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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 4r2, 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