The Creation of Earth and Oceans

Dear Paul,

Thank you very much for your comment of my earlier article at:

http://scientificlogic.blogspot.com/search?q=how+much+heat+to+boil++off+oceans

I appreciate it very much.  You have asked a very good question, the answer to which I was from the very beginning tried to avoid because I really do not have the answers.

 My article was actually meant to be a very simple one for the sake of scientific interest for ordinary readers. It was never meant to go into the intricacies of various unknown reactions and inter reactions such as whether or not the water will condense back to Earth when it boils off high up into troposphere, stratosphere, mesosphere, the exosphere and beyond into the magnetosphere, solar wind towards the Sun and outwards into the Solar System. That was never my intention. 

Mine article was just to ask a simple question as to how much heat is needed to convert ice into water, and from water into steam till they all boil off. I wrote it mainly for school science students and ordinary readers just to generate an interest in popular science. 

I naturally assume we will be dealing with water and ice when brought to heat under ordinary atmospheric conditions, say in a lab or in the kitchen when we boil water, or that some ice from our refrigerator. 

If that was our intention, then using simple basic equations in physics most people learn in school, then the question how much heat would be needed if we were to boil off all the water on planet Earth will still hold true and remains very valid.

But since you have asked, I have elaborate the article with a little more additions which I tried to avoid as it was just meant for lay readers. My aim is to make the complexities of Science, Mathematics and Technology reachable to the lay masses in a popular way. It is so different from publishing a research paper in a journal. 

So let me stretch my initial article just by a little bit more. Hopefully, it will answer your question.  Many thanks again fort your comments. 

We know that in order to thaw a kilogram of pure ice under ordinary conditions, the heat of fusion (melting or freezing) of ice is always 334 kJ /kg. Thus if we decide to melt 4.3 trillion (4.3 x 10 ¹²)  tonnes or 4.3 x 10 ¹⁵ kg of ice from Greenland, Antarctica, the glaciers and ice caps  then the heat  energy required would be:

E (Energy) = 334 x (4.3 x 10 ¹²) x 1000 = 1.436 x 10 ¹⁸ kilojoules.  (1 tonne = 1000 kg).  This equation does not change under ordinary physical conditions. 

Similarly, the amount of heat energy required to change temperature of one gram of liquid water by one degree Celsius (at 15 ⁰ C) will require 1 calorie. This is equivalent to 4.184 joule. Thus if we were to change 4.3 trillion (4.3 x 10¹²) tons or 4.3 x 10¹⁵ kg of melted ice  from 0⁰ C  to its boiling point at 100⁰ C  under atmospheric pressure, then  the amount of heat required will be:

4.184 J x (4.3 x 10¹⁵) tonnes x 1000 kg x 1000 gm = 1.8 x 10 ²² joules or 18,000,000,000 terajoules (TJ)

A terajoule (TJ) is equal to one trillion (10¹²) joules.

In the same way, the amount of heat required to boil off   4.3 x 10¹⁸  kg of pure water  at 100 ⁰ C under standard  sea-level pressure = (4.3 x 15¹⁸) kg  x 2260 = 9.72 x 10 ²¹ Joules since it requires 2260 KJ of heat to boil off 1 kg of water at 100 ⁰ C under STP, This is the latent heat of evaporation of water.

You might ask where then can Earth get the energy to boil off all the waters from its surface? Is this possible?

There are only two possible sources. 

·          Geothermal energy

·          The Energy of the Sun

We shall discuss these two possibilities shortly. But first allow me to list out the content of this follow-up article.

Content of this Article:

1.     The Hydrosphere
2.     How much water
3.     Global Ice Mass
4.     How much heat ice to water
5.     Temperature of Oceans
6.     How much heat needed to boil water?
7.     How much heat required from water to steam?
8.     The enthalpy of vaporization x mass of water:  
9.     Geothermal Energy
10.  The Sun
11.  Mass of Sun
12.  The Sun, a Main Sequence Star:
13.  Energy Output of Sun
14.  The Death of the Sun
15.  Second life:
16.  Geothermal Energy
17.  Total Energy of the Sun
18.  What happens if the Sun becomes a supernova?
19.  The Milky Way Galaxy
20.  What Happens if Sun Explodes
21.  Supernova
22.  Volume of Earth-Sun Sphere
23.  Light time
24.  Solar Energy Received:
25.  Solar Energy:
26.  Distance of Sun:
27.  Theoretical Calculations vs Satellite Measurements:
28.  The Total Energy of the Sun:
29.  Calculated vs. Estimated Energy Output:
30.  Gravitation Force
31.  The Volume of the Oceans:
32.  Vaporized Water and Waterless World: The Pull
33.  Geosynchronous Satellite
34.  Height of Satellite:
35.  Speed of a Geostationary Satellite:
36.  Why does a geostationary satellite not fall down?
37.  The Separation of the Waters: A Cloud of Waters:
38.  Clouds float, Rain falls:
39.  Water Condensing:
40.  Making it Easier:
41.  A Narration Summary from the Calculations:
42.  The Searing Scenario:
43.  Scaling it down:
44.  How much water then?
45.  A Mini Supernova:
46.  Gravity acts only…? 
47.  Neither would there be a merry-go-round:
48.  A Barren Earth:
49.  The Creation of Earth:
50.  Chemical marriage of hydrogen and oxygen:
51.  The Waters Above and Below:
52.  Torrential Rains for A Millennium:
53.  The Birth of the Oceans:
54.  The Earth and the Oceans  How much hydrogen and oxygen: 
55. The Chemical Calculations: 
56. How shall approach this problem?  
57. Mass of Hydrogen atom: In Deep Space: 
58. Background Radiation: 
59. How Much? 
60. What a Horrendous Volume 
61. How Big was The Gas Cloud?
62. Light Time
63. Bigger or Smaller
64. How far between Sun to Pluto and back?  
65. Genesis
66. The following equivalents were used in the calculations for the above article:

The Hydrosphere:

The hydrosphere described in physical geography is the combined mass of water found on, under, and over the surface of a planet.

Scientists estimate total mass of the Earth's hydrosphere is about 1.4 × 10¹⁸ metric tons (1.4 x 10 ²¹ kg) or about 0.023% of the Earth's total mass. The Earth’s atmosphere contains about  20 × 10¹² tonnes of water. Water has a density of one tonne of water per cubic metre.  Some 75% of the Earth's surface is covered by water. This represents an area of about 361 million square kilometers, or 139.5 million square miles. 

Since the total amount of water on Earth is estimated at 1.4 x 10 ²¹ kg, let us assume there may be another 10 % of it unaccounted for, hidden somewhere on the planet. This put the total amount to be about 1.4 x 10 ²¹ kg + 10 % hidden amount = 1.54 X 10 ²¹ kg

How much water:

The total amount of ice + liquid water + water vapour + 10% extra allowance for hidden and unaccounted water on planet Earth is estimated to be:

1.4 x 10 ²¹ kg + 10/100 (1.4 x 10 ²¹) = 1.54 x 10 ²¹ kg

Mass of Earth = 5.974 x 10²⁴ kilograms 
Mass of Earth devoid of all water = Mass of Earth = 5.974 x 10²⁴ kilograms - 1.54 x 10 ²¹ kg

= 5.9726 x 10 ²⁴ kg

Mass of Earth devoid of all water = Mass of Earth = 5.974 x 10²⁴ kilograms - 1.54 x 10 ²¹ kg

= 5.97 x 10 ²⁴ kg

Global Ice Mass:

It has been estimated that total global ice mass loss from Greenland, Antarctica, glaciers and ice caps was about 4.3 trillion tons since the turn of the century a NASA  study found. 

That amount has raised the global sea level by 0.5 inches (12 millimeters). How much ice is that? It would cover the entire United States 1.5 feet deep, scientists said.

How much heat:  ice to water?

Hence the amount of heat required to raise 1.54 X 10 ²¹ kg of water from an average ocean temperature of 17 ⁰ C to 100 ⁰ C as given by:

Q =cm Δ t

(where, Q = amount of heat required, c =specific heat, m = mass,  Δ t = change in temperature)

= (100-17) x 4.184 (specific heat of water) x (1.54 x 10 ²¹) kg x 1000 (joules into kilojoules)  = 5.35 x 10 ²⁶ kilojoules (ok)

(The average temperature of all the ocean waters is 17 ⁰ C).

Temperature of the Oceans:

Water temperature in the deepest parts of the ocean averages about 36°F (2°C). The average temperature of ocean surface waters = 17 degrees Celsius.

The average water temperature worldwide was 62.6 degrees Fahrenheit  (17 C) according to the National Climatic Data Center.

Every summer the hottest places for water are usually the red sea and the Persian Gulf. The Gulf of Mexico is pretty hot. However the water at Abu Dabi in the UAE is usually the hottest. It can get up to 94/95 deg F (34 / 35 deg C) degrees in summer. The coolest recorded ocean temperature was 15.17 degrees C (59.3 degrees F) in December 1909.

How much heat needed to boil water?

A calorie is defined as an amount of heat required to change temperature of one gram of liquid water by one degree Celsius.

Thus, the amount of heat energy required to change temperature of one gram of liquid water by one degree Celsius (at 15 ⁰ C) will require 1 calorie. This is equivalent to 4.184 joule. Thus if we were to change 4.3 trillion (4.3 x 10¹²) tons or 4.3 x 10¹⁵ kg of melted ice  from 0⁰ C  to its boiling point at 100⁰ C  under atmospheric pressure, then  the amount of heat required will be:

4.184 J x (4.3 x 10¹⁵) tonnes x 1000 kg x 1000 gm  = 1.8 x 10 ²² joules or 18,000,000,000 terajoules (TJ)

(A terajoule (TJ) is equal to one trillion (10¹²) joules).

In the same way, the amount of heat required to boil off   4.3 x 10¹⁸  kg of pure water  at 100 ⁰ C under standard  sea-level pressure = (4.3 x 15¹⁸) kg  x 2260 = 9.72 x 10 ²¹ Joules since it requires 2260 KJ of heat to boil off 1 kg of water. This is the latent heat of evaporation of water.

How much heat required from water to steam?

Now having reached all the waters to standard boiling point at sea-level STP, the amount of heat required to boil them all off will require:

The enthalpy of vaporization x mass of water:  

= (2257 kJ kg⁻¹) x (1.54 x 10 ²¹) = 3.48 x 10²⁴ kilojoules or 3.48 x 10 ²⁷ joules

Hence the total heat requirements to change all the average ocean and global waters estimated  at 17 ⁰ C  into ordinary steam at 100 ⁰ C under normal sea-level pressure,  would be:

5.35 x 10 ²⁶ + 3.48 x 10 ²⁴ = 5.38 x 10 ²⁶ kilojoules (okay)

(heat of fusion of ice is 334 kJ/kg)

This is the latent heat of evaporation of water is 2257 kilojoules / kg, meaning, it requires 2257 kilojoules of heat to boil off 1 kg of water at 100 ⁰ C under STP,

In the same way, the amount of heat required to boil off   4.3 x 10¹⁸  kg of pure water  at 100 ⁰ C under standard  sea-level pressure = (4.3 x 10 ¹⁸) kg  x 2257 = 9.7 x 10 ²¹ kilojoules  (9.7051 x 10 ²⁴) Joules since it requires 2257 KJ of heat to boil off 1 kg of water. This is the enthalpy of vaporization of water, or the latent heat of evaporation of water.  This applies provided there is no change in the state of other physical conditions such as changes in the pressure as the water is evaporated off into the colder outer space re- condensation of the water back to Earth, increase of heat to super heat the steam, etc.

However, if the total amount of water in the entire hydrosphere estimated to be 1.54 x 10 ²¹ kg were to be boiled off, then the amount of heat required would be:

(1.54 x 10 ²¹ kg) x (100-17) x

Mass of Earth devoid of all water = Mass of Earth = 5.974 x 10²⁴ kilograms - 1.54 x 10 ²¹ kg  = 5.97 x 10 ²⁴ kg

(Mass of Earth = 5.974 x 10²⁴ kilograms)

(Mass of total waters = 1.54 x 10 ²¹ kg) 


 
The enthalpy of vaporization, sometimes called,  the  heat of vaporization or heat of evaporation, is the energy required to change a given quantity of a substance from a liquid into a its gaseous stage at a given pressure (often atmospheric pressure).


Normally it is measured at the normal boiling point of a substance example water into steam at 100 ⁰ C.  

Geothermal Energy:

Now let us answer the question where would it be possible to get so much heat to boil off all the waters of the oceans and elsewhere hidden on Earth? One nearest and tremendous source of heat naturally lies just below our feet – in the Earth’s core.

The heat from the Earth’s core is extremely high. The temperature of the interior of Earth is estimated  between 5650 to 7,000 Kelvin. 

The temperature of the interior of Earth is estimated to be between 5650 and 7,000 Kelvin.  Total heat loss from the earth’s interior to the surface is estimated at 44.2 TW (4.42 × 10¹³ watts). This is about 1/10 watt/square meter on average.  This amount is about 1/10,000 of the energy from the Sun.

The total heat locked up in the interior of earth is estimated at 10³¹ joules.  But it only requires 1.8 x 10 ²² Joules to change 4.3 trillion (4.3 x 10¹²) tons or 4.3 x 10¹⁵ kg of melted ice on Earth  from 0⁰ C  to its boiling point at 100⁰ C  under atmospheric pressure, and another 9.72 x 10 ²¹ Joules to boil off a total of 4.3 x 10¹⁸  kg of water. 

We have shown that the total heat energy required to boil off all the ice + water on Earth is:

1.8 x 10 ²² + 9.72 x 10 ²¹ = 2.77 x 10 ²² Joules

 (2260 KJ of heat required to boil off 1 kg of water).

But the heat locked away as geothermal energy in the Earth’s interior is 10³¹ Joules. This is a whooping 361 million times more heat in the interior of Earth than it would be required to boil off every drop of water from the Earth’s surface if the Earth were to crack open into many parts (very, very unlikely), and all the waters on the Earth’s surface drains right into the interior, or if all the geothermal heat were to suddenly leak out onto oceans beds, seas, rivers, and lakes, and everywhere in the hydrosphere.

Of course for the Earth to crack open to gush out all its trapped heat, or for all the waters on the surface o=f Earth to gush into the interior of Earth can never happen, but just for the sake of argument and interest in imagination for this article, let us assume it happens. 

If this possibility is just in a fringe of our imagination, let us then consider the energy from the Sun. 

 

The Energy of the Sun:

The Sun was formed from the collapse of part of a giant molecular cloud of hydrogen and helium about 4.57 billion (4,570,000,000) years ago.  (1 billion is 10⁹ or 100 million). The Sun pours out a horrendous amount of energy each second, and it can continue to generate its stupendous energy for another 5,000 million years. But before that, let’s briefly talk something about the Sun. 

Mass of Sun:

The mass of our Sun is 2×10³⁰ kilograms. This is 330,000 times that of Earth.

The mass of the Sun (solar mass) makes up 99.86% of the total mass of the Solar System. Hydrogen is the principal composition of the Sun. It accounts for 75 % of the Sun's mass, the rest is mostly helium

As in most stars, our Sun is called a main-sequence star, generating its energy through nuclear fusion by converting its hydrogen nuclei into helium. In its core, the Sun fuses 620 million metric tons of hydrogen into helium each second.

Such a reaction, known as a proton–proton chain occurs around a whooping tune of 9.2×10³⁷ times per second within the core of the Sun. Four free protons of hydrogen nuclei) are being used in the conversion, and about 3.7×10³⁸ protons are being converted into alpha particles of helium nuclei every second.   

There are about 8.9×10⁵⁶ free protons in the Sun, or about 6.2×10¹¹ kg of hydrogen nuclei are being converted each second.

About 0.7% of the fused mass is released in the mass-energy conversion, the rate of conversion is in the order of 4.26 million metric tons per second, yielding 3.846×10²⁶ W of energy. This is the power of 9.192×10¹⁰ megatons of TNT each second.

The Sun is a Main Sequence Star:

The Sun, like most stars in the Universe, is one the main sequence stage of life.

Sun is middle-aged about 4.57 billion (4,570,000,000) years old The Sun is about halfway through its main-sequence evolution.  . At this rate, the Sun has so far converted around 100 Earth-masses of matter into energy. 

The Sun will spend a total of 10 billion years as a main-sequence star.

Energy Output of Sun:

Every second, 600 million tons of hydrogen is converted into helium in the Sun’s core, generating 4 x 10²⁷ Watts of energy, and neutrinos. This process has been going on for 4.57 billion years  since its formation.  But there isn’t an unlimited amount of hydrogen in the core of the Sun. In fact, it’s only got another 7 billion years worth of fuel left. 

The Death of the Sun:

Approximately 6000 million years from now, the current chapter of our Sun will come to a close.  The Sun will run out of hydrogen fuel. When this occurs,  the helium in the conversion from  hydrogen will built up. This will make it unpredictable and unstable.  It will then collapse under its own gravitational pull. This immense pressure within the Sun’s core as it collapses will cause the interior to heat up making it becomes denser. 

A Red Giant:

It will then begin to bloat up to become a red giant as part of stellar evolution. 

In the scenario that follows the red giant phase, extreme thermal pulsations will cause the Sun to fling off its outer layers, forming a planetary nebula. What remains after the outer layers are evicted is the exceedingly hot stellar core. This will slowly cool and fade into a white dwarf over billions of years. It will be a scenario of a low to medium mass stars 

The expanding red giant Sun will devour the orbits of Mercury Venus,  the Earth, and perhaps into the outer Solar System as well. The intense heat of an expanding and dying red Sun will burn Earth and boil off all the waters within. Nothing can survive then. The Sun will then end its life as a main sequence star

Second life:

But that’s not the final end.  When it enters its stage as a red giant, there will be a built up heat and pressure in the heart of the red giant. The red giant Sun will trigger off a another fusion, but this time it will convert helium into carbon instead of hydrogen into helium. 

 It will continue this fusion for another 100 million years until all helium fuel is exhausted. In its death breath, the envelope of helium becomes unbalanced causing the Sun to pulse strongly. It will then blow off several portions of its atmosphere into space; many of these pulses will reach Earth and the outer planets taking all the other planets into the grave with it.

Total Energy of Sun: 

The Sun is much farther away than what lies in the interior of Earth below our feet. So we can expect to get much less energy. But is this so? Let us have a look at some facts and figures. 

If the Sun's current energy output is (3.839 × 10 ²⁶)  Joules per second, and if we assume that the Sun has a 10, 000 million (1 X 10 ¹⁰)  year main sequence lifetime, then we can expect it to have a total energy output equivalent to 3.839 × 10 ²⁶ J/s, multiply by (1 x 10 ¹⁰) x 31557600 = 1.2 x 10 ⁴⁴ Joules. This is a horrendously huge amounts of energy beyond imagination. 

(1 solar year = 86400 sec. a day x 365.25 solar days a year = 31,557,600 seconds).

Even then, how does this energy output compare with the energy released by a supernova? Let us have a look.

Give or take a factor 10 the total energy output of a supernova is 10^46 J, one percent of which is visible light, and 99% neutrinos. The solar mass of a supernova is 10 – 100 times that of our Sun. So the total energy output of a supernova ought to be 10 – 100  times as great as that of the Sun should our Sun decides to become a supernova, namely between 10⁴⁵ – 10⁴⁶ Joules per second. 

Once again, rest assured our Sun is not massive enough to become a supernova. We need not worry.

If the Sun's current energy output is (3.839 × 10 ²⁶)  Joules per second, and if we assume that the Sun has a 10, 000 million (1 X 10 ¹⁰)  year main sequence lifetime, then we can expect it to have a total energy output equivalent to 3.839 × 10 ²⁶ J/s, multiply by (1 x 10 ¹⁰) x 31557600 = 1.2 x 10 ⁴⁴ Joules.

(1 solar year = 86400 sec. a day x 365.25 solar days a year = 31,557,600 seconds). 

But what happens if the Sun becomes a supernova?

Stars with a solar mass of ten or greater can explode into a supernova as their inert iron cores collapse into an tremendously dense neutron star or even a black hole.

The Sun does not have enough mass to qualify into a supernova. Instead, in about 5 billion years just as its hydrogen fuel runs out, it will enter a red giant phase. Its outer shell will expand as the last remaining hydrogen fuel in its core is consumed.

The Giant Sun Expanding:

The red giant Sun will steadily expand until the core temperature reaches around 100 million Kelvin, The core will begin producing carbon. Intense thermal pulsations will cause the Sun to eject its outer layers forming a planetary nebula leaving only an extremely hot core. This will slowly cool and fade into a white dwarf over tens of billions of years. This shall be the fate of our Sun. (OKAY)

In a supernova event should the Sun decides to explode (not possible) then heat and light may arrive on Earth in 8 min 19 sec, but its shock waves sweeping out an expanding shell of solar materials consisting of hydrogen, helium, neutrinos and other elements traveling at 30, 000 km per second ( 10 % the speed of light) may take about 83 minutes or more to arrive.

The Energy Outshines the Entire Galaxy:

But since the energy output of a supernova outshines the energy output of an entire galaxy like our Milky Way Galaxy of 150,000 million stars, then if we presume the Sun is going to behave like that then we can expect the energy output of the Sun will suddenly increase by 200–400 thousand million times. 

Let us take a very modest figure of just 150 000 000 000 stars for our Milk Way Galaxy, each star like our own Sun releasing  3.846 × 10²⁶ J of energy per second, then we can expect a sudden release in the order 5.8 x 10³⁷ J each second. However, once again allow me to remind my gentle readers that our Sun is not a super massive star, and it can never become a supernova. 

Instead it will slowly outlive its life in another about 7 billion (7000 000 000) years  before becoming a red giant star to engulf the Mercury, Venus, Earth and most of the outer planets beyond. It is very unlikely the Sun will explode like a supernova. 

Such a Rare Event:

In fact a super nova is so rare that there were only about supernovae recorded in the entire history of mankind, the last prominent one was the Crab Nebula.

So rest assured, nothing will happen to our Sun. But just for the sake of this article, let us assume it will go supernova. In that case the Sun will suddenly release 5.8 x 10³⁷ J of energy each second. With such stupendous energy output, not just all the water on Earth, but the entire Earth and all its elements within will be annihilated completely in a blink of a second. 

There is no question of asking how much heat energy will be required to boil off all the waters on this planet.

The Milky Way Galaxy:

Our Milky Way is a barred spiral galaxy, estimated of between 100,000–120,000 light-years across. It is a normal galaxy the population is between 200 to 400 billion stars (200 000 million – 400 000 million). Our Sun is just one of the average stars in our Milky Way Galaxy. 

The number of stars in smaller galaxies is about 100 billion (100,000,000,000).

What Happens if Sun Explodes?

If the Sun decides  to explode into a supernova (not possible) the energy releases would be  6.8 x 10 ²⁹ Joules.

This is 195 times more than is needed to boil off the entire waters found and hidden on planet Earth

Supernova:

The Sun can never become a supernova. It will have to be a super massive Sun with some 10 solar masses before it can explode into a supernova.

A supernova is an explosion of a massive super giant star. It may shine with the brightness of 10 billion suns! The total energy output may be 10⁴⁴ joules, as much as the total output of the sun during its 10 billion year lifetime.

Should it explode, it can last only about 2 weeks during which it will radiate as much energy as the Sun would emit over its entire life span of 100 billion years.  The explosion will throw off much or all of a star's material to the tune of 10 % the speed of light, namely at 30,000 km/s sending off shock waves into the surrounding deep space.  This shock wave sweeps up an escalating shell of gas and dust called the remnants of a supernova. 

Even though no supernova has been seen in the Milky Way since 1604, supernovae remnants indicate that on average, a supernova event occurs approximately only once every 50 years in our own Milky Way Galaxy. The supernova seen in the year AD 1054 gave rise to the Crab Nebula, and the last SN remnants seen in the year 1604 was the Kepler's supernova remnant.

The cloudy remnants of SN 1054 are now known as the Crab Nebula.

Supernova plays an important role by enriching the interstellar medium with higher mass elements. Additionally, the expanding shells of shock waves bursting out from supernova explosions can activate the birth of new stars. 

Volume of Earth-Sun Sphere:

Let us do some simple calculations before narrating further.

The Astronomical Unit is the distance between the Earth and the Sun. As the oribit of the Earth round the Sun is an ellipse, the mean Earth-Sun distance is 149,597,870.700 kilometres  or 1.495979 x 10 ¹¹ metres. This is the Earth-Sun mean distance.

Since the volume (V) of a sphere is given by: 

V = 4/3 πr³

Hence, the volume of the sphere in space enclosed by the Sun to Earth’s distance as the radius would be:

 1.402 x 10 ³⁴ cubic metres.

The surface area of such a sphere with a radius of 1 A.U. will have a volume equals to:

 A = π r²

= 7 x 10²² square metres 

Area of Sphere of Earth:

The surface area of a sphere is given by :

A = π r²

The radius of the Earth varies between 6,353 and 6,384 km. (3,947–3,968 miles). The mean radius is 6,371 km (6371,000 metres) 

Applying this formula, surface area of Earth would be about 5.1 x 10 ¹⁴ square metres if we ignore the tiny bumps of mountains and hills, the dips of oceans and seas and other uneven terrain here and there all over the skin thin Earth’s crust

Light time:

The time taken for light from the Sun to reach Earth is:

Sun-Earth distance ÷ Speed of light

=1.495979 x 10 ¹¹ metres ÷ 299 792 458 m /s

= 499 seconds (8 min 19 sec)

Solar Energy Received:

The Earth surface area is   = 5.1 x 10 ¹⁴ square metres. This is (5.1 x 10 ¹⁴) ÷ (7 x 10²²) x 100 = 7.29 x 10^-7 % or 0.0 000 007 % of the Sun-Earth spherical area.

Since the Sun emits 3.846 × 10²⁶ watts (Joules) per second (1 watt second = 1 J)

Hence each second the Earth theoretically should receive;

Energy output of Sun per second ÷ Surface area of Sun-Earth sphere x Surface area of Earth ÷ ½ (sunlit surface) 

= (3.846 x10²⁶) ÷ (7 x 10²²) x (5.1 x 10 ¹⁴) ÷ 2 = 1.4 x 10¹⁸ Joules per second or (1.4 x 10¹⁸) watt-second for the sunlit part at any time.

Solar Energy:

The energy output of Sun is 3.8×10²⁶ joule  per second or 3.8×10²⁶  watt second

 (1 watt second = 1 joule)

Hence in its life span of 10 billion years (3.15576 x 10¹⁷   seconds), the Sun will have expended (3.846 × 10²⁶) x (3.15576 x 10¹⁷) = 1.2 x 10 ⁴⁴ Joules of energy. If it decides to go supernova and releases all these horrendous amount of energy in, say 15 days (1,296,000 seconds). Hence each second, the amount of energy emitted will be 9.26 x 10 ³⁷ J.

Distance of Sun:

In order for us to calculate out how much solar energy the Earth receive at least in theory, we need to know how far the Sun is away from Earth, and how much of energy is spread out into space over a sphere sustained  by the Earth-Sun distance.

The mean distance of the Sun from the Earth is at an average distance of 149.6 million kilometers called an Astronomical Unit (1 AU).  This distance varies as the Earth moves from perihelion when it is closest to the Sun at 147,098,291 km in January to aphelion when it is farthest from the Sun at 152,098,233 km in July as Earth describes an elliptical orbit round the Sun. 

Over this distance, light from the Sun, takes 8 minutes and 19 seconds to reach Earth.

Thus if the energy from the Sun is spread out over a sphere of 7 x 10²² square metres sustained by the Sun-Earth distance, then Earth will get part of the energy determined by the surface area of Earth.  

The surface area of Earth or a sphere is given by : 

A = π r²

The radius of the Earth varies between 6,353 and 6,384 km. (3,947–3,968 miles). The mean radius is 6,371 km (6371,000 metres) 

Applying this formula, surface area of Earth would be about 5.1 x 10 ¹⁴ square metres if we ignore the tiny bumps of mountains and hills, the dips of oceans and seas and other uneven terrain here and there all over the skin thin Earth’s crust

Since the surface area of Earth is 5.1 x 10 ¹⁴ square metres, Earth will receive:

Energy output / Total area of Sun-Earth sphere x Area of Earth

= 9.365 x 10 ³⁷  ÷ 7 x 10²² x 5.1 x 10 ¹⁴ = 6.8 x 10 ²⁹ J (okay)

The Earth surface area is   = 5.1 x 10 ¹⁴ square metres. This is (5.1 x 10 ¹⁴) ÷ (7 x 10²²⁾ x 100 = 7.29 x 10^-7 % or 0.0 000 007 % of the Sun-Earth spherical area.

The surface area of a sphere can be calculated by:

A = ∏ r²

where r, is the mean radius of Earth at  6,371 km (6371,000 metres) between 6,353 and 6,384 km. (3,947–3,968 miles), the surface area of Earth would be about 5.1 x 10 ¹⁴ square metres if we ignore the tiny bumps of mountains and hills, the dips of oceans and seas and other uneven terrain here and there all over the skin thin Earth’s crust

The Earth receives 174 petawatts (PW) of solar radiation at the upper atmosphere. Approximately 30% of this is reflected back to space while the rest is absorbed by clouds, oceans and land masses

The solar constant embraces all types of solar radiation, not just the visible light. It is measured by satellite to be roughly 1.361 kilowatts per square meter (kW/m²) at solar minimum (measurements done by satellite)

Let us now calculate the solar energy Earth receives at maximum solar output using  first principle:

Energy output of Sun per second ÷ Surface area of Sun-Earth sphere x Surface area of Earth ÷ ½ (sunlit surface)

= (3.846 x10²⁶) ÷ (7 x 10²²) x (5.1 x 10 ¹⁴) ÷ 2 = 1.4 x 10¹⁸ Joules per second or (1.4 x 10¹⁸) watt-second

= 1.4 x 10 ¹⁸ watt-second (max solar output) ÷ 5.1 x 10 ¹⁴ (surface area of Earth) ÷ 1000 (watts into kilowatts) x 0.7 (70 % absorbed, 30 % reflected back into space) = 1.92 KW / sq. metre

Theoretical Calculations vs Satellite Measurements: 

·          1.9 kilowatts per square metre (calculated out by first principle at solar maximum)

·          1.361 kilowatts per square meter (kW/m²) as measured by satellite at solar minimum

The Beauty of Mathematics:

Please note the very close match, one at maximum solar output (determined by simple calculations), and the other (by actual satellite measurements) at minimum solar output.

The above calculation using simple school-level mathematics demonstrates the beauty of mathematic.

The Total Energy of the Sun:

There energy output of Sun in its  entire life of 10 billion years = (3.15576 x 10¹⁷) seconds x 3.846 × 10²⁶ J = 1.2 x 10⁴⁴ Joules ok

Since the Sun emits 3.846 × 10²⁶ watts (Joules) per second (1 watt second = 1 J),  its energy output in one second in its 15 days of it being a supernova will be 1.2 x 10⁴⁴   Joules ÷ 1,296,000 seconds = (9.26 x 10 ³⁷) .

Another way of putting it, if there are 250 000 million stars in our Milky Way, and if a supernova in it explodes, its very short life-span of just a few days will outshines the energy output of an entire galaxy as claimed by astronomers and astrophysicists and then the total energy output for the entire galaxy =

(3.846 × 10²⁶) x 2.5 x 10 ¹¹ = 9.615 x 10 ³⁷ Joules

Calculated vs. Estimated Energy Output:

1.     9.26 x 10 ³⁷) J (calculated out here by first principle)

2.     9.615 x 10 ³⁷ J (estimated by the consensus of scientists elsewhere)

The Beauty of Mathematics Again:

We can see the remarkably similar results of total energy of our Sun as calculated out above using first principles, and the total energy output of all 250 000 million stars in a galaxy as scientists said ‘the energy output of a supernova will outshines the total energy output of an entire galaxy’. What a beautiful statement!

Gravitation Force:

Let us now talk on something else, but related.

The gravitation force of attraction between two bodies is proportional the product of their masses and inversely proportional to the square of their distances.

Let us take an example. Applying the law of universal gravitation, the gravitation force (F) between the Earth and Moon is proportional to the product of their masses (m₁ and m₂), and inversely proportional to the square of the distance (d) between them:

F = G. m1m2 / d²

= 6.67300 × 10^-11 x (5.974 x 10²⁴) x (7.36x10²²) kg ÷ (384,400,000 metres)²

= 1.9856 x 10 ²⁰ Newton (ok)

Where, G (Gravitational Constant) = 6.67300 × 10^-11 m³ kg^-1 s^-2 (Newton)

M1 = mass of Earth = 5.974 x 10 ²⁴ kg

M2 = mass of the Moon = 7.36x10 ²² kg

d = average distance between Earth and elliptical orbit of Moon = 384,400 kilometers (384,400,000 metres)

The Volume of the Oceans:

The mass of the oceans is in the region of 1.35×10¹⁸ metric tons (1.35 x 10 ²¹) kg. This is roughly 1/4400 of the total mass of the Earth. In terms of coverage, the oceans cover up an area of 3.618×10⁸ km² with a mean depth of 3,682 m (3.682 km). An estimated volume of the oceans is 1.332×10⁹ km³.

Most of the water (97.5%) is salt water; the remaining 2.5% is fresh water. Of the fresh water, about 68.7%, is in the form of ice.

Vaporized Water and Waterless World: The Pull

Let us now imagine the world is now completely devoid of water due to say a sudden surge of heat from the Sun causing all the ice to melt off, and all the water to suddenly vaporized away. But before we proceed, I have to admit this can never happen even in the foreseeable future because our Sun is not massive enough for it to suddenly explode into a supernova. 

It has to be a super massive Sun or star with 10 times in mass in order this to happen. But for the sake of this article, let us assume our Sun can go bang like a supernova. 

Let us assume all the waters on Earth have vaporized in a puff. Let us imagine the heat is just sufficient to send all the waters into space, say the same height as a geostationary satellite 35900 km above the Earth’s  surface.

What then will be the pull then between the waterless Earth and all the water 35900 km (35900000 metres) above the surface of a waterless Earth.   

Let us calculate:

Mass of Earth devoid of all water = Mass of Earth = 5.974 x 10²⁴ kilograms - 1.54 x 10 ²¹ kg  = 5.97 x 10 ²⁴ kg (ok)

Mass of Earth = 5.974 x 10²⁴ kilograms

Mass of total waters = 1.54 x 10 ²¹ kg

Force of gravity between a waterless Earth and the mass of water separated out from it  by heat of boiling  to a distance of say, a  geosynchronous orbits of around 37, 000 km (23,000 miles) away.

Applying the law of universal gravitation, the attractive force (F) between the waterless Earth and all the waters separated from it to a distance of  35,900 km (35,900,000 metres)  away from Earth centre will be  proportional to the product of their masses (m₁ and m₂), and inversely proportional to the square of the distance (d) between them:

 F = G.m1m2 / r2

= 6.67300 × 10^-11 x (5.9726 x 10 ²⁴ x 1.4 x 10 ²¹) kg ÷ (3.59 x 10⁷ + radius of Earth ) ²

= 6.67300 × 10^-11 x (5.9726 x 10 ²⁴ x 1.4 x 10 ²¹) kg ÷ (3.59 x 10⁷ + 6.37 x 10⁶ metres) ²

 

= 5.58 x 10 ³⁵ ÷ 1.79 x 10 ¹⁵ = 3.12 x 10 ²⁰ Newton

But for the sake of our imagination, let us assume under almost impossible situation, the Sun does for just a fraction of a second and suddenly it goes back to normal stability again. 

Force of gravity between a waterless Earth and the mass of water separated out from it  by heat of boiling  to a distance of say, a  geosynchronous orbits of around 37, 000 km (23,000 miles) away. 

Applying the law of universal gravitation again, the attractive force (F) between the waterless Earth and all the waters separated from it to a distance of 37,000 km (37,000,000 metres)  away from Earth centre will be  proportional to the product of their masses (m₁ and m₂), and inversely proportional to the square of the distance (r) between them:

 F = G.m1m2 / r2

= 6.67300 × 10^-11 x (5.9726 x 10 ²⁴ x 1.4 x 10 ²¹) kg  ÷ (37,000,000) ²

⁼ 4.075765 x 10 ²⁰ Newton

Geosynchronous Satellite:

Now let us take the example of a satellite that is neither pulled down by Earth’s gravity, nor fly off at a tangent into space.  We shall apply this knowledge later in this article. So kindly be patient, even though we may think it has nothing to do with the ocean waters being boiled away.

A geosynchronous satellite orbits the earth with an orbital period of 24 hours. This synchronizes with the period of the earth's rotational motion. Thus a geostationary satellite appears permanently fixed over the same point on the Earth. 

We can calculate the height of such a satellite as it orbits in the same time as the rotation of Earth in 24 hours

In order to cut short a lengthy explanation and its calculation, allow me to give the final equation. We can derive the results by first principle for the motion of a satellite around Earth as: 

R³ = (T² x G x  M) ÷ (4 x pi²)

R³ = (86400 s)² x (6.673 x 10^-11 N m²/kg²) x (5.98x10²⁴ kg) ÷ (4 x (3.1415)²

= 2.979 x 10 ²⁴ ÷ 39.476

= 7.54 x 10²² m³

(where, T = orbital period (86400 sec),  G = gravitational constant  (6.673 x 10^-11 N m²/kg²), M = Mass of Earth (5.98x10²⁴ kg).

The approximate radius of orbit of the satellite from the centre of Earth would be the cube root of R³ (7.54 x 10²²)  metres. 

R = 4.23 x 10⁷ m (42,300 km) from the Earth’s centre

Height of Satellite: 

Since R (4.23 x 10 ⁷ m ) is the distance of the satellite  from centre of   the Earth we need to subtract the mean radius of  Earth (6.37 x 10⁶ m)   to find the height of the satellite above the surface of Earth

Thus, (4.23 x 10 ⁷ m) – (6.37 x 10⁶ m)   = 35,930,000 m = 35930 km

Speed of a Geostationary Satellite:

The angular speed ω of the satellite can also be found by dividing the angle travelled in one revolution (360° = 2π rad) by the orbital period (the time it takes to make one full revolution). In the case of a geostationary orbit, the orbital period is one sidereal day, or 86,164.09054 seconds). This would be:

 ω = 2 π rad / 86164 = 7.2921 x 10^-5 rad / sec

By multiplying this angular velocity by its radius of orbit (4.23 x 10 ⁷) its orbiting speed would be about 3084.6 metres / sec or 11,104.6 km / hr.

The radius of orbit indicates the distance that the satellite is from the center of the earth. Now that the radius of orbit has been found, the height above the earth can be calculated. 

Since the earth's surface is 6.37 x 10⁶ m from its center (that's the mean radius of the earth), the satellite must be 4.23 x 10⁷ m – 6.37 x 10⁶ = 35,930,000 metres (35,930 km) above the Earth’s surface.  More accurately it is 35,786 km above the Earth’s equator.

Why does a geostationary satellite not fall down like rain?

Why does a satellite stays stationary up there not fall down due to gravity? A satellite stays up there in a circular orbit because of its centrifugal force outwards just balances against the pull of gravity (centripetal force) that is directed inwards towards the centre of the circular motion. If not for gravity the satellite will be flung off at a tangent towards outer space. 

It is these two forces that keep a satellite in a stable orbit around the Earth. But eventually a satellite will still fall down when it loses its speed and kinetic energy as it collides with myriads of micrometeorites in space. 

The Separation of the Waters: A Cloud of Waters:

What about if all the waters were separated from Earth and remain in the same orbit height of a geosynchronous satellite?  Would it not all condense, and fall down back to Earth?

In fact we have calculated that even if all 5.974 x 10 ²⁴  kg  of total amount of water on earth were to be boiled off to the height of say,  a geostationary satellite, the force of attraction between the waters that have separated and the waterless Earth below will still have a whooping force of: 

3.12 x 10 ²⁰ Newton

This is even 1.57 times the force of attraction between the Earth and the Moon at 1.9856 x 10 ²⁰ Newton.

Of course the Moon is much farther away at 384,400 kilometers (384,400,000 metres), whereas the separated water was assumed at 4.23 x 10⁷ m (42,300 km). This is over 9 times farther away, whereas the force of gravitational attraction falls off inversely as the square of their distances, namely, the further away the two objects, the greater the diminishing fall of attraction.  But the gravity is still there no matter how far the objects.

  

Clouds float, Rain falls:

A cloud floats high up, and does not fall down even by the action of gravity. Why is this so?

A typical cloud droplet has a size of 10 micrometres, and it would take 15 million cloud droplets to form an average raindrop. The terminal velocity of a rain droplet is about  0.3 cm/s or about 10 metres per hour.

The fall of rain droplets from the cloud at 3500 m at this speed would take 350 hours.  However, when the droplets coalesce do they form heavier rain drops, some 300 times larger The rain clouds  will then fall down  quickly

The rain droplets moreover is being pushed upwards against gravity by the warmer air below so much so they are actually rising up instead of falling down.  That does not mean clouds are so light that gravity cannot act on it. 

If the water droplets, no matter how light, will accelerate down to Earth in the same rate  of 9.81 m/s²  as lead or a brick had there been no air resistance in a vacuum.   Of course, a droplet caught in a downdraft will descend at the rate of the downdraft plus its terminal velocity.

So in the same vein it does not mean that if all the waters on Earth have evaporated and  stayed  up there at  the same height of say,  a geosynchronous satellite at 35,786 km above the Earth’s equator, it cannot fall down like a  satellite because of ‘zero’ gravity up there.

This is not true at all. Gravity is universal. Its force, albeit very weak is the only force that permeates across the entire Universe, except that it falls off inversely as the square of its distance.

Water Condensing:

Of course, even if all the waters have been boiled off, and there is no more heat left, the super heated steam, steam or water vapour will be exposed to the frigid temperature of outer space. This will cause them to condense and cause them to fall down to Earth. 

There will be 5.974 x 10 ²⁴  kg  of them up there, and there will no air, no resistance in a vacuum to block them to rapidly condense and fall down to Earth if there is no more heat left to keep them apart from Earth. Eventually they will cool and fall down as torrential rains for years until all the oceans, seas, lakes, rivers, and land are filled up again with water.

Making it Easier:

In order to make it easier for us to visualize astronomical dimensions, we need to scale down our Sun and Earth and their distance. 

·           If the Earth’s mean radius of 6.37 x 10⁶ m is shrunk down to just 1.0 mm,  its volume  would be: V = 4/3 πr³  = 4.19 cubic mm

·          Similarly, the Sun with a radius of 6.955 x 10 ⁸  metre will  have to be  a scaled down by:

1/(6.37 x 10⁶) x 6.955 x 10 ⁸ = 109.18 mm. It volume would be 4/3 π (109.18)³ = 5451523.38 cubic mm = 5.45 litres.

·          The Earth and Sun will be separated by a distance of:

1/(6.37 x 10⁶) x 149,597,870,700 = 23484.75 mm = 23.48 metres apart

·          Scale down speed of the supernova explosion: 23.48 metres / 4990 sec = 4.7 x 10 -3 metres or 0.47 cm per second.

A Narration Summary from the Calculations:

The Searing Scenario:

Let us now imagine we now have a scaled-down that is only Earth 4.19 cubic mm in volume.  Imagine we now use the hottest chemical available on Earth – the torch of an oxy-acetylene flame at 3000 0 C, a flame that can easily cut through steel.   

The flame volume from the jet of an oxyacetylene torch is just about 5 cubic milliliter. Let us now direct this blast of flame over 4.19 cubic mm of water.  What happens? Obviously all the water will vapourize instantly in a flicker of a second.

Scaling it down:

Let us now take the second scenario of an extremely searing flame, one that represents the plasma remnants of a star or Sun that has exploded into a supernova by scaling down the size of the Sun and Earth. 

On such a scale, the Sun would be 5.45 litres in size, and the Sun-Earth’s distance would be 23.48 metres (2348 cm) apart.   It has exploded into a supernova where the temperature is known to reach about 100 billion (1 x 10 ¹²) kelvin (100 gigakelvin).

This is 6000 times hotter than the core temperature of the Sun, or 333 million times hotter than an oxyacetylene flame at 3,000 ⁰ C, the hottest available chemical flame on Earth.

Since it has been measured the speed of ejection of stellar materials from a supernova can reach a velocity 10 % the speed of light at 30,000 km/s (30,000,000 m /s) will take 4986.59569 sec = 83.10992817 min = 1.38516547 hours = 1 hr 23 mins

On such a scale, the speed of the supernova ejection will be 2348 cm / 4986.59 sec = 0.47 cm / sec

 I leave this scenario what will happen to the entire Earth to my readers to imagine. I am talking about the entire planet Earth, and not just the oceans, and all the waters found on Earth.

How much water then?

We must not forget the Earth is like an apple, most of it mass is in its interior with its very hot molten core and the Earth’s crust containing all the water on Earth is just like a thin apple skin with some drops of water on it. is just Remember even if we give it the most generous estimate of 1.54 x 10 ²¹ kg of all the visible, invisible and hidden waters on planet Earth compared to the mass of Earth, it is just:

Mass of water / mass of Earth x 100 %

= 1.54 x 10 ²¹ / 5.974 x 10²⁴ kilograms x 100 = 0.026 %

Since the density of water is 1 kg per litre (1 gm /ml), the total water would be 1.54 x 10 ²¹ litres. But the volume of our scaled down Earth is 4.19 cubic mm = 4.19 x 10 ^-3 ml.  (1 ml =1000 cubic millimeter). Hence with 0.026 % water, the actual amount of water is 1.09 x 10 -3 (0.001 ml.)

A Mini Supernova:

Now, I want you to imagine a horrendously searing ball of nuclear fire 5.45 litres in volume representing the Sun suddenly releasing some 5.8 x 10³⁷ Joules of energy each second, at an unspeakable temperature of 100 billion (1000, 000,000,000) Kelvin, 333 million times hotter than the interior of the Sun, inching its way at a rate of 0.47 cm / sec towards a miniaturized Earth just 4.19 cubic mm in volume, and containing just 0.001 ml of water.

What happen to that Earth and all the waters then? Need I say more? I leave this to the imagination of my gentle readers to ponder over during coffee, dinner or bed time.

Gravity acts only…? 

If we have a situation where there is just sufficient heat to boil off every drop of water on Earth and send them hurling up to the height of 40,000 km where geosynchronous satellites are placed, and let us now assume  there is  suddenly no more heat left to push them any further, we can expect the pull of Earth’s gravity albeit weaker at this height,  to bring the waters down again. That 40,000 km it is far beyond the where air is. 

It is almost a vacuum up there.  There will be no air resistance to drag the entire mass of water down. With that kind of heat that can boil off the entire ocean, seas, lakes, rivers glaciers and atmospheric waters, even the entire atmosphere will be torn apart and pushed away.

Even if there is no more heat, both water and air would not stay as a layer over the Earth because they will continue to fly away outwards as nothing could stop them. They will obey Newton first law of motion which states that every object will continues in its state of rest, or of uniform motion in a straight line, unless acted upon by an external force.   

High up at the altitude of over 35, -40,000 km, Earth’s  gravitation force may not be sufficient to pull the waters back to Earth if the momentum is great enough. It will escape gravity completely. 

Neither would there be a merry-go-round:

But if it remains motionless, gravity will act no matter how weak. But there is no reason it will stop moving outwards into space. There is no external force to act against its motion.  There is also no reason to suppose the mass of water will go into orbit round the Earth like a satellite. 

There is nothing to push it round and round in orbit.  If it does the centripetal pull of gravity will just balance the opposite centrifugal force outwards and the waters will stay put in orbit round the Earth. 

Why should this be so?  Why should it orbit the Earth in the first place? We would expect the exceedingly hot super heated steam to be ejected straight outwards into space. It will probably go in puffs of rings  and shells like puffs of steam into the air.

A Barren Earth:

The Earth will be completely barren, of water. It would be similar to the other planets in the Solar System. The Earth would remain very hot then for any water to condense back once it was blown off into the reaches of outer space. I suspect the heat radiation would keep pushing all those waters further, and further away. 

Once at a distance, the Earth’s gravity will rapidly weaken inversely as the square of its distance.  The gravity would no longer be effective then.

The Creation of Earth:

As the formless mass of dark clouds accreted due to inter-atomic and gravitation crush,  the heat generated would be so immense that no ordinary chemical reaction can take place. The heat itself will energize the hydrogen and oxygen atoms to keep them apart to form water. 

In other words, the hydrogen and oxygen atoms will be kept as separate elements to prevent ordinary chemical combustion.  There was completely no water then. The Earth will also be molten, formless, and devoid of water and life.

Chemical marriage of hydrogen and oxygen:

But as the Earth cools over the eons, the hydrogen and oxygen enveloping Earth would allow them to combine in stages for the first time. But they still cannot come down as rain as the Earth was still exceedingly hot. Any attempt to come down as rain, the steam generated by the very hot Earth, would instantly cause them to boil off once again on contact to rejoin the superheated steam still enveloping Earth in space.  

But over the millions of years, the heat would slowly be loss due to continuous cycles of rains and instant boiling off, and remaining heat radiated out into space. The Earth then began to cool. Each cyclic event will require a stupendous amount of heat to boil and cool off. It is an energy transfer from heat into mechanical energy of cyclic rains.  There would be cycles of heat losses then. 

The Waters Above and Below:

The water enveloping the Earth would exist as a separate filament above the Earth, from what was below. By and by, they would then be dragged down to Earth by gravity.  Over the millions of years, the waters existing as a separate filament of superheated steam would cool rapidly at a frigid temperature of near zero Kelvin in space.

Torrential Rains  for A Millennium: 

As both Earth and water cooled, torrential rains will pour down onto Earth for the first time. The rains would probably last for hundreds, if not thousands of years to form the oceans for the earliest time.  The filament above will then be separated from the filament below.

The Birth of the Oceans:

It is my belief during the initial formation of Earth from a large mass of interstellar clouds containing an abundance of hydrogen and oxygen the Earth was exceedingly hot due to the accretion of a mass of the dark clouds coming together.  This dark mass of cloud would be void, and without any form.

The Earth and the Oceans:

The dark formless mass of interstellar cloud without any division that came and created Earth and eons later, water and the oceans were formed.  It was initially just one dark formless mass devoid of anything living.

Although I do not have the scientific evidence to base my hypothesis, it is my very personal and private belief based on my humble understanding of astronomy and astrophysics.

How much hydrogen and oxygen:

The question that just ran into my mind as I pen the last sentence above was, how much hydrogen and oxygen were needed during the genesis of all the waters found in the oceans, seas, lakes, rivers, ice-caps, glaciers, and throughout the entire hydrosphere on Earth?

In order for us to understand that, we need to know how much water was synthesized out of hydrogen and oxygen from the passing gas cloud during the Creation of Earth? To answer that, we must first understand basic chemistry with this equation:

2 H₂ + O₂ → 2 H₂O

In order to calculate how much hydrogen and oxygen were needed to form water; we need to know the mole ratio of hydrogen and oxygen. In chemistry this is technically called the stoichiometric ratio which is the amount of one reactant with the amount of another reactant in the reaction.

In our case the reactants are hydrogen and oxygen as shown in the above chemical reactions. This means, for every two moles of hydrogen, two moles of water are produced. The mole ratio between H₂ and H₂O is 1 mol H₂ ÷ 1 mol H₂O.

The Chemical Calculations:

Now, we need to calculate out the theoretical yield of water to form all the ocean waters, the seas, lakes, rivers, the ice caps, the glaciers, water in the atmosphere, and any hidden waters elsewhere in the entire hydrosphere on this planet Earth.

How shall approach this problem?

We shall do this:

Convert the molar mass of hydrogen into moles

1.     Apply the mole ratio between hydrogen and water

2.     Apply the molar mass of water to change moles water to mass of water

This applies for all reactants in a chemical reaction, not just hydrogen and oxygen to form water. In our case, it is the synthesis of water.

Thus,

Wt. of water produced = wt. of reactant x (1 mol reactant ÷ molar mass of reactant) x (mole ratio product ÷ reactant) x (molar mass of product ÷ 1 mol product)

In order to calculate the amount of oxygen required, we need to know the mole ratio of oxygen to water. The equation tells us, that for every mole of oxygen used, 2 moles of water are produced. The mole ratio between oxygen and water is 1 mol O² ÷ 2 mol H₂O.

 

 Thus the amount of oxygen required to generate the entire mass of water on this planet estimated at 1.54 x 10 ²¹ kg (1.54 x 10 ²⁴ gm) would require

= 1.54 x 10 ²⁴ gm H₂O x (1 mol H₂O ÷ 18 g) x (1 mol O₂ ÷ 2 mol H₂O) x (32 g O₂ ÷ 1 mol H₂)

 

1.37 x 10 ²⁴ gm or 1.37 x 10 ²¹ kg of oxygen

Hence, the mass of hydrogen required would be:

=  (1.54 x 10 ²¹ kg) – (1.37 x 10 ²¹ kg) = 1.7 x 10 ²⁰ kg hydrogen

 

Mass of Hydrogen atom:

But the mass of a hydrogen atom consists of:

A proton of mass = 1.7 x 10 ^-27 kg and an electron of mass = 9.1 x 10 ^-31 kg. Hence, the total mass is = 1.7 x 10 ^{– 27} kg.

But  since there are 1.7 x 10 ²⁰  kg of hydrogen in all the oceans and the total waters elsewhere on this planet, then the number of hydrogen atoms in all of them would be

= 1.7 x 10 ²⁰ ÷ 1.7 x 10 ^{– 27} = 1 x 10 ⁴⁷

In Deep Space:

Now, let us now look at the amount of hydrogen in deep space.

There is no absolute vacuum in space. There is always a molecule or two even in between the horrendously vast chasms of space between the galaxies.


Matter is very, very tenuously distributed throughout the Universe.  The average density of gas in our Milky Way galaxy is about one atom per cubic centimeter. This is by far a much more better vacuum than scientists can achieve in a laboratory. But this can mean a lot of matter inside a galaxy. 

Throughout this vast universe, the most abundant element is hydrogen (90 %), followed by helium (9 %) with oxygen, silica and the rest of the elements in the Mendeleev's Periodic Table at 1 %.   In between the galaxies (intergalactic space) the amount of hydrogen is even more tenuous, probably with just a hydrogen atom per cubic metre of space. But there is always something there no matter how feeble and thin.

In short, there is no absolute vacuum throughout the entire universe.  All the elements on the Earth came from a massive gas cloud that collapsed to form the Sun and the planets including Earth.

Background Radiation:

Even the temperature in throughout the universe is not spared. In every direction, there is a very low energy and very uniform radiation permeating the entire Universe. Astronomers found there is a left over of near 3 degree Kelvin Background Radiation.

Sometimes, this is called Cosmic Background Radiation, or the Microwave Background. These were termed because this radiation is basically a black body with temperature slightly less than 3 degrees Kelvin (about 2.76 K).

So we can see there are always some remnants of something, whether temperature or matter left over from a Big Bang during the creation of the Universe. There is no such thing as an absolute vacuum, or an absolute zero temperature anywhere in the Universe.

How Much?

But how much matter, especially the hydrogen, oxygen and the silicon are there in the gas cloud that created this Earth? We can never know for sure because we were never there when the Sun, Earth and the rest of the Solar System was created.  But we can make a hypothetical guess without going into the statistical complexity of explaining Null Hypothesis and Probabilities.

Assumption:

Let us assume instead of just one hydrogen atom per cubic metre as in intergalactic space, let us multiply that density conservatively to just a million times more. Compare this with a calculation using Avagadro's constant that tells us there are 6.02 x 10 ²³ molecules in a mol of gas. To cut a calculation short for my gentle readers, there would be 2.69 x 10 ²⁵ molecules of air per cubic metre at STP.

Hence if we assume that there were a million hydrogen atoms in each cubic metre of the gas cloud that created the oceans on Earth, then it can be shown that the volume of gas cloud needed to accommodate  1 x 10 ⁴⁷  number of hydrogen turns out to be 10 ⁴⁷  / 10 ⁶ = 10 ⁴¹ cubic metres.

What a Horrendous Volume:

But the volume of Earth and Sun is only 1.082 x 10 ²¹ m³   and 1.409 x 10 ²⁷ m³ respectively. This implies that the volume of gas cloud containing 1 atom of hydrogen per cubic metre required to form the Earth’s oceans was approximately 1 x 10 ²⁰ times (100 quintillion or 1 followed by 20 zeros) and 1 x 10 ¹⁴ (100 trillion or 1 followed by 14 zeros) times larger in volume than those of the Earth and Sun respectively.

What a bizarre volume of hydrogen and oxygen needed just to create the ocean and other waters on Planet Earth.

How Big was The Gas Cloud?

The next question we would like to ask is, how far would such a ball of hydrogen gas stretch in order to create all the oceans and waters on Planet Earth during Creation?

Let us calculate again.

Volume of a ball of Gas Cloud = 4 π r ³ ÷ 3

Its radius would be:

3V = 4 π r ³

r³   = 3V ÷ 4 π

r   = ³√ (3V ÷ 4π)

=   ³√ ( 3  x  10 ⁴¹ ÷ 4 π)

= ³√ 2.4 x 10 ⁴⁰ = 2.9 x 10 ¹³ metres in radius

Hence diameter of a ball of gas cloud = 5.8 x 10 ¹³ metres across

(Where, 10 ⁴¹ cubic metres was the volume of the amount of hydrogen needed at a density of 1 million atoms  per cubic metre)

Light Time:

An unexcited hydrogen atom has a radius called the Bohr radius.  This is a physical constant, approximately equal to the distance between the proton and electron in an unexcited hydrogen atom at ground state. It is named after physicist Niels Bohr, and is called the Bohr model of an atom.

The radius of a hydrogen atom or the Bohr radius has a value of 5.2917721092 × 10⁻¹¹ m. That’s 0.000 000 000 05292 of a second, or 52.92  picosecond.

We know that the speed of light in a vacuum is = 299 792 458 m / s or nearly 300 000 km per second.

Hence it takes 1.765 x 10^-19 second (0.1765 attosecond) for light to cross the diameter of a hydrogen atom at ground state.

Since the volume (V) of a sphere is 4 π r³ ÷ 3

Its radius (r) will be:

r =    ³√ 3V÷4 π

Thus, a sphere with a volume of 1 cubic metre will have a radius of:

r =   ³√ 0.2387324146 

= 0.6203504909 metres, or a diameter of 1.240700982 metres.

If we were to line up all the hydrogen atoms side-by-side across a sphere of 1 cubic metre in volume, there will be 2.3 x 10 ¹⁰ (23 billion) hydrogen atoms all in.  

This means, it takes light 4.1 x 10 ^-9 seconds (4.1 nanosecond) to cross 2.3 x 10 ¹⁰ atoms lying across the diameter of a sphere one cubic metre in volume.

If we assume the Gas Cloud was a sphere with a diameter of 5.8 x 10 ¹³ metres across, containing 10 ⁴⁷ hydrogen atoms  with a density of a million atoms per cubic metre,  it would take light travelling at nearly 300 000 km per second,  53 hrs 44 min 27 seconds to cross. 

Bigger or Smaller:

How large or small the Gas Cloud depends on the numbers of hydrogen atoms per cubic metre.  It may be very much smaller if the hydrogen density was far greater and much larger if it was more tenuous.

We do not know exactly, but as we work it out, for a Cloud of this magnitude, there must be at least a million hydrogen atoms per cubic metre. 

But conventional chemistry tells us, there must be at least 10 ⁴¹ hydrogen atoms to create all the waters on the hydrosphere of  his planet 

  

This would be the size of the Cloud upon which all the waters on Earth was created.

How far between Sun to Pluto and back?

The distance of Pluto from the Sun is between 30 to 49 AU (4.4–7.4 billion km ( 4400 000 000  to  7400 000 000 km  or 4.4 x 10 ¹² – 7.4 x 10 ¹² metres).

This means the Gas Cloud that created all the oceans and waters on Earth must have stretched up to nearly 8 times the distance of Pluto from the Sun.  

In other words, its diameter could enclose the distance between the Sun and Pluto to and fro 4 times over.

What an horrendous distance of hydrogen needed at 1 atom per cubic metre needed to create the oceans.

What an Almighty Hand God has in His Creation! 


Genesis:

And the earth was without form, and void; and darkness was upon the face of the deep. And the Spirit of God moved upon the face of the waters. (Genesis 1: 2). This verse speaks about the lifeless and molten formless Earth.

And God said, Let there be a firmament in the midst of the waters, and let it divide the waters from the waters

And God made the firmament, and divided the waters which were under the firmament from the waters which were above the firmament: and it was so (Genesis 1: 6-7)

Verses 6 and 7 speak about how water was divided into two stages on probably an exceedingly hot Earth, the very hot steam above, and the partially cooler, and condensed water below on the surface of Earth. 

Of course in geological time scale this took millions of years in human time scale, but in God’s eternal time scale, it was ‘just the second day’ to Him.

For a thousand years in your sight are like a day that has just gone by, or like a watch in the night (2 Peter 3:8)

The following equivalents were used in the calculations for the above article:

1.     1 solar day: 60 x 60 x 24 = 86400 sec.

2.     1 Solar Year: 365.25 solar days = 31,557,600 seconds

3.     15 solar days:  86400 x 15 = 1,296,000 seconds

4.     4.57 billion years = 4570,000,000   years = 1.44 x 10 ¹⁷ seconds

5.     10 billion years (10¹⁰):  10¹⁰ x 31557600 = 3.15576 x 10¹⁷   seconds

6.     1 watt hour = 3600 J

7.     1 kilowatt hour = 3.6×10⁶ J (or 3.6 MJ)

8.     1 watt second = 1 Joule

9.     Speed of light = 299 792 458 m / s

10.  Age of the Sun: 4.57 billion (4,570,000,000) years old

11.  Lifespan of our Sun: 10, 000 million (1 X 10 ¹⁰) years (main sequence lifetime)

12.  Mass of Sun: 1.98892 × 10³⁰ (nearly two nonillion kilograms

13.  Mass of Earth: 5.974 x 10²⁴ kilograms

14.  Sun is 332929.4 times more massive than Earth

15.  Mass of Moon:  7.36x10 ²² kg

16.  Radius of the Sun: 6.955 x 10 ⁸  metre = 6.955 x 10 ¹¹ mm (695,500 km)

17.  Mean Radius of Earth: 6.37 x 10⁶ m (6.37 x 10 ⁹ mm)

18.  Sun: Earth radius ratio =109:1

19.  1 Astronomical Unit (Earth-Sun distance) : 149,597,870,700 metres

20.  Volume (V) of a sphere is: V = 4/3 πr³

21.  Volume of Sun = 1.409 x 10 ²⁷ m³

22.  Volume of Earth = 1.082 x 10 ²¹ m³

23.  Sun /Earth volume ratio: 1,302,218 : 1

24.  Gravitational Constant = 6.67300 × 10^-11 m³ kg^-1 s^-2 (Newton)

25.  1000 mm = 1 metre

26.  1 metre = 1000mm

27.  1 litre = 1 000 000 cubic millimeter

28.  1 metre = 100 cm

29.  1 litre = 1 000 000 cubic millimeter

30.  1 cubic metre = 1 000 000 000 cubic millimeter

31.  1 cubic metre = 1000 000 cc = 1000 litres

32.  Mass of ocean waters: 1.35×10¹⁸ metric tons (1.35 x 10 ²¹) kg

33.  Mass of water in hydrosphere: 1.4 × 10¹⁸ tonnes (1.4 x 10 ²¹ kg)

34.  Mass of hidden water at 10 % of hydrosphere:  1.4 x 10 ²⁰ kg

35.  Mass of total water: 1.54 x 10 ²¹ kg

           

36.  Mass of Earth: mass of total water : 3879:1

37.  Mass of total water on Earth:  1.54 x 10 ²¹ kg

38.  Mass of total water: mass of earth = 2.57 x 10 ^-4 (0.025 %)

39.  Zero Kelvin = - 273.15 ⁰ C

40.  (1 x 10 ¹²) Kelvin = (1 x 10 ¹²) ⁰ C

41.  (1 x 10 ¹²) Kelvin = 333 million times hotter than an oxyacetylene flame at 3,000 ⁰ C

42.  Average temperature of ocean waters: 17 ⁰ C (National Climatic Data Center)

43.  Heat of fusion of ice is 334 kJ/kg

44.  Specific heat of water: 1 calorie/gram °C = 4.186 joule/gram °C

45.  Total geothermal energy locked up:  10³¹ joules

46.  Energy output of a Supernova =1.2 x 10⁴⁴ Joules

47.  Temperature of a supernova is about 100 billion (1 x 10 ¹²) kelvin (100 gigakelvin). This is 6000 times the core temperature of the Sun

48.   Energy output of Sun = (3.839 × 10 ²⁶)  Joules per second

49.  Energy Total of the Sun: 1.2 x 10⁴⁴ Joules

50.  Solar Constant (calculated): 1.92 KW. m^-2

51.  Solar Constant (measured by satellite):  1.361 KW. m^-2

52.  No. of stars in Milky Way Galaxy: 200 000 million – 400 000 million (average no: 250 000 million)

53.  Diameter of Milky Way Galaxy: 100,000–120,000 light-years across

54.  1 tonne = 1000 kg

55.  One  Newton (N) in SI unit for force required to accelerate a mass of one kilogram at a rate of one metre per second squared

56.  N = kg. m /s²

57.  Heat of fusion of ice is 334 kJ/kg

58.  Enthalpy of vaporization of water :  2257 kJ kg⁻¹

59.  1 calorie = 4.1868 joule
1 J = 0.238846 cal

60.  Radius of the Sun: 6.955 x 10 ⁸  metre = 6.955 x 10 ¹¹ mm (695,500 km)

61.  Mean Radius of Earth: 6.37 x 10⁶ m (6.37 x 10 ⁹ mm)

62.  1 Astronomical Unit (Earth-Sun distance) : 149,597,870,700 metres

63.  Speed of light = 299 792 458 m / s

64.   Sun-Earth travel light-time = 499 seconds = 8.3167  min =  8 min. 19 se

65.  Sun-Earth supernova travel time = 4990 sec = 83.16746397 min = 1.3861244 hours =

1 hr 23 min 10 sec

  

 Mathematics is the Science of Logic. Facts and Figures do not lie. 

Thank you for your patience in reading.

lim ju boo

Founder and First President

The Astronomical Society of Malaysia

Dated: 23rd July, 2012