Establishment Lackeys, such as I are often accused of accepting the figures that we read in textbooks. It is because we tend believe in the basics - because they work. Nevertheless, it is sometimes nice to verify things for oneself.

I was busy spouting off about the energy which was available as the Earth was formed by the accretion of a large volume of bits of the Solar System into one lump. This energy would then have been dissipated by radiative cooling after melting all the bits together.

So I did the following calculation.

I think the logic of the following is OK but would appreciate any errors being pointed out.

Start off with a large sphere of material of low density (and uniform density to make it simple). Work out the Gravitational Potential for the initial, large, sphere and the GPE of the final sphere (i.e. the present Earth). The difference between the two will be the energy available to heat the Earth up, melt it together etc. and, the heat which has been lost since it was formed.

The GPE of a piece of the sphere at distance r from the centre is provided by the material nearer to the centre (the effect of the material further out all cancels out).

The mass of a shell, radius r and thickness dr will be

4πr^{2}*ρ*dr.

where ρ is the average density

The mass of the sphere, inside that shell is

(4/3)πr^{3}ρ

GPE = -G (Mass inside shell)(Mass of shell)/r

=-4Gπr^{3}ρ4πr^{2}ρdr/3r

where G is the Gravitational Constant

The density is the total mass / the volume of the sphere

so

ρ = 3M/(4πD^{2}π)

where D is the radius of the whole sphere.

This gives the GPE of the shell as:

-3Gr^{4}M^{2}dr/4D^{6}

which gets rid of the ρ.

You can simplify this into

GPE of the shell = -3Gr^{4}M^{2}dr/4D^{6}

To find the total GPE of the sphere you have to integrate over all the shells - this gives you

Total GPE = -3GM^{2}/20D

This looks OK as the units of the final result are in energy!

It also gives the GPE of the Earth with its present radius (say d)

The difference between the two will be

=-3GM^{2}/20D +3GM^{2}/20d

If D is, say, a thousand times d (it must have been at least as much)then the formula becomes

-3GM^{2}/20d

Putting in the numbers, you get the available energy as 6e31J.

That's a lot of energy! Easily enough to do all the melting and fusing and stuff. It's about ten million year's worth of Solar energy.

Cooling. If the Earth had the same albedo as it has today and assuming that the surface temperature was similar to present day (300k; you've got to start somewhere), using Stefan's law tells you the rate the Earth will radiate energy if the surface were matt black. That's about 10e17W.

At that rate, the energy would last 6e31/10e17seconds - only about 10 million years.

It must imply that, for billions of years, the surface temperature must have been dominated by the Sun's radiation. I'm ignoring the fact that it would cool exponentially and that the initial rate would be much higher.

Interesting, though!

There is the other issue of how fast heat can travel through thousands of km of mantle and crust.

[mod - please try to make the subjects of new posts into questions as this makes the forum more intuitive to use. Thanks. CS]