How much heat was generated in the formation of the earth?

I.e., the gravitational potential energy (GPE) of the Earth's mass. Assuming, as you pose it, a constant density, you can do it by integrating, by spherical shells, the GPE of bringing each successive shell from ∞ to the surface, as you build that surface up from r=0 to r=R, the Earth's radius. This yields:

Universal gravitational constant: G = 6.674*10^-11 m^3 /(kg s^2)

Earth (mean) radius: R = 6.371*10^6 m

Mean density of Earth: ρ = 5.515*10^3 kg/m^3

Mass of shell at r: M(r,dr) = 4πρr^2 dr

Mass of sphere interior to r: M(r) = 4/3 πρr^3

Gravitational potential at r: GM(r)/r

Mass of Earth: M = M(R) = 4/3 πρR^3 = 5.9737*10^24 kg

GPE = G∫[r=0,R] M(r,dr) M(r)/r

= 16/3 π^2 G ρ^2∫[r=0,R] r^2 r^3 /r dr

= 16/3 π^2 G ρ^2∫[r=0,R] r^4 dr

= 16/15 π^2 G ρ^2 R^5

= 3/5 G M^2 /R

= 2.2430*10^32 J

Heat equivalent:

1 kg-cal (kcal) = 4182 J (thermochemical calorie)

GPE = 5.3635*10^27 kcal

To estimate the temperature to which this would raise the essentially 0K matter, requires some estimate of its heat capacity. This will be substantially less than that of water. In any case, with a specific heat value somewhere in the range, (0.1 to 0.5) kcal/kg-K, the temperature of the newly condensed Earth would come to (10,000 to 2,000)K. But this is without considering phase changes.

Additional considerations:

A more accurate value of GPE could be obtained by modeling the Earth in 3 (or more) radial zones, each with its own range of r values and its own density (this approach will necessarily increase the computed GPE, because deeper levels are denser, so that more of the mass has fallen in the gravitational field):

• the core, mostly iron;

• the mantle, mostly oxides of silicon and magnesium, and

• the crust, mostly oxides of silicon and aluminum

Also important in working out the heating of the early Earth would be the very significant role played by radioactive decay of primordial uranium and its decay products, which is responsible for most of the heat still present in the Earth's core. George Gamow did some important work on this point in the 1940's.

P.S. I gather that the TD's are not for your benefit, but rather, intended as feedback to "drive-by" answerers. Not that THAT necessarily does any good, either.

EDIT:

• I've replaced "a" with "R" throughout (hope I haven't missed any of them!).

• A much messier calculation shows that, if the density, ρ, is linear in r, dropping to 0 at the surface, the result increases to:

GPE = 26/35 GM^2/R = (3/5 + 1/7)GM^2/R = 2.7708*10^32 J; not a gigantic increase.

• Which is a special case (with α=0) of the even messier calculation with ρ linear in r, dropping to ρ[S] at the surface:

GPE = (26/35 - 13/70 α + 3/70 α^2)GM^2 /R = [3/5 + 1/7 (1 - 3/10 α)(1 - α)]GM^2 /R,

where α = 4πR^3 ρ[S]/3M = ρ[S]/ρ[ave]

The original result is also a special case of this, with ρ[S] = ρ[ave], so that α = 1.

In this regime, α is restricted to the range from 0 (making ρ[S]=0) to 4/3 (making ρ[0]=0). The multiplier on G M^2 /R is:

26/35 for α=0, 3/5 for α=1, 4/7 for α=4/3

We are leaving out a ton of factors, but

I cooked up one answer using thermodynamics.

Thinking that this might be wrong, I attempted

using the virial theorem, and to my surprise, I

got exactly the same answer. Am I on the right

track?

Edit:

Thank you.

I should have been more clear. We had a go with Part 1

of this question awhile back I believe. So you can kick my

butt out of this. I am interested in the bonus.

Part1:

The energy required (absorbed) to remove uniform differential masses

of particles from a points of Gaussian potential (r) to infinity is:

ρ=5473.68752 kg/m^3

r=6387100 m

dE = G M(r)*dm(r) / r

dE = G (4/3)πr^3ρ*4πr^2ρdr / r

dE = G (16/3) π^2 ρ^2 r^4 dr

Total gravitational energy required to take the system apart.

E = G (16/3) π^2 ρ^2 ∫ r^4dr [0,r]

E = G (16/15) π^2 ρ^2 r^5 |0,r

E = 2.24*10^32 J

Part2:bonus:

I'll Hold'em for now ;-)

I need more time to think about this.

You rascal FGR!

Edit:

For the time being: I believe:

The rate of accretion is key. If it was equilibrated,

to slow, gravitational energy could have been lost

to electromagnetic radiation. Some other source of

energy would have then been required to account for

earth's differentiation. ie: radioactive decay, etc.

On the other hand, if the collapse was relatively fast via QM/

viral theorem/thermodynamics/what have you, there is more

than enough gravitational thermal energy to account for earth's

differentiation.

Consequently, earth's density gradient could have been created before,

after, or during formation, and needless to say, I was not around to

take measurements.

Rather than give an estimate to this interesting bonus, I'll keep what I have for

some later question as FGR has, I think, suggested. I have attacked this question

using everything from quantum mechanics to Newton's law of cooling, and I am not

yet satisfied with my analysis.

Nice question. But I think Heat would be require in forming earth, anyway..

We have to ignore so many variables like increasing mass of earth, External gravitation force from any other star or planet, particles were brought without changing KE and so on, so by this method only we can get approximate value.

Actually, Gravitation is a conservative field. The heat require in formation is the amount of work done in bringing a atom from infinity to a point.

Avg Denisty of Earth is 5.52gm/cc, Which contain

iron (32.1%)= 1.771 gm/cc

oxygen (30.1%)= 1.66gm/cc

silicon (15.1%), and so on

magnesium (13.9%),

sulfur (2.9%),

nickel (1.8%),

calcium (1.5%)

aluminium (1.4%)

It would be more tedious if we take so many elements, because then we have to integrate for every element radius in work done, so i am assuming earth is purely made of iron atoms.

Then, Molecular mass of Fe= gm = 6.023*10^23 atom

or 5.52 gm Fe has = 5.953*10^22 atoms

so avg separtation between Fe atom is 9.2711*10^-25m

We can also think inverse, work done in collapse and this will be same as formation.

dW=∫ GM/r² dv

M=mass of 1 Fe atom= 1.3284*10^-23 gm= 1.3284*10^-26kg

G=Gravitation constt

where we will use volume integral and integrate it from v to ∞ as its volume is decreasing constantly, where v=π [r²x-x³/3]dx from r=1.32824*10^-23 to Radius of earth that is

r²=x²+y²

At any given x, a right-angled triangle connects x, y and r to the origin, as in sphere.

This will be the heat generated by 1 atom of Fe, assuming Fe as it is chief element.

Total atoms= 6.023*10^23*Mass Earth(5.9742 × 10^24 kg)/0.055kg

=6.5422*10^49

dW=-dH in 1 atom; ignoring all other non conservative forces, friction, or gravitation due to other planet or star.

Total heat= Heat by 1 atom*6.5422*10^49

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Solving Roughly, I got

Heat generated = -6.2525*10^37 Joule (approx)

Heat generated by one ton of TNT is equal to 4.184 × 10^9 J so,

"THINK THIS WOULD BE 1494375853000000000000000000 TIMES GREATER IN MAGNITUDE THAN HEAT PRODUCE IN 1 TON EXPLOSION OF TNT"

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|dH(formation)|=|dH(collapsing)|

Same magnitude of heat will generate in collapsing if we ignore mass changing with time, because stable state is free state and this is exothermic reaction, heat will emit out. That is the reason universe is expanding.

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Plz reply is it correct, or if wrong then where i am?

Unfortunately, VERY difficult to ascertain. One main variable would be the atmosphere's formation at a particular time in the Earth's history, which greatly affects the amount of heat energy lost to space. Early on, there was no atmosphere, therefore heat dissipated more easily, then later our Earth developed it's 'protective blanket'. Another variable is the radioactive decay going on constantly, thereby heating the mantle and the like. Different elements are going to decay at different rates. Sorry I couldn't be of more help.

I don't think thermometer was invented at that time.

a lot. like more than 100

I wouldn't know I wasn't there