In the process, it generates 3.827 x 10 26 joules of heat and light per second. We have seen it requires 3.6 x 10 27 joules of energy to vaporize all the oceans, seas, ice, lakes, river, and all the waters from Planet Earth (see calculation above). In order to supply this amount of energy to boil off all the waters, including all the glaciers and all the permanent ice on Earth, the Sun will have to destroy (5 x 10 9) (3.827 x 10 26) x (3.1 x 10 27) =
4.05 x 10 10 kg of matter
This will yield a stupendous 3.6 octillion joules.
The Second Nightmare!
Imagine 3.6 octillion joules of heat are needed to vaporize all the water on Earth until it is bone-dry to the ocean floors. And how long will that take?
(3.1 x 10 27) (3.827 x 10 26) =
8 seconds
Eight seconds, that’s all it takes if Earth were to be thrown into the Sun. And 1.3 Million Earths can jolly well be thrown into the Sun – A Colossal Lake of Fire that Will Burn with A Searing Temperature of
15 million °C
This nuclear fire will burn if not throughout
All Eternity, but for at least another
5,000 Million Years More
Save Your Souls if you ever get inside!
The Sun’s heat falls off inversely as the square of its distance. Hence, if Earth were to stay put in her orbit at a respectable mean distance of 149.6 million km away around the Sun, then that will take:
(3.1 x 10 27) joules (6.917 x 10 17) joules per second (watts) = 4 479 768 786 seconds
But one Solar Year = 365.25 = 60 x 60 x 24 x 365.25 = 31 557 600 seconds. Hence it will take 4 479 768 786 31 557 600 =
141.95 years to empty all the oceans
The Third Nightmare
This means, not a drop to drink, let alone a drop to wash my hands
Save my OCD (Obsessive Compulsive Disorders)
This obsessive No Water thought is astronomically far, far too much for me to bear.
Now anyone cares for a drink? Get it now before 150 years
Beginning From
The Era of
Total Green House Effect
An Era When No Heat Can Escape
From Planet Earth
This is a Major TRAUMA psychologically for me with my OCD and washing hands obsession at the touch of every drop of blood!
But that chap who skidded off his motorbike this morning was relatively was just a minor drop of blood in the ocean compare to this psychological trauma. It would be easier to handle a much bigger traumas and MVA (motor vehicle accidents) in our streets. But not this one when all the oceans all boiled off.
A Warning Comment
Not So Easy As That!
Having said all that, the scenario of what will actually happen, and the time it will for all the water to disappear from the surface of Earth is not a picture as easy to paint from some simple calculations as that. The picture is going to be far more complex. The time taken, and the amount of heat needed may differ as much as 10 fold or more (or less) ? Many things can happen which no one can predict. The only way to know is to wait until it happens, and then take measurements. That is the only actual practical way to monitor the progress. Using updated measurements every now and then is the only approach to evaluate and reevaluate the status again and again, and only after much observation can we forecast and project the trend theoretically using statistical analysis. There is no easy way out. That’s the only way to tell. We can’t assume that 150 years is all that is needed to dry up all the oceans after a total green house effect has taken place? It is not as simple as that.
However, the laws of nature and physics have to be obeyed no matter what the scenario. They will not change, and they will never change even if Earth came to an end. The amount of heat requires to melt ice, the amount of heat needed to raise a certain amount of pure water from a certain temperature to another temperature, and the amount of heat needed to boil off boiling water completely into steam under a certain a pressure are all governed by physical laws. These values are all fixed, and they can never change. It is on these physical laws that we are basing our calculations. That is provided other confronting factors such as changing pressures; water-vapour equilibrium, concentration of the solute, etc don’t interfere, and may alter the initial physical constants. So our estimates are based on simple straight forward physic that other factors within the system don’t change or interfere. If they don’t, then those results holds true, accurate, and predictable. But they probably will not be as far as we can see.
Samples of many of the things that can happen to greatly alter the calculations are given below:
The Current Status of the Atmosphere:
An ocean of air envelops the surface of the Earth. It is estimated that this mantle of air that surrounds the Earth weighs some 5000 million, million (5 x 10 15) tons which exerts an atmospheric pressure (force / surface area) of 1013 mill bars or 760 mm of mercury Since air can be compressed, it is not surprising to find that most of the mass of air is concentrated at the lower altitudes. A quarter of the mass of the atmosphere lies below 2,000 metres, half of the mass below 5,000 metres, and around 75 % or three quarter at an altitude of around the height of Mt Everest, which is 8850 metres (29035 feet).
Despite the mass of the atmosphere, it is actually just only 0.3 % of the mass of the water in the ocean, so we can imagine what happens if all the water and ice on the surface of Earth were to evaporate and mixed up with the air above.
We can of course easily calculate out the additional volume being added One kilogram of water at 00 C occupies a space of just one litre (density of water is 1 kg per litre), but if it was be brought to boil, and converted into steam and maintained at 100 0 C, then for every kilogram of it, it will occupy a space of 1.673 cubic metres, and exert a pressure of 101.3 kPa (1.013 bars). But that’s under ideal lab conditions.
Even using the present parameters of just 0.0009 % of all the water available on Earth floating in the air, the pressure, density, composition and temperature varies differently at varying attitudes. What if all the waters and ice were to evaporate and mixed up with the air, plus the vastly increased in carbon dioxide levels from total greenhouse effect? What will these parameters be, and how would they affect and change the atmospheric scenarios? Even then, we are only assuming that the temperature will be maintain at 100 0 C and no higher. The Sun’s energy being trapped into the altered atmosphere is not going to stop there. The continuous penetration of solar energy into the altered atmosphere will continue to expand the super-heat steam-air-carbon dioxide mantle to a state where no scientist can predict.
What then will be the nett effect on the balance of temperatures, pressures, densities, layers of atmosphere, composition among others? Even right now it is impossible to give a definite upper limit of our atmosphere. For instance, the “lapse-rate” meaning the rate of which temperature decreases with increases altitude above sea level is 1.6 0 C for every 300 metres. There maybe temporary “inversion” or increase in temperature at low level. But generally the lapse rate does not vary very much. At about an altitude of about 11 km the lapse rate dropped to an almost constant value of – 55 0 C. We see these changing values even for our normal and “stabilized” atmosphere currently. What if the whole scenario is upset by entirely new dynamics, with much more heat plus all the oceans waters added in.
At the present moment we have the lowest layer (troposphere) where all the normal clouds and weather occurs, and where commercial jets fly. After this, comes the stratosphere where the ozone layer is, and where supersonic, hypersonic and probably surveillance, spy and military jets cruise. This stratosphere extends up to some 30 km up before coming to the ionosphere laying some 150 –250 km above the Earth’s surface. This is the layer of ionic discharge where radio signals are reflected back to Earth.
Above the ionosphere lies the exosphere, which lies with the border of interplanetary space. It lies between 720 upwards to some 2,500 km where it merges with the interplanetary medium. Hence it is not possible to give a definite answer to the upper boundary to the atmosphere. There is just no sharp border. The mass and density decrease until it merges with the density of interplanetary space.
So what would the picture like if all the waters on Earth just evaporate away? How far will the atmosphere be? As said earlier, one kilogram of water at 00 C occupies a space of just one litre (density of water is 1 kg per litre), but if it was be brought to boil, and changed into steam, then an additional space of 1.673 cubic metres is needed, and this will exert a pressure of 101.3 kPa (1.013 bars). This is provided it is maintained at 100 0 C. But it is not ?
Then again we are assuming that it is clean fresh water from the oceans all the time as it is being boiled off. We forget as the oceans are being evaporated off, and no rain can come down to replace the loss, the salt get gets more and more concentrated, and as it does so, boiling point also rises. This means more and more energy is needed as the ocean waters get more and more concentrated. This result is longer and longer time for the oceans to dry up since the Sun’s energy is constant per unit area. This affects the results of the simplistic calculations above. Of course in the initial stages, the melting glaciers will dilute the concentration of the salts. But this cannot last too long as the ice is only 1.77 % of all the waters on Earth. Soon that too will have to evaporate off.
Even if all the water is now boiled off, initially we may think it would be easy to use just straight logic on how much addition volume will be added to the air using the water-steam conversion data. We can even determine the addition thickness the atmosphere would be generated by the additional steam and even the new surface area over the steam-air mantle. Calculations like that are very simple and straightforward, but in reality, the problem is far from being as easy as that.
First of all the atmosphere is greatly going to continue to expand due to the continuous injection of heat from the Sun. The surface area of the new water-air mantle will continue to change, and so are all the parameters within. This cannot be worked out using straightforward mathematics. The problem is going to be very, very complex indeed, as it has to take into considerations other variables that cannot be handled by simplistic mathematical methods. Second, we do not even know how would all these superheated mixtures of gasses, water, including the heavier carbon dioxide have on the thermodynamics of all these gasses. Carbon dioxide is added only from just from the burning of fossil fuels, but much large qualities of them are also released from the oceans due decreased solubility by increasing ocean temperatures. We know the solubility of gasses in water decreases with increasing temperature, but what happens when additional vast quantities are released into the atmosphere, and also when the volume of water decreases as dissolved gasses in the oceans are also discharged into the atmosphere.
The vicious cycle will even be more disastrous and complex then we can think, let alone attempt to calculate. My imagination tells me nothing will be left over this hard and hot planet. Nothing volatile like superheated steam and highly energetic gases can remain. They will all escape the clutches of the gravity of Earth. The velocity of escape from Earth is currently about 10,000 meters per second. Any highly heated particle, whether air or water molecules, will exceed that speed and they will be thrown out into space. The Earth will probably be a dead world, devoid of any, not just more water, but air as well. It will be just like we see them in the rest of the other planets in our Solar System.
We may be wondering if we could then treat this Earth as a whole as if it was some kind of a black body emitting some electromagnetic radiation such as infrared. We might then apply the laws of thermodynamics of “black body radiation” to determine just exactly how much of this radiation heat managed to escape from the planet to be in thermal equilibrium with the surroundings Heat radiation enclosed in an isolated cavity such as within the walls of the atmospheric mantle with a temperature T may be used. We are thinking of a hypothetical Earth that emits such radiation in conformity with the classical thermodynamic system. But this works in our case where the atmosphere may undergo adiabatic expansion of the radiation. where T V –1/3. From there we might even be tempted to work out Wein’s law, and Planck’s law of radiation. But can it apply? I am unsure, and I am not even going to try.
JB Lim
An Amateur Astronomer in Thought
Dear Captain,
Okay, it was my mistake spelling your name wrongly. I was trying to type as fast as I could, and never went back to check on anything, not even on my own articles. I must have made a lot of other mistakes there also. Of course I know your name very well, and the very fact I spelt your initial one moment as LK, and then as KL only showed I was just careless with my fingers that’s all. I am not a professional typist you know. My thousand apologies to you.
I shall post this and your letter into my blog. But meantime I am cc a copy of this letter to the rest. Hope it is okay with you.
Where have you been flying over the last few months? One day I must buy a ticket specially to go to Hainan Island with you if you are still flying there. But if you are not flying anywhere soon, I shall phone you and we shall go for seafood in Port Klang. We shall take a KTM train to Port Klang, and then take a motorized sampan across the delta to Bagan Hylam where I know a Hainanese family who runs a seafood restaurant there. We shall not fly there in your Jumbo Jet unless you want to crash land there. Ha! ha!
I shall ring you first, and if confirmed you are free, I shall ring the family to prepare some good Hainanese sea food for us. There is no hairy crab there. Maybe you can fly me to West Lake in Hangzhou, China in your AirAsia to have that type of crab.
Cheers to you.
Regards
JB Lim
On Fri, Apr 23, 2010 at 10:28 PM, Capt KH Lim wrote:
Hi JB Lim,
Thanks for publishing my site. However, the link to my site is wrong. It is http://www.askcaptainlim.com/ Clicking my site (Now renamed to 'Just About Flying' instead of 'Ask Captain Lim' on your link only goes back to your Blog. (http://scientificlogic.blogspot.com/)
Further, you have mentioned my name as Captain LK Lim in your Blog and Captain KL Lim in this email. In fact, it is KH Lim (Lim Khoy Hing). Never mind, it's a slip of the fingers!J
All the best!
Lim Khoy Hing
Okay, it was my mistake spelling your name wrongly. I was trying to type as fast as I could, and never went back to check on anything, not even on my own articles. I must have made a lot of other mistakes there also. Of course I know your name very well, and the very fact I spelt your initial one moment as LK, and then as KL only showed I was just careless with my fingers that’s all. I am not a professional typist you know. My thousand apologies to you.
I shall post this and your letter into my blog. But meantime I am cc a copy of this letter to the rest. Hope it is okay with you.
Where have you been flying over the last few months? One day I must buy a ticket specially to go to Hainan Island with you if you are still flying there. But if you are not flying anywhere soon, I shall phone you and we shall go for seafood in Port Klang. We shall take a KTM train to Port Klang, and then take a motorized sampan across the delta to Bagan Hylam where I know a Hainanese family who runs a seafood restaurant there. We shall not fly there in your Jumbo Jet unless you want to crash land there. Ha! ha!
I shall ring you first, and if confirmed you are free, I shall ring the family to prepare some good Hainanese sea food for us. There is no hairy crab there. Maybe you can fly me to West Lake in Hangzhou, China in your AirAsia to have that type of crab.
Cheers to you.
Regards
JB Lim
On Fri, Apr 23, 2010 at 10:28 PM, Capt KH Lim
Hi JB Lim,
Thanks for publishing my site. However, the link to my site is wrong. It is http://www.askcaptainlim.com/ Clicking my site (Now renamed to 'Just About Flying' instead of 'Ask Captain Lim' on your link only goes back to your Blog. (http://scientificlogic.blogspot.com/)
Further, you have mentioned my name as Captain LK Lim in your Blog and Captain KL Lim in this email. In fact, it is KH Lim (Lim Khoy Hing). Never mind, it's a slip of the fingers!J
All the best!
Lim Khoy Hing
Friday, April 23, 2010
Ask Captain Lim and also Just About Flying websites
Dear relatives, colleagues and friends,
May I suggest you and visitors to this blog "Scientific Logic" of mine also visit my friend website? His name is Captain LH Lim, and he is a Senior Pilot with Air Asia. He has hosted a fantastic website which I always enjoy reading. In fact my first love is on aviation, or anything that moves at very fast speeds such as jet planes, super fast bullet trains (like those world fastest trains in China), nuclear particles, cosmic rays, radio waves, speed of light, etc. These ‘moving subjects’ are actually my first love. My love is not about medicine, food and nutrition, health-care or biomedical research, and those sissy subjects. They are very boring to me. Unfortunately these are areas where I am trained in, and I have already spent my entire professional career on them. I am now retired from them. Life is unfortunate.
My real interest:
My interest in aviation, astronomy, physics, especially particle and nuclear physics, mathematics, and rocket science started even when I was in school, and they have still not gone away. I have very little interest in biological sciences like nutrition, health care, medicine, and all the biomedical sciences you care to name. Unfortunately an ill wind kept blowing me against my direction and wishes. The powerful hurricane forced me to make an emergency landing into the wrong airport. So I landed up as a medical research scientist and not a pilot. I wanted to be a pilot like Captain KH Lim. That was why I find his website extremely fascinating, and I am still reading his new entries every now and then. Life is like that!
Captain Lim and Dr Lim website link:
You may access Captain Lim’s website by just keying in these words "Ask Captain Lim" and also "Just About Flying" into the Google Search Engine bar and you will get him there. I have also written a lot of articles on various subjects which are linked to Captain Lim's website. But you may also access them by merely keying into Google Search these words: "Dr JB Lim's Corner". However, if you are interested only on my other articles related to health and medicine, then please go into this site below:
http://www.di2u.com.my/english/index.php?option=com_content&task=view&id=371&Itemid=1
Non-science articles:
I have actually written many other articles; ranging my thoughts on spiritual matters and art subjects like music (I am a violinist). I have attempted to link the un-measurable entities (non-science) to life sciences. Strange isn’t it for a scientist like me. I have analyzed my thought to answer exactly what is this entity called ‘life and the spirit’, and how measurable entity like science is actually linked to a un-measurable and mysterious dimension called ‘vital force’. I shall post the article into this blog later as it was 47 pages long. It was written inside the Newsletter of The Astronomical Society of Malaysia on January, 1987. It was written using a typewriter as there was no computer then. They need to be edited and retyped once again into a computer for uploading.
I have also written articles on music, mathematics, biomedical sciences, nutrition, food sciences, philosophy, and many other thoughts. Unfortunately a lot of people borrow a lot of my articles and put them into their websites and blog because they wrote nothing inside their own sites. They just took mine and put them inside theirs. But it is okay. They have asked my permission.
Scattered hither and thither:
Hence my articles are actually scattered all over various websites of other people, and it is very time consuming and difficult to recover them into one place. I intend to have a more elaborate non-commercial website instead of a personal blog site. For the moment I shall put my thoughts here.
Even this blog was created by me some 3 years ago. I think it was in 2007, with just a few of my articles inside as an introduction. Then I left them in cold storage after that because no body contributed anything inside to keep it alive. Instead, others ‘stole’ (with permission of course) my articles and conveniently put mine inside their websites, blog, twitter, MySpace, etc. But I don’t mine. Knowledge is about sharing, not keeping them only for ourselves.
Empty U-tube:
You are also very welcomed to contribute your ideas into my blog. I also have a U tube, but it is empty there. I have to upload the thousands of photos of my travels around the world if I want to activate my U-tube, and I won’t bother.
Your articles welcomed:
But your articles must be academic and educational in nature if you want them into my blog. Any ideas, opinion in Science & Technology (ST) preferably original thoughts are welcomed. ST means engineering and medicine also which are all applied sciences, and not just pure basic sciences.
Send them to me by e-mail, and if they are not offensive, and original in thoughts, I shall post them inside. The articles will be credited under your name. So provide me your full name, titles, and degrees (if any), and your position.
But please do not send me articles about politics, or condemning another race or religion, or any offensive materials. My blog is not meant for that. Mine is entirely on academic and educational materials.
Thank you.
JB Lim
May I suggest you and visitors to this blog "Scientific Logic" of mine also visit my friend website? His name is Captain LH Lim, and he is a Senior Pilot with Air Asia. He has hosted a fantastic website which I always enjoy reading. In fact my first love is on aviation, or anything that moves at very fast speeds such as jet planes, super fast bullet trains (like those world fastest trains in China), nuclear particles, cosmic rays, radio waves, speed of light, etc. These ‘moving subjects’ are actually my first love. My love is not about medicine, food and nutrition, health-care or biomedical research, and those sissy subjects. They are very boring to me. Unfortunately these are areas where I am trained in, and I have already spent my entire professional career on them. I am now retired from them. Life is unfortunate.
My real interest:
My interest in aviation, astronomy, physics, especially particle and nuclear physics, mathematics, and rocket science started even when I was in school, and they have still not gone away. I have very little interest in biological sciences like nutrition, health care, medicine, and all the biomedical sciences you care to name. Unfortunately an ill wind kept blowing me against my direction and wishes. The powerful hurricane forced me to make an emergency landing into the wrong airport. So I landed up as a medical research scientist and not a pilot. I wanted to be a pilot like Captain KH Lim. That was why I find his website extremely fascinating, and I am still reading his new entries every now and then. Life is like that!
Captain Lim and Dr Lim website link:
You may access Captain Lim’s website by just keying in these words "Ask Captain Lim" and also "Just About Flying" into the Google Search Engine bar and you will get him there. I have also written a lot of articles on various subjects which are linked to Captain Lim's website. But you may also access them by merely keying into Google Search these words: "Dr JB Lim's Corner". However, if you are interested only on my other articles related to health and medicine, then please go into this site below:
http://www.di2u.com.my/english/index.php?option=com_content&task=view&id=371&Itemid=1
Non-science articles:
I have actually written many other articles; ranging my thoughts on spiritual matters and art subjects like music (I am a violinist). I have attempted to link the un-measurable entities (non-science) to life sciences. Strange isn’t it for a scientist like me. I have analyzed my thought to answer exactly what is this entity called ‘life and the spirit’, and how measurable entity like science is actually linked to a un-measurable and mysterious dimension called ‘vital force’. I shall post the article into this blog later as it was 47 pages long. It was written inside the Newsletter of The Astronomical Society of Malaysia on January, 1987. It was written using a typewriter as there was no computer then. They need to be edited and retyped once again into a computer for uploading.
I have also written articles on music, mathematics, biomedical sciences, nutrition, food sciences, philosophy, and many other thoughts. Unfortunately a lot of people borrow a lot of my articles and put them into their websites and blog because they wrote nothing inside their own sites. They just took mine and put them inside theirs. But it is okay. They have asked my permission.
Scattered hither and thither:
Hence my articles are actually scattered all over various websites of other people, and it is very time consuming and difficult to recover them into one place. I intend to have a more elaborate non-commercial website instead of a personal blog site. For the moment I shall put my thoughts here.
Even this blog was created by me some 3 years ago. I think it was in 2007, with just a few of my articles inside as an introduction. Then I left them in cold storage after that because no body contributed anything inside to keep it alive. Instead, others ‘stole’ (with permission of course) my articles and conveniently put mine inside their websites, blog, twitter, MySpace, etc. But I don’t mine. Knowledge is about sharing, not keeping them only for ourselves.
Empty U-tube:
You are also very welcomed to contribute your ideas into my blog. I also have a U tube, but it is empty there. I have to upload the thousands of photos of my travels around the world if I want to activate my U-tube, and I won’t bother.
Your articles welcomed:
But your articles must be academic and educational in nature if you want them into my blog. Any ideas, opinion in Science & Technology (ST) preferably original thoughts are welcomed. ST means engineering and medicine also which are all applied sciences, and not just pure basic sciences.
Send them to me by e-mail, and if they are not offensive, and original in thoughts, I shall post them inside. The articles will be credited under your name. So provide me your full name, titles, and degrees (if any), and your position.
But please do not send me articles about politics, or condemning another race or religion, or any offensive materials. My blog is not meant for that. Mine is entirely on academic and educational materials.
Thank you.
JB Lim
Dear Captain KH Lim,
A Journey of Science Fantasy from Kuala Lumpur to Macau in an Air Bus A320
(Hind thought from ‘Alice in Wonderland’)
By: JB Lim
I know you are extremely busy with your work, and also having to answer so many seemingly endless questions from air-travelers posted in your blog. So I did not want to trouble you with unnecessary questions. Instead, I begin to dream of a journey from Kula Lumpur to Macau, and finally I decide to cross the almost infinite dimension of our Universe in an Air Bus.
So instead of troubling you, I decided to search the Internet myself for some basic facts, and from there I began to build up my calculations into a world of fantasy.
This was some very basic but interesting information about the Airbus A320 from which I made some conclusions, albeit only assumptions about my flight from Kuala Lumpur International Airport (KLIA) to Macau International Airport (MIA). I also made some assumptions about the cost of an air ticket.
Fuel consumption of an Airbus A320 is:
665 Imperial gallons (3,025 litres / 2,420 kg) per hour
Cruising speed:
530mph (853 kph / Mach 0.78) at 35,000ft (10,668m)
Range between KLIA and Macau International Airport:
2321 miles: (3717km):
This was based purely on theoretical calculation using coordinated geometry along a Great Circle as I do not have the benefit of using an onboard plane computer. I only have a scientific calculator, and some knowledge on spherical geometry.
However, a website gave a typical range with 150 passengers for the A320-200 as about 2,900 nautical miles (5,400 km). Yet another website gave the range as 2,650 miles (4,900km). I do not know which is correct, so I rely solely on my calculations which I think is safer.
Engine thrust:
Powered by two CFMI CFM56-5 or IAE V2500 with thrust ratings between 25,500 to 27,000 pounds force (113 kN to 120 kN). Another website said it is powered by two IAE V2527-A5 power plants with a thrust of 26,500lb (117.8 k N) per engine. I don’t know which is which?
Take off and landing speed:
160 mph (258 kph)
Seating Capacity:
150 passengers (Air Asia gave the figure as 180 passengers)
From the above Internet info, we can derive the following:
Density of jet fuel = 0.799 kg per litre
There are many grades of aviation fuels for jet engines. But I am unsure what type of fuel was used to power the jet engines of an Airbus A320? The general info I got from a website concerning the energy value of jet fuel for Boeing 767, the one that crashed into the New York World Trade Centre, the net calorific value of the jet fuel is between 42-44 MJ / kg.
Energy value of jet fuel:
Other sites gave the energy values of jet fuel (aviation kerosene) as between 43.28-43.71 MJ / kg (average 43.49 MJ / kg). So the energy values do not differ very much from each grade. Let us use the average caloric value of 43.49 MJ / kg for the different grades of jet fuels.
This means, the total energy used by the Airbus A320 to fly the plane from Kuala Lumpur to Macau was 43.49 MJ x 2,420 kg x 3.346944 hours (3 hrs 20 min 49 sec) = 352252 Mega joules. This is equivalent to 8099 kg of fuel used up.
Of course this was only an average estimate since the actual amount depended entirely on the weight of the plane with the number of passengers and their luggage, the amount of fuel that has to be burnt to provide the necessary thrust exerted by the engines to lift it up into the air, the forward motion at the required speed against the varying wind velocities blowing against, or helping it from behind, the air resistance at varying attitudes – probably much less energy needed at cruising heights where the air density, humidity and oxygen availability were much less.
Fuel consumption:
But if we assume the average fuel consumption at 2420 kg per hour remained the same throughout the journey with a full passenger load, and that the power of the engines were also constant (117.8 k N per engine) to keep the plane afloat at cruising height, then the only thing that will be affected, as far as I can see is, the velocity of plane due to varying wind resistance for reasons given above.
Fuel needed from KLIA to Macau:
This means that the total amount of fuel used up for my flight from Kuala Lumpur International Airport to Macau International Airport must have been around:
2420 kg x 3 hrs 20 min 49 sec (actual time) = 8099 kg. This is equivalent to a volume of:
Volume of fuel (v) = weight of fuel used up (w) ÷ density of fuel (d) = 10136 litres (10.17 cubic metres) calculated from first principle
This is almost similar to the 10125 litres (10.13 cubic metres) figure calculated directly from the info given in the Internet. I do not know the shape of the fuel tank, but it is equivalent to a cube of at least 2.16 meters on each side to store just sufficient fuel for the entire journey.
All the above calculated assumptions are based on whatever basic info I could get about A320 from the Internet, but in practice it may vary, and I do not know. Only Captain KH Lim can tell us exactly.
Now, the Air Fare, Cost of Transportation and Revenue:
In recent years, air fares offered by a lot of carriers are becoming ridiculously cheap. Air Asia operating in this region is one of them. The air fare from Kuala Lumpur to Macau by Air Asia is 319.99 MYR per person one way, as sited in Air Asia’s websites. If Air Asia can take in 180 passengers, and if we assume a maximum passenger load of 180 per flight, then the maximum fare collected is RM 57,598.
However, normally I found there a lot of empty seats on each flight.
But let us assume Air Asia got only 150 passengers, which is the seating capacity for an A 320 operating in European routes, then the revenue ought to be 47,998 MYR for the distance between KL and Macau (3 hours 20 minutes) for 1356 nautical miles (1561 statute miles = 2512 km). This distance was what you gave. This means it is just 20.5 cents per passenger per statute mile (12.74 cents per kilometer per passenger). Even the taxi fare in Kula Lumpur is already RM 3.00 for the first two kilometers and 10 cent for every 200m there after. This means if we were to travel 20 km by taxi, the fare will already cost us some RM 12.
What if we drive from Kuala Lumpur to Macau?
Even if we use the national car, say the Proton Savvy that recorded a fuel consumption rate of about 24 km / L (or about RM 0.08 / km of fuel), making the car as the most fuel-efficient Malaysian car as verified by Malaysian Book of Records. At the current price of petrol in Malaysia, this means that even for a Proton Savvy, the most fuel saving Malaysian car, for it to travel all the way to Macau (assuming theoretically possible) 2512 km away, the petrol itself would have already cost RM 200, without counting the cost of wear and tear, servicing, and the maintenance to the car along the way. But Air Asia charges only RM 319.99 for the same distance, and 10 times faster too without any stop. At some destinations say to Johore Bahru, they charge only a ridiculous RM 9.99? How did Air Asia do it? I always wonder?
Aviation Fuel Prices:
I really do not know the price of aviation fuel (high grade kerosene?) used by Air Asia compared to car petrol in this country. I have tried to search, but could not find the answer on current price. There was not much information about the prices of aviation jet fuel, their grades, and the one actually used by the engines of Air Bus A320. So it is not possible for us to give exact figures.
But I know that jet fuel prices have fallen by about 26% since the record high of US$93 (325.17 MYR) per barrel last August. But Air Asia said the high cost of jet fuel remains a concern, yet their air fares are so low. How do they then make money? Was it by imposing fuel surcharges on passengers?
Based on oil prices averaging US$47 (164.335 MYR) a barrel for the year (Taipei Times, AFP, Singapore reported on Monday, Sep 05, 2005 – rather outdated figure due to lack of current info, I admit), Malaysia Airlines imposed a fuel surcharge on all international routes from 1 June last year due to the “surging price of jet fuel”. I really do not know what all this means to a poor and ignorant passenger like me – whether they are really offering low air fares but ‘with extra fuel surcharges’ added or what? But all I know after all the taxes, the fare is no longer cheap.
But I know there are 42 US gallons or 159 litres in a barrel of oil. Since it takes about 10,125 litres (63.68 barrels) of jet fuel to fly from KL to Macau, the fuel itself would have cost the airline RM 10,465 per flight over the distance of KL to Macau at a price of US $47 (RM 164.335) per barrel of oil. 1 USD = 3.4965 MYR, if this is aviation fuel at that price.
Other Expenses:
Then what about other operating costs – high salaries of the management staff, the pilots, air and ground crew, the engineers, the office staff, the rentals of office, the agents, the mechanics, maintenance staff, the engineering maintenance of the aircraft, the landing fees, etc, etc? From revenue of 47,998 MYR from 150 passengers per flight, there will be a surplus of RM 37,533 to pay for all these other expenses. But how much are there left after deducting all other operating expenses? But Air Asia claims they are making money, while Malaysian Airline which charges far more for their air-fares claims they are operating at a loss. I believe both the carriers’ stories. But if this is true, there is something seriously wrong somewhere, but I do not know what and where.
Beyond A Scientist to Answer:
This question is far beyond me and Science to answer. I need to pass this question to the Business Management people in the airline industry. Maybe my friend Captain KH Lim can answer as he is a Senior Pilot with AirAsia even though not in the management division to manage policy matters.
This is a puzzle a scientist cannot solve. Only the smarter business people knows how to tackle problem without incurring losses, albeit sometimes with disastrous results
My imagination:
More pleasant thoughts for a scientist than trying to figure out commercial and financial headaches would be to answer my own original question as to how long it will take for a A320 commercial jet to transverse across the diameter Universe estimated to be 40,000 million light years, or 3.78 x 1023 km (378 sextillion km) across. Of course we will have to imagine that was possible as if it was flying in an atmosphere like ours with unlimited fuel.
The answer is 4.2 x 1020 hours at a maximum speed of 900 km per hour. This would be 4.79 x 1016 or 47.9 quadrillions (47.9 pentallions) Earth Solar years. But how much fuel it would use up, as if it was flying through atmosphere 10,000 metres high where the air pressure is 26.5474 kPa (26.157 % of 1 atmosphere), a density of 0.414403 kg/m3 and an air viscosity of 1.45787e-05 kg / m.s
The answer is 1.0164 x 1024 kg (1.0164 x 1021 metric tons) or 1 sextillion metric tons of aviation kerosene will be needed. The mass of Earth is 5.97 x 1024 kg. Thus the amount of fuel required is 17 % that of the mass of Earth.
Just a dreamy fantasy:
Of course all these are just not possible. A scientist fantasizes just for theoretical fun sake only. In the outer reaches of space, in between the galaxies it is almost an absolute vacuum, containing just one hydrogen atom per cubic metre of space. There is almost no matter to encounter any resistance that will require fuel and energy. In fact there will be just about 4 x 1026 hydrogen atoms the air-craft or space ship will encounter once it is in deep outer space between the stars and the galaxies (inter-stellar and inter-galactic space). With our Solar System there are much more matter to encounter between the planets, particularly in the asteroid belt between Mars and Jupiter with the myriads of asteroids, meteors, meteoroids, inter-planetary dust, micro-meteorites, particles from solar winds, and other space debris floating there. Once the plane leaves our Solar System, there will be just darkness, emptiness, and complete void with just one hydrogen atom from interstellar dusts to encounter for every cubic meter of space. The density of air at 10,000 metres at minus 50 degrees Celcius – the cruising height of a jet liner is about 0.38696 kg/m3.
Cruise along without fuel:
With almost no resistance to slow it down, the plane will just have to obey Newton First Law of Motion, provided it does not accelerate to near the speed of light, that it will remain in that state of motion in a ‘straight’ line at a uniform velocity of 900 km per hour till it reaches the end of the observable Universe from end-to-end.
At sea level and at 20 °C dry air has a density of approximately 1.2 kg/m3 varying with pressure and temperature. Air density and air pressure decrease with increasing altitude. The density of dry air at sea level is about 1/800th the density of water." The density of air at sea level is about 1.25 kg / m3 (1.25 g/L) at 10 km, d is about 1/4 its sea-level value.
Amount of molecules per cubic metre:
A cube meter of space at ground (sea) level contains about 45 moles of air. Since 1 mole or Avogadro’s number = 6.022 x 1023, hence at ground level, there will be 2.7099 x 1025 ‘air molecules’ per cubic meter.
Bear in mind there is no such specific entity as ‘air molecules’ since air is roughly 78% nitrogen (normally inert except upon electrolysis by lightning), 21% oxygen, 0.93% argon, 0.04% carbon dioxide, and trace amounts of other gases, in addition to about 3% water vapor. This mixture of gases is commonly known as air. But for the sake of simplicity let us call a mixture of these molecules as ‘air molecules’
Since the density of air at ground level is 1.25 kg / m3 (1.25 g/L), each ‘air molecules’ will have a mass of 4.6127 x 10-26 kg. That same cube at 10 km altitude will contain just over 13 moles of air (7.8286 x 10 24 air molecules). Therefore the density of air at 10,000 will be 0.36 kg. m3. or just about 28.9 % that at sea level. This figure varies of course depending on humidity, temperature and pressure up there. It can be as high as ¼ that of sea level. Clearly, number density declines with altitude.
The hydrogen atom:
The hydrogen atom consists of a proton of mass mp=1.7 x 10-27kg + an electron of mass m e= 9.110-31kg. Hence the mass of a neutral hydrogen atom = 1.70091 x 10-27 kg. This means an ‘air molecule’ will have an average mass 27 times that of a single hydrogen atom. A hydrogen atom is 0.0369 times lighter than an air molecule based on the average air density. But there is no such thing as an ‘air molecule’ since air is a mixture of gasses, namely: nitrogen 78.084%, oxygen 20.946%), argon 0.9340%, carbon dioxide 0.0387%, neon 0.001818%, helium 0.000524%, and numerous other trace gasses. But just to make this story simple we shall call it ‘air molecules’.
Volume of air sucked in by a jet-engine:
I do not know how much of air is being sucked in by an engine of an Airbus A320, but from an information I got some years ago from a newspaper, it was reported that an RR Trent Engine empties 945.12 cubic meters or 1.2 Imperial tons (1.2192 tonne / metric tons) of air per second during take off.
Size of Malaysian residential houses:
Houses in Malaysia come in all shapes, sizes and prices. They range from mansions, istana-like, bungalows, apartments, terrace and lined houses to small squatter living spaces. But most urbanites in the middle income group live in linked or terrace houses, either double or single storey. Whether double or single the land area is about the same except the height.
From several architectural as well as real estate websites, sales and purchase agreement with floor plan measurements, and actual measurements from all these sources, and all the data put together for statistical analysis to derive their mean values, below is a summary of the dimensions of most of the houses in Malaysia.
Approx total land space = 1600 square feet
Length & breadth of house inside = 48 ft x 20 ft = 960 sq ft
Length & breadth of backyard = 8 ft x 20 ft = 160 sq ft
Area of car porch = 11 ft x 12 ft = 132 sq ft
Area of driveway = 24 ft x 20 ft = 480 sq ft
Hence total area = 960 + 160 + 480 = 1600 sq feet
Height of floor to ceiling = 130 inches (10.8 feet)
Volume of built up indoor area of house = 960 x 10.8 = 10368 cubic ft = 293.589 cubic metres
Hence, the dimension inside the house of an average single storey or a double storey terrace house in Malaysia is 48 ft x 20 ft of floor space, with a standard indoor ceiling height of 130 inches (10.8 feet). Hence the average volume of either a single storey house or the volume of either downstairs or upstairs of a double-storey house within its enclosed area is only 10368 cubic feet or 293.589 cubic metres.
How fast and how much can it sucks?
This mean a single RR Trent Jet Engine of a plane on take off can completely empty all the indoor air of 3.22 houses put together in one second just to get enough oxygen to ignite its fuel to provide the thrust it requires for lift-off.
But a plane has two engines. The emptying rate will be 6.44 single-storey houses per second on take off. It is like having nearly 7 houses imploding together when all its air inside is emptied within one second - if the air is not replaced fast enough from outside. Quite a thought!
The atmosphere at sea level will exert a pressure of 101.3 kPa (kilopascals) = 14.7 psi (pounds per square inch) = 760 torr = 29.9 inches of mercury on all its exterior walls and roof and instantly cause all the walls and the roof to collapse instantly (implode inwards) if air flow into the house through its doors and windows is not fast enough to replace what was sucked out by the jet engines. Just imagine the power of the jet engines and fancy that. I never thought of this without this simple calculation.
Let us now assume two RR Trent Engines were used. The density of air at 10,000 meters = 0.36 kg / m3 At this density, there are 13 moles or 7.8286 x 10 24 of air molecules. Both the engines would have sucked in (2 x 1.2192 metric tons x 1000 kg x 7.8286 x 10 24) ÷ 0.36 kg = 5.30 x 1028 air molecules per second. The kinetic energy generated by the plane’s two engines per second as they strike against 5.30 x 1028 molecules, each with a mass of 4.6127 x 10-26 kg. will be ½ mv2 = ½ (5.30 x 1028 x 4.6127 x 10-26) x (236.9 meter per second)2 = 137 202 239 joules (137 megajoules) provided the molecules are stationary before being sucked in – which is not possible of course.
Fluid dynamics: Bernoulli's principle
This is the minimum energy expenditure against the two engines since we are assuming that the speed at which the air molecules were being sucked in is the same as the speed of the plane at 853 km. per hour, which of course is not true. The speed of the air intake has to be a lot faster than the forward thrust velocity of the whole plane which also has to suffer the impact of other air molecules covering the much larger surface of the entire plane. I do not know how large the surface area of the plane is, so we cannot determine how many more molecules the plane will have to strike. The ejection of the mass of gasses behind the jets has to be equal to the thrust forward against air resistance (Newton 3rd Law of Motion)
Furthermore, an increase in the speed of the fluid occurs simultaneously with a decrease in pressure or a decrease in the fluid's potential energy (Bernoulli's principle).
We can easily calculate this out if we have some very basic info from Captain KH Lim about the plane’s surface area especially the wings which give it the thrust forward and lift upwards against gravity. But Captain Lim is such a busy pilot, and I do not wish to trouble him. So I like to work things on my own. Of course I was only conducting medical research all along, but this should not hinder me from trying to figure out problems in aeronautical or molecular physics.
We can guess:
However we can make a guess indirectly. Since the fuel consumption of an Airbus A320 is 665 Imperial gallons (3,025 litres / 2,420 kg) per hour or 0.672 kg per second, and since the energy values of jet fuel (aviation kerosene) given as between 43.28-43.71 MJ / kg (average 43.49 MJ / kg), the energy expenditure is 29.225 MJ per second. The net velocity of the plane is 236.9 meters per second, and the mass of each air molecule is 4.6127 x 10-26 kg, the total number of air molecules the plane has strike in other areas excluding those sucked in by the engines each second, can be determined by:
E = ½ mv2
2E = mv2
m= 2E / v2 = 2 x 29 225 000 joules / 236.92 = 1041.48 kg of air molecules
= 2.2578 x 1028 molecules per second.
Much slower:
But we know the velocity of the plane is very much slower than that of the air-intake and the velocity of the jets of gases ejected behind.
The calculations I gave showed that the engines alone suck in 5.30 x 1028 molecules or 2444.7 kg of air per second, not counting other air molecules encountered by the wings and fuselage. Hence the plane will require 0.6722 kg of fuel per second at the rate of 2420 kg per hour at a cruising height and speed of 10 km per second and 853 kph respectively to be equivalent to the striking kinetic energy against the non-engine parts of the plane.
Let us use the figure 2,2578 x 1028 molecules or about 1040 kg of air per second against the moving plane in still air. This is equivalent to encountering about 6.12 x 1029 hydrogen atoms per square meter in deep intergalactic space. We assume below there are about just 10 hydrogen atoms per cubic meter in deep intergalactic space.
This means our plane can afford to travel for 6.12 x 1026 km or 6.469 x 1013 light years before encountering the same mass of resistance and energy usage as a plane traveling for one second or to a distance of about 237 meters in our Earth’s troposphere (1 light year = 9,460,730,472,580.8 km).
No more fuel needed:
In short, we assume our A320 has already left the Solar System, and is now cruising without the need of anymore fuel in the emptiness of deep space except an encounter with just 10 hydrogen atoms per cubic meter or per square meter of space it scooped up. It will continue in that state in a ‘straight’ line at 530mph (853 kph / Mach 0.78) for all eternity as if it was flying at 35,000ft (10,668m) in our own atmosphere, unless acted by an external force such as interstellar and intergalactic dust and molecules to resist that state as prescribed by Newton First Law of Dynamics.
The void of interstellar space:
The interstellar medium (ISM) is usually extremely tenuous, with densities ranging from a few thousand to a few hundred million particles per cubic meter, and an average value in the Milky Way Galaxy of a million particles per cubic meter. Other estimates gave it as 300,000 atoms per cubic meter. The elemental composition of interstellar clouds is much like that of the sun, about 90 percent of hydrogen, and 9.99 percent helium. The heavier elements make up the remaining 0.01 percent. The average density of the Universe is just 10 to 100 hydrogen atoms per cubic meter. But deep between the galaxies, there may be just one hydrogen atom most of the time.
In the Cold Neutral Medium (CNM) of space where the temperature is just 50 – 100 Kelvin, there are just 1 - 103 neutral hydrogen atoms per cubic cm of space. The density at different locations of the Universe varies enormously. It is very much denser within a galaxy than in between the galaxies, much, much more dense in the centre of a galaxy where the black holes are, than in the peripheries where the stars are scattered apart, and the interstellar densities are so much more tenuous. Cosmology and astrophysics is a very complicated subject and the data on densities and amount of matter from the dark matter to the derived varies so greatly. In the above calculation, we assumed we only encounter an average of just 10 hydrogen atoms per cubic metres or per sq. metre of our plane.
Interstellar density:
In the solar neighborhood, the stellar density is about one star per cubic parsec (one parsec is 3.26 light-years). At the Galactic core, around 100 parsecs from the Galactic center, the stellar density has risen to 100 per cubic parsec, crowded together because of gravity, let alone where ‘neutron stars’ exist. These stars have a radius of only 10 km, and the density is about 100 million tons per cubic centimeter. This is insignificant compared to a black hole or a super black hole where volume and mass collapsed into a singularity. Here the density is infinite. Because of all these variations let us steer our plane far, far away from these grotesque cosmic events, well away from any event horizons Let our plane drift through the immense intergalactic valley, end to end, an immense abyss of space spanning 40,000 million light years, or 3.78 x 1023 km (378 sextillion km) across.
An eternal cruise:
For that, it will take the plane almost 48000 million, million Earth years to achieve. Fancy that! I salute the super-pilot who can live and endure that kind of journey. To solve that, he may have to marry abroad, bear children over 1.6 x 1015 (1600 million, million) generations to take over the piloting once each child attains the age of 30 years, taken as the span of one generation.
A better idea:
But I have a better idea. This is not possible. But it is possible for his sperm and his wife’s eggs be frozen in liquid nitrogen as they are left to drift into the frigid coldness and darkness of space where the temperature is almost 0 Kelvin. The pilot may either remain back on Earth, or kept in suspended animation if he wishes to follow his genes aboard. A robot is programmed to take over which will only be activated towards the end of the journey. The awaken robot will then take out the sperm-ovum in deep freeze, fertilize them. It will then nurse them, and bring them up. It will teach them where they came from, their language, culture and civilization, and what their world looked like.
A voyage guided and narrated by a robot:
It will tell him or her purpose of their voyage, their fate, destination, and destiny. They will pictures and images of their world, their parents, other humans, plants, animals and all other living things. All the images of this world will be beamed towards the plane for their benefit from the day it left Earth and the Solar System. All scenes of Earth will be continuous with time, such that the entire length of 40,000 million years of history from the beginning to the edge of this Universe could be shown.
The TV transmissions will not be broadcast the usual way. This will ‘dilute’ the energy of transmission over a wider and wider volume into space as the plane leaves planet Earth. The entire energy of the signals will have to be concentrated into just a very narrow beam in the direction of the plane in order to focus the pictures clearly on arrival without being spread out. It would be like a laser beam.
The transmission should be continuous so that there is no gap in time in receiving the images. Even then, towards the end of the journey, all the images would have been at least 40,000 millions years out of date for the far-away pilot. He will perhaps never be able to learn of his / her origin, and the world he came from.
Time needed even by light to cross the chasm:
The distance across the horrendous chasm of the Universe from one end to the other is: 299792458 metres ÷ 1000 (to change into km / sec) x 60 sec (to change to km per min) x 60 min (to change to km per hour) x 24 hr in a day x 365.25 days in a year x 40,000, 000,000 years = 3.78 x 1023 km (378 sextillion km).
This will take light 4 x 1010 (40 billion) years to cross this grotesque time
space corridor. How long would my Air Bus A320 take?
‘Ask Captain KH Lim’ in his fantastic website I highly recommend you to visit if you ‘Ctrl & Click’ on my articles above. They are linked to his.
http://askcaptainlim.com/index.php?option=com_content&view=category&id=74&Itemid=89
What a journey? Fancy that! Have a safe trip.
JB Lim
A Journey of Science Fantasy from Kuala Lumpur to Macau in an Air Bus A320
(Hind thought from ‘Alice in Wonderland’)
By: JB Lim
I know you are extremely busy with your work, and also having to answer so many seemingly endless questions from air-travelers posted in your blog. So I did not want to trouble you with unnecessary questions. Instead, I begin to dream of a journey from Kula Lumpur to Macau, and finally I decide to cross the almost infinite dimension of our Universe in an Air Bus.
So instead of troubling you, I decided to search the Internet myself for some basic facts, and from there I began to build up my calculations into a world of fantasy.
This was some very basic but interesting information about the Airbus A320 from which I made some conclusions, albeit only assumptions about my flight from Kuala Lumpur International Airport (KLIA) to Macau International Airport (MIA). I also made some assumptions about the cost of an air ticket.
Fuel consumption of an Airbus A320 is:
665 Imperial gallons (3,025 litres / 2,420 kg) per hour
Cruising speed:
530mph (853 kph / Mach 0.78) at 35,000ft (10,668m)
Range between KLIA and Macau International Airport:
2321 miles: (3717km):
This was based purely on theoretical calculation using coordinated geometry along a Great Circle as I do not have the benefit of using an onboard plane computer. I only have a scientific calculator, and some knowledge on spherical geometry.
However, a website gave a typical range with 150 passengers for the A320-200 as about 2,900 nautical miles (5,400 km). Yet another website gave the range as 2,650 miles (4,900km). I do not know which is correct, so I rely solely on my calculations which I think is safer.
Engine thrust:
Powered by two CFMI CFM56-5 or IAE V2500 with thrust ratings between 25,500 to 27,000 pounds force (113 kN to 120 kN). Another website said it is powered by two IAE V2527-A5 power plants with a thrust of 26,500lb (117.8 k N) per engine. I don’t know which is which?
Take off and landing speed:
160 mph (258 kph)
Seating Capacity:
150 passengers (Air Asia gave the figure as 180 passengers)
From the above Internet info, we can derive the following:
Density of jet fuel = 0.799 kg per litre
There are many grades of aviation fuels for jet engines. But I am unsure what type of fuel was used to power the jet engines of an Airbus A320? The general info I got from a website concerning the energy value of jet fuel for Boeing 767, the one that crashed into the New York World Trade Centre, the net calorific value of the jet fuel is between 42-44 MJ / kg.
Energy value of jet fuel:
Other sites gave the energy values of jet fuel (aviation kerosene) as between 43.28-43.71 MJ / kg (average 43.49 MJ / kg). So the energy values do not differ very much from each grade. Let us use the average caloric value of 43.49 MJ / kg for the different grades of jet fuels.
This means, the total energy used by the Airbus A320 to fly the plane from Kuala Lumpur to Macau was 43.49 MJ x 2,420 kg x 3.346944 hours (3 hrs 20 min 49 sec) = 352252 Mega joules. This is equivalent to 8099 kg of fuel used up.
Of course this was only an average estimate since the actual amount depended entirely on the weight of the plane with the number of passengers and their luggage, the amount of fuel that has to be burnt to provide the necessary thrust exerted by the engines to lift it up into the air, the forward motion at the required speed against the varying wind velocities blowing against, or helping it from behind, the air resistance at varying attitudes – probably much less energy needed at cruising heights where the air density, humidity and oxygen availability were much less.
Fuel consumption:
But if we assume the average fuel consumption at 2420 kg per hour remained the same throughout the journey with a full passenger load, and that the power of the engines were also constant (117.8 k N per engine) to keep the plane afloat at cruising height, then the only thing that will be affected, as far as I can see is, the velocity of plane due to varying wind resistance for reasons given above.
Fuel needed from KLIA to Macau:
This means that the total amount of fuel used up for my flight from Kuala Lumpur International Airport to Macau International Airport must have been around:
2420 kg x 3 hrs 20 min 49 sec (actual time) = 8099 kg. This is equivalent to a volume of:
Volume of fuel (v) = weight of fuel used up (w) ÷ density of fuel (d) = 10136 litres (10.17 cubic metres) calculated from first principle
This is almost similar to the 10125 litres (10.13 cubic metres) figure calculated directly from the info given in the Internet. I do not know the shape of the fuel tank, but it is equivalent to a cube of at least 2.16 meters on each side to store just sufficient fuel for the entire journey.
All the above calculated assumptions are based on whatever basic info I could get about A320 from the Internet, but in practice it may vary, and I do not know. Only Captain KH Lim can tell us exactly.
Now, the Air Fare, Cost of Transportation and Revenue:
In recent years, air fares offered by a lot of carriers are becoming ridiculously cheap. Air Asia operating in this region is one of them. The air fare from Kuala Lumpur to Macau by Air Asia is 319.99 MYR per person one way, as sited in Air Asia’s websites. If Air Asia can take in 180 passengers, and if we assume a maximum passenger load of 180 per flight, then the maximum fare collected is RM 57,598.
However, normally I found there a lot of empty seats on each flight.
But let us assume Air Asia got only 150 passengers, which is the seating capacity for an A 320 operating in European routes, then the revenue ought to be 47,998 MYR for the distance between KL and Macau (3 hours 20 minutes) for 1356 nautical miles (1561 statute miles = 2512 km). This distance was what you gave. This means it is just 20.5 cents per passenger per statute mile (12.74 cents per kilometer per passenger). Even the taxi fare in Kula Lumpur is already RM 3.00 for the first two kilometers and 10 cent for every 200m there after. This means if we were to travel 20 km by taxi, the fare will already cost us some RM 12.
What if we drive from Kuala Lumpur to Macau?
Even if we use the national car, say the Proton Savvy that recorded a fuel consumption rate of about 24 km / L (or about RM 0.08 / km of fuel), making the car as the most fuel-efficient Malaysian car as verified by Malaysian Book of Records. At the current price of petrol in Malaysia, this means that even for a Proton Savvy, the most fuel saving Malaysian car, for it to travel all the way to Macau (assuming theoretically possible) 2512 km away, the petrol itself would have already cost RM 200, without counting the cost of wear and tear, servicing, and the maintenance to the car along the way. But Air Asia charges only RM 319.99 for the same distance, and 10 times faster too without any stop. At some destinations say to Johore Bahru, they charge only a ridiculous RM 9.99? How did Air Asia do it? I always wonder?
Aviation Fuel Prices:
I really do not know the price of aviation fuel (high grade kerosene?) used by Air Asia compared to car petrol in this country. I have tried to search, but could not find the answer on current price. There was not much information about the prices of aviation jet fuel, their grades, and the one actually used by the engines of Air Bus A320. So it is not possible for us to give exact figures.
But I know that jet fuel prices have fallen by about 26% since the record high of US$93 (325.17 MYR) per barrel last August. But Air Asia said the high cost of jet fuel remains a concern, yet their air fares are so low. How do they then make money? Was it by imposing fuel surcharges on passengers?
Based on oil prices averaging US$47 (164.335 MYR) a barrel for the year (Taipei Times, AFP, Singapore reported on Monday, Sep 05, 2005 – rather outdated figure due to lack of current info, I admit), Malaysia Airlines imposed a fuel surcharge on all international routes from 1 June last year due to the “surging price of jet fuel”. I really do not know what all this means to a poor and ignorant passenger like me – whether they are really offering low air fares but ‘with extra fuel surcharges’ added or what? But all I know after all the taxes, the fare is no longer cheap.
But I know there are 42 US gallons or 159 litres in a barrel of oil. Since it takes about 10,125 litres (63.68 barrels) of jet fuel to fly from KL to Macau, the fuel itself would have cost the airline RM 10,465 per flight over the distance of KL to Macau at a price of US $47 (RM 164.335) per barrel of oil. 1 USD = 3.4965 MYR, if this is aviation fuel at that price.
Other Expenses:
Then what about other operating costs – high salaries of the management staff, the pilots, air and ground crew, the engineers, the office staff, the rentals of office, the agents, the mechanics, maintenance staff, the engineering maintenance of the aircraft, the landing fees, etc, etc? From revenue of 47,998 MYR from 150 passengers per flight, there will be a surplus of RM 37,533 to pay for all these other expenses. But how much are there left after deducting all other operating expenses? But Air Asia claims they are making money, while Malaysian Airline which charges far more for their air-fares claims they are operating at a loss. I believe both the carriers’ stories. But if this is true, there is something seriously wrong somewhere, but I do not know what and where.
Beyond A Scientist to Answer:
This question is far beyond me and Science to answer. I need to pass this question to the Business Management people in the airline industry. Maybe my friend Captain KH Lim can answer as he is a Senior Pilot with AirAsia even though not in the management division to manage policy matters.
This is a puzzle a scientist cannot solve. Only the smarter business people knows how to tackle problem without incurring losses, albeit sometimes with disastrous results
My imagination:
More pleasant thoughts for a scientist than trying to figure out commercial and financial headaches would be to answer my own original question as to how long it will take for a A320 commercial jet to transverse across the diameter Universe estimated to be 40,000 million light years, or 3.78 x 1023 km (378 sextillion km) across. Of course we will have to imagine that was possible as if it was flying in an atmosphere like ours with unlimited fuel.
The answer is 4.2 x 1020 hours at a maximum speed of 900 km per hour. This would be 4.79 x 1016 or 47.9 quadrillions (47.9 pentallions) Earth Solar years. But how much fuel it would use up, as if it was flying through atmosphere 10,000 metres high where the air pressure is 26.5474 kPa (26.157 % of 1 atmosphere), a density of 0.414403 kg/m3 and an air viscosity of 1.45787e-05 kg / m.s
The answer is 1.0164 x 1024 kg (1.0164 x 1021 metric tons) or 1 sextillion metric tons of aviation kerosene will be needed. The mass of Earth is 5.97 x 1024 kg. Thus the amount of fuel required is 17 % that of the mass of Earth.
Just a dreamy fantasy:
Of course all these are just not possible. A scientist fantasizes just for theoretical fun sake only. In the outer reaches of space, in between the galaxies it is almost an absolute vacuum, containing just one hydrogen atom per cubic metre of space. There is almost no matter to encounter any resistance that will require fuel and energy. In fact there will be just about 4 x 1026 hydrogen atoms the air-craft or space ship will encounter once it is in deep outer space between the stars and the galaxies (inter-stellar and inter-galactic space). With our Solar System there are much more matter to encounter between the planets, particularly in the asteroid belt between Mars and Jupiter with the myriads of asteroids, meteors, meteoroids, inter-planetary dust, micro-meteorites, particles from solar winds, and other space debris floating there. Once the plane leaves our Solar System, there will be just darkness, emptiness, and complete void with just one hydrogen atom from interstellar dusts to encounter for every cubic meter of space. The density of air at 10,000 metres at minus 50 degrees Celcius – the cruising height of a jet liner is about 0.38696 kg/m3.
Cruise along without fuel:
With almost no resistance to slow it down, the plane will just have to obey Newton First Law of Motion, provided it does not accelerate to near the speed of light, that it will remain in that state of motion in a ‘straight’ line at a uniform velocity of 900 km per hour till it reaches the end of the observable Universe from end-to-end.
At sea level and at 20 °C dry air has a density of approximately 1.2 kg/m3 varying with pressure and temperature. Air density and air pressure decrease with increasing altitude. The density of dry air at sea level is about 1/800th the density of water." The density of air at sea level is about 1.25 kg / m3 (1.25 g/L) at 10 km, d is about 1/4 its sea-level value.
Amount of molecules per cubic metre:
A cube meter of space at ground (sea) level contains about 45 moles of air. Since 1 mole or Avogadro’s number = 6.022 x 1023, hence at ground level, there will be 2.7099 x 1025 ‘air molecules’ per cubic meter.
Bear in mind there is no such specific entity as ‘air molecules’ since air is roughly 78% nitrogen (normally inert except upon electrolysis by lightning), 21% oxygen, 0.93% argon, 0.04% carbon dioxide, and trace amounts of other gases, in addition to about 3% water vapor. This mixture of gases is commonly known as air. But for the sake of simplicity let us call a mixture of these molecules as ‘air molecules’
Since the density of air at ground level is 1.25 kg / m3 (1.25 g/L), each ‘air molecules’ will have a mass of 4.6127 x 10-26 kg. That same cube at 10 km altitude will contain just over 13 moles of air (7.8286 x 10 24 air molecules). Therefore the density of air at 10,000 will be 0.36 kg. m3. or just about 28.9 % that at sea level. This figure varies of course depending on humidity, temperature and pressure up there. It can be as high as ¼ that of sea level. Clearly, number density declines with altitude.
The hydrogen atom:
The hydrogen atom consists of a proton of mass mp=1.7 x 10-27kg + an electron of mass m e= 9.110-31kg. Hence the mass of a neutral hydrogen atom = 1.70091 x 10-27 kg. This means an ‘air molecule’ will have an average mass 27 times that of a single hydrogen atom. A hydrogen atom is 0.0369 times lighter than an air molecule based on the average air density. But there is no such thing as an ‘air molecule’ since air is a mixture of gasses, namely: nitrogen 78.084%, oxygen 20.946%), argon 0.9340%, carbon dioxide 0.0387%, neon 0.001818%, helium 0.000524%, and numerous other trace gasses. But just to make this story simple we shall call it ‘air molecules’.
Volume of air sucked in by a jet-engine:
I do not know how much of air is being sucked in by an engine of an Airbus A320, but from an information I got some years ago from a newspaper, it was reported that an RR Trent Engine empties 945.12 cubic meters or 1.2 Imperial tons (1.2192 tonne / metric tons) of air per second during take off.
Size of Malaysian residential houses:
Houses in Malaysia come in all shapes, sizes and prices. They range from mansions, istana-like, bungalows, apartments, terrace and lined houses to small squatter living spaces. But most urbanites in the middle income group live in linked or terrace houses, either double or single storey. Whether double or single the land area is about the same except the height.
From several architectural as well as real estate websites, sales and purchase agreement with floor plan measurements, and actual measurements from all these sources, and all the data put together for statistical analysis to derive their mean values, below is a summary of the dimensions of most of the houses in Malaysia.
Approx total land space = 1600 square feet
Length & breadth of house inside = 48 ft x 20 ft = 960 sq ft
Length & breadth of backyard = 8 ft x 20 ft = 160 sq ft
Area of car porch = 11 ft x 12 ft = 132 sq ft
Area of driveway = 24 ft x 20 ft = 480 sq ft
Hence total area = 960 + 160 + 480 = 1600 sq feet
Height of floor to ceiling = 130 inches (10.8 feet)
Volume of built up indoor area of house = 960 x 10.8 = 10368 cubic ft = 293.589 cubic metres
Hence, the dimension inside the house of an average single storey or a double storey terrace house in Malaysia is 48 ft x 20 ft of floor space, with a standard indoor ceiling height of 130 inches (10.8 feet). Hence the average volume of either a single storey house or the volume of either downstairs or upstairs of a double-storey house within its enclosed area is only 10368 cubic feet or 293.589 cubic metres.
How fast and how much can it sucks?
This mean a single RR Trent Jet Engine of a plane on take off can completely empty all the indoor air of 3.22 houses put together in one second just to get enough oxygen to ignite its fuel to provide the thrust it requires for lift-off.
But a plane has two engines. The emptying rate will be 6.44 single-storey houses per second on take off. It is like having nearly 7 houses imploding together when all its air inside is emptied within one second - if the air is not replaced fast enough from outside. Quite a thought!
The atmosphere at sea level will exert a pressure of 101.3 kPa (kilopascals) = 14.7 psi (pounds per square inch) = 760 torr = 29.9 inches of mercury on all its exterior walls and roof and instantly cause all the walls and the roof to collapse instantly (implode inwards) if air flow into the house through its doors and windows is not fast enough to replace what was sucked out by the jet engines. Just imagine the power of the jet engines and fancy that. I never thought of this without this simple calculation.
Let us now assume two RR Trent Engines were used. The density of air at 10,000 meters = 0.36 kg / m3 At this density, there are 13 moles or 7.8286 x 10 24 of air molecules. Both the engines would have sucked in (2 x 1.2192 metric tons x 1000 kg x 7.8286 x 10 24) ÷ 0.36 kg = 5.30 x 1028 air molecules per second. The kinetic energy generated by the plane’s two engines per second as they strike against 5.30 x 1028 molecules, each with a mass of 4.6127 x 10-26 kg. will be ½ mv2 = ½ (5.30 x 1028 x 4.6127 x 10-26) x (236.9 meter per second)2 = 137 202 239 joules (137 megajoules) provided the molecules are stationary before being sucked in – which is not possible of course.
Fluid dynamics: Bernoulli's principle
This is the minimum energy expenditure against the two engines since we are assuming that the speed at which the air molecules were being sucked in is the same as the speed of the plane at 853 km. per hour, which of course is not true. The speed of the air intake has to be a lot faster than the forward thrust velocity of the whole plane which also has to suffer the impact of other air molecules covering the much larger surface of the entire plane. I do not know how large the surface area of the plane is, so we cannot determine how many more molecules the plane will have to strike. The ejection of the mass of gasses behind the jets has to be equal to the thrust forward against air resistance (Newton 3rd Law of Motion)
Furthermore, an increase in the speed of the fluid occurs simultaneously with a decrease in pressure or a decrease in the fluid's potential energy (Bernoulli's principle).
We can easily calculate this out if we have some very basic info from Captain KH Lim about the plane’s surface area especially the wings which give it the thrust forward and lift upwards against gravity. But Captain Lim is such a busy pilot, and I do not wish to trouble him. So I like to work things on my own. Of course I was only conducting medical research all along, but this should not hinder me from trying to figure out problems in aeronautical or molecular physics.
We can guess:
However we can make a guess indirectly. Since the fuel consumption of an Airbus A320 is 665 Imperial gallons (3,025 litres / 2,420 kg) per hour or 0.672 kg per second, and since the energy values of jet fuel (aviation kerosene) given as between 43.28-43.71 MJ / kg (average 43.49 MJ / kg), the energy expenditure is 29.225 MJ per second. The net velocity of the plane is 236.9 meters per second, and the mass of each air molecule is 4.6127 x 10-26 kg, the total number of air molecules the plane has strike in other areas excluding those sucked in by the engines each second, can be determined by:
E = ½ mv2
2E = mv2
m= 2E / v2 = 2 x 29 225 000 joules / 236.92 = 1041.48 kg of air molecules
= 2.2578 x 1028 molecules per second.
Much slower:
But we know the velocity of the plane is very much slower than that of the air-intake and the velocity of the jets of gases ejected behind.
The calculations I gave showed that the engines alone suck in 5.30 x 1028 molecules or 2444.7 kg of air per second, not counting other air molecules encountered by the wings and fuselage. Hence the plane will require 0.6722 kg of fuel per second at the rate of 2420 kg per hour at a cruising height and speed of 10 km per second and 853 kph respectively to be equivalent to the striking kinetic energy against the non-engine parts of the plane.
Let us use the figure 2,2578 x 1028 molecules or about 1040 kg of air per second against the moving plane in still air. This is equivalent to encountering about 6.12 x 1029 hydrogen atoms per square meter in deep intergalactic space. We assume below there are about just 10 hydrogen atoms per cubic meter in deep intergalactic space.
This means our plane can afford to travel for 6.12 x 1026 km or 6.469 x 1013 light years before encountering the same mass of resistance and energy usage as a plane traveling for one second or to a distance of about 237 meters in our Earth’s troposphere (1 light year = 9,460,730,472,580.8 km).
No more fuel needed:
In short, we assume our A320 has already left the Solar System, and is now cruising without the need of anymore fuel in the emptiness of deep space except an encounter with just 10 hydrogen atoms per cubic meter or per square meter of space it scooped up. It will continue in that state in a ‘straight’ line at 530mph (853 kph / Mach 0.78) for all eternity as if it was flying at 35,000ft (10,668m) in our own atmosphere, unless acted by an external force such as interstellar and intergalactic dust and molecules to resist that state as prescribed by Newton First Law of Dynamics.
The void of interstellar space:
The interstellar medium (ISM) is usually extremely tenuous, with densities ranging from a few thousand to a few hundred million particles per cubic meter, and an average value in the Milky Way Galaxy of a million particles per cubic meter. Other estimates gave it as 300,000 atoms per cubic meter. The elemental composition of interstellar clouds is much like that of the sun, about 90 percent of hydrogen, and 9.99 percent helium. The heavier elements make up the remaining 0.01 percent. The average density of the Universe is just 10 to 100 hydrogen atoms per cubic meter. But deep between the galaxies, there may be just one hydrogen atom most of the time.
In the Cold Neutral Medium (CNM) of space where the temperature is just 50 – 100 Kelvin, there are just 1 - 103 neutral hydrogen atoms per cubic cm of space. The density at different locations of the Universe varies enormously. It is very much denser within a galaxy than in between the galaxies, much, much more dense in the centre of a galaxy where the black holes are, than in the peripheries where the stars are scattered apart, and the interstellar densities are so much more tenuous. Cosmology and astrophysics is a very complicated subject and the data on densities and amount of matter from the dark matter to the derived varies so greatly. In the above calculation, we assumed we only encounter an average of just 10 hydrogen atoms per cubic metres or per sq. metre of our plane.
Interstellar density:
In the solar neighborhood, the stellar density is about one star per cubic parsec (one parsec is 3.26 light-years). At the Galactic core, around 100 parsecs from the Galactic center, the stellar density has risen to 100 per cubic parsec, crowded together because of gravity, let alone where ‘neutron stars’ exist. These stars have a radius of only 10 km, and the density is about 100 million tons per cubic centimeter. This is insignificant compared to a black hole or a super black hole where volume and mass collapsed into a singularity. Here the density is infinite. Because of all these variations let us steer our plane far, far away from these grotesque cosmic events, well away from any event horizons Let our plane drift through the immense intergalactic valley, end to end, an immense abyss of space spanning 40,000 million light years, or 3.78 x 1023 km (378 sextillion km) across.
An eternal cruise:
For that, it will take the plane almost 48000 million, million Earth years to achieve. Fancy that! I salute the super-pilot who can live and endure that kind of journey. To solve that, he may have to marry abroad, bear children over 1.6 x 1015 (1600 million, million) generations to take over the piloting once each child attains the age of 30 years, taken as the span of one generation.
A better idea:
But I have a better idea. This is not possible. But it is possible for his sperm and his wife’s eggs be frozen in liquid nitrogen as they are left to drift into the frigid coldness and darkness of space where the temperature is almost 0 Kelvin. The pilot may either remain back on Earth, or kept in suspended animation if he wishes to follow his genes aboard. A robot is programmed to take over which will only be activated towards the end of the journey. The awaken robot will then take out the sperm-ovum in deep freeze, fertilize them. It will then nurse them, and bring them up. It will teach them where they came from, their language, culture and civilization, and what their world looked like.
A voyage guided and narrated by a robot:
It will tell him or her purpose of their voyage, their fate, destination, and destiny. They will pictures and images of their world, their parents, other humans, plants, animals and all other living things. All the images of this world will be beamed towards the plane for their benefit from the day it left Earth and the Solar System. All scenes of Earth will be continuous with time, such that the entire length of 40,000 million years of history from the beginning to the edge of this Universe could be shown.
The TV transmissions will not be broadcast the usual way. This will ‘dilute’ the energy of transmission over a wider and wider volume into space as the plane leaves planet Earth. The entire energy of the signals will have to be concentrated into just a very narrow beam in the direction of the plane in order to focus the pictures clearly on arrival without being spread out. It would be like a laser beam.
The transmission should be continuous so that there is no gap in time in receiving the images. Even then, towards the end of the journey, all the images would have been at least 40,000 millions years out of date for the far-away pilot. He will perhaps never be able to learn of his / her origin, and the world he came from.
Time needed even by light to cross the chasm:
The distance across the horrendous chasm of the Universe from one end to the other is: 299792458 metres ÷ 1000 (to change into km / sec) x 60 sec (to change to km per min) x 60 min (to change to km per hour) x 24 hr in a day x 365.25 days in a year x 40,000, 000,000 years = 3.78 x 1023 km (378 sextillion km).
This will take light 4 x 1010 (40 billion) years to cross this grotesque time
space corridor. How long would my Air Bus A320 take?
‘Ask Captain KH Lim’ in his fantastic website I highly recommend you to visit if you ‘Ctrl & Click’ on my articles above. They are linked to his.
http://askcaptainlim.com/index.php?option=com_content&view=category&id=74&Itemid=89
What a journey? Fancy that! Have a safe trip.
JB Lim
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