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I Need Some Help, Please

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Zone

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The question is what would be the radius of a spherical shell that is 1 cm thick that had enough water to cover the world with 8 km of water.

The answer ain't 3.
The Earth's total land surface area = 1.5 x 10[sup]8[/sup] sq km (i.e., 57,500,000 sq miles).

Therefore r = 1.5 x 10[sup]8[/sup] x 8 x 1 = 1.2 x 10[sup]9[/sup]
 
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Anyone want to calculate the energy of that much water falling back to Earth and the consequences thereof?
Hi The Barbarian,
I would like to calculate it for everyone.

Water = H[sub]2[/sub]O, (i.e., HOH)

1 Oxygen molecule = 1,313.9kJ·mol[sup]−1[/sup]
2 Hydrogen molecule = 2,624kJ·mol[sup]−1[/sup]
1,313.9kJ·mol[sup]−1[/sup] + 2,624kJ·mol[sup]−1[/sup] = 3937.9kJ·mol[sup]−1[/sup] ( = 3937.9kJ·mol[sup]−1[/sup] x 1.2 x 10[sup]9[/sup] = 4.72548 x 10[sup]12[/sup])
 
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TemperateSeaIsland

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Anyone want to calculate the energy of that much water falling back to Earth and the consequences thereof?

Calculating the velocity would be tricky. I'll then just give it a velocity of 100 m/s, no idea how accurate that will be I suspect much much slower than in reality.

The water will have a mass of 4.1 x 10[sup]21[/sup] Kg

So the kinetic energy will be E = 1/2 . 4.1 x 10[sup]21[/sup] . 100[sup]2[/sup]

Giving 2.05 x 10[sup]25[/sup] Joules of energy, which if wiki is to believed about 4 times the energy we recieve from the Sun over a year.

I think it would be bad.
 
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Hespera

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when glaciers melt there is isostatic rebound. The interior of greenland is below sea level, from the weight of the ice.

IF one piled five miles of water all over the surface of the earth THEN something funny would result from all that pressure, and then the release of it.

Wind would drive enormous waves, that would sweep all the way around the world.

All the plants would die because someone forgot to put those on the boat.
What would happen to temperatures and ocean currents is beyond imagining.

Well it is all pretty silly. if you want to say that all rules of physics and reality in general were suspended for a year, fine. As an actual event that could have happened in the universe as it exists, it is of course impossible. The conversation is basiccaly about whether magic is real. Some think it is. So where do you go from there other than ask for a demonstration..?
 
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Zone

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Calculating the velocity would be tricky. I'll then just give it a velocity of 100 m/s, no idea how accurate that will be I suspect much much slower than in reality.

The water will have a mass of 4.1 x 10[sup]21[/sup] Kg

So the kinetic energy will be E = 1/2 . 4.1 x 10[sup]21[/sup] . 100[sup]2[/sup]

Giving 2.05 x 10[sup]25[/sup] Joules of energy, which if wiki is to believed about 4 times the energy we recieve from the Sun over a year.

I think it would be bad.
Hi TemperateSeaIsland,
Here's something that you may find interesting..... (pertaining to E = m) According to your Wiki-link:

The total energy (i.e., E = m) output of the Sun, our yellow-dwarf Star (even though the E = m is 1 x 10[sup]6[/sup] years old) each day the Sun produces 3.34 x 10[sup]31[/sup]J x 365 = 1.2191 x 10[sup]34[/sup]J
 
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TemperateSeaIsland

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Hi TemperateSeaIsland,
Here's something that you may find interesting..... (pertaining to E = m) According to your Wiki-link:

The total energy (i.e., E = m) output of the Sun, our yellow-dwarf Star (even though the E = m is 1 x 10[sup]6[/sup] years old) each day the Sun produces 3.34 x 10[sup]31[/sup]J x 365 = 1.2191 x 10[sup]34[/sup]J

Yup and thankfully we only recieve a fraction of the total energy output of the Sun. To quote wiki...

5.5×10[sup]24[/sup] J, the total energy from the Sun that strikes the face of the Earth each year
 
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Zone

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Yup and thankfully we only recieve a fraction of the total energy output of the Sun. To quote wiki...
See all of that (E = m, or) energy that actually reaches the Earth's surface (5.5 ×10[sup]24[/sup]J) absolutely amazing. :doh:
 
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Zone

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Calculating the velocity would be tricky. I'll then just give it a velocity of 100 m/s, no idea how accurate that will be I suspect much much slower than in reality.

The water will have a mass of 4.1 x 10[sup]21[/sup] Kg

So the kinetic energy will be E = 1/2 . 4.1 x 10[sup]21[/sup] . 100[sup]2[/sup]

Giving 2.05 x 10[sup]25[/sup] Joules of energy, which if wiki is to believed about 4 times the energy we recieve from the Sun over a year.

I think it would be bad.
Some many various complexities of (E = m, or) energy:
  • Mass energy
  • kinetic energy
  • Joules of energy
  • Potential energy
  • Etc, etc, etc...... :thumbsup:
 
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atomweaver

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Hi The Barbarian,
I would like to calculate it for everyone.

Water = H[sub]2[/sub]O, (i.e., HOH)

1 Oxygen molecule = 1,313.9kJ·mol[sup]−1[/sup]

1314 kJ/mol looks familiar, but... Ionization potential is a measurement of the amount of energy it takes to remove one mole of electrons from one mole of a gaseous element.
Your calculation quantifies the amount of energy one would need to remove 3 moles of electrons per mole of water, after the water has been separated into its atomic, gaseous constituents.

Which is to say; Whiskey..? Tango..? Foxtrot..? Your calculation has nothing to do with the question asked.
 
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Hespera

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1313 kJ/mol-1 looks familiar, but... Ionization potential is a measurement of the amount of energy it takes to remove one mole of electrons from one mole of a gaseous element.
Your calculation quantifies the amount of energy one would need to remove 3 moles of electrons per mole of water, after the water has been separated into its atomic, gaseous constituents.

Which is to say; Whiskey..? Tango..? Foxtrot..? Your calculation has nothing to do with the question asked.


What, after all, do you expect from Captain Cut n' Paste? You didnt imagine he understand s any of the stuff he posts, do you?
 
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atomweaver

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What, after all, do you expect from Captain Cut n' Paste? You didnt imagine he understand s any of the stuff he posts, do you?

Oh, not at all. This time, I recognized what it was that he was calculating, even if he didn't... Quite funny stuff.
 
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sfs

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Calculating the velocity would be tricky. I'll then just give it a velocity of 100 m/s, no idea how accurate that will be I suspect much much slower than in reality.
Don't calculate the velocity -- just calculate the difference in potential energy for that mass at the different radii (U = -GMm/r). The result is 2.5 x10[sup]29[/sup] J. That's a lot of energy.
 
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TemperateSeaIsland

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Don't calculate the velocity -- just calculate the difference in potential energy for that mass at the different radii (U = -GMm/r). The result is 2.5 x10[sup]29[/sup] J. That's a lot of energy.

Good thinking...

That is an insane amount of energy. I calculated using my vastly smaller estimate that the energy was equivalent to a 9.5 Megatonne nuke going off on every sq km of the earth’s surface. Your more accurate estimate makes that look like nothing.
 
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To all those here with the mathematical savvy to put this all together, thanks for sharing your expertise. Your results are very interesting. :thumbsup:



To all those here who can't do much more than respond with childish invectives, thanks for sharing the limits of your intellect and integrity. :thumbsup:
 
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