I guess the important thing to realise is that the size atoms is irrelevant to the problem. If we considered two atoms joined together to be a new “atom” then they would be twice as heavy, so the forces would be four times as strong, but there would be only half as many atoms, so there would be four times fewer pairs.
So the answer is just the integral as r and r’ range over the interior of the earth of G ρ(r) ρ(r’)/(r-r’)^2, where ρ(r) is the density. We can assume constant density, but I still can’t be bothered to do the integral.
The earth has mass 5.97*10^24 kg and radius 6.37*10^6 m, G = 6.674*10^-11 m^3 kg^-1 s^-2 and we want an answer in Newtons = m kg s^-2. So by dimensional analysis, the answer is about G M^2/r^2 = 5.86*10^25.
I guess the important thing to realise is that the size atoms is irrelevant to the problem.
That doesn’t seem right, though? Imagine a one dimensional version of the problem. If a stick of length 1 is divided into n atoms weighing 1/n each, then each pair of adjacent atoms is distance 1/n apart, so the force between them is 1. Since there are n such pairs, the total force grows at least linearly with n. And it gets even worse if some atoms are disproportionately closer to others (in molecules).
Cool insight. We’ll just pretend constant density of 3M/4r^3.
This kind of integral shows up all the time in E and M, so I’ll give it a shot to keep in practice.
You simplify it by using the law of cosines, to turn the vector subtraction 1/|r-r’|^2 into 1/(|r|^2+|r’|^2+2|r||r’|cos(θ)). And this looks like you still have to worry about integrating two things, but actually you can just call r’ due north during the integral over r without loss of generality.
So now we need to integrate 1/(r^2+|r’|^2+2r|r’|cos(θ)) r^2 sin(θ) dr dφ dθ. First take your free 2π from φ. Cosine is the derivative of sine, so substitution makes it obvious that the θ integral gives you a log of cosine. So now we integrate 2πr (ln(r^2+|r’|^2+2r|r’|) - ln(r^2+|r’|^2-2r|r’|)) / 2|r’| dr from 0 to R. Which mathematica says is some nasty inverse-tangent-containing thing.
Okay, maybe I don’t actually want to do this integral that much :P
EDIT: On second thoughts most of the following is bullshit. In particular, the answer clearly can’t depend logarithmically on R.
I had a long train journey today so I did the integral! And it’s more interesting than I expected because it diverges! I got the answer (GM^2/R^2)(9/4)(log(2)-43/12-log(0)). Of course I might have made a numerical mistake somewhere, in particular the number 43⁄12 looks a bit strange. But the interesting bit is the log(0). The divergence arises because we’ve modelled matter as a continuum, with parts of it getting arbitrarily close to other parts.
To get an exact answer we would have to look at how atoms are actually arranged in matter, but we can get a rough answer by replacing the 0 in log(0) by r_min/R, where r_min is the average distance between atoms. In most molecules the bond spacing is somewhere around 100 nm. So r_min ~ 10^-10, and R = 6.37*10^6 so log(r_min/R) ~ −38.7, which is more significant than the log(2)-43/12 = −2.89. So we can say that the total is about 38.7*9/4*GM^2/R^2 which is 87GM^2/R^2 or 5.1*10^27.
[But after working this out I suddenly got worried that some atoms get even closer than that. Maybe when a cosmic ray hits the earth it does so with such energy that it gets really really close to another nucleus, and then the gravitational force between them dominates the rest of the planet put together. Well the strongest cosmic ray on record is the Oh-My-God particle with mass 48J. So it would have produced a spacing of about h_barc/48, which is about 6.6\10^-28. But the mass of a proton is about 10^-27, so Gm^2/r^2 is about G, and this isn’t as significant as I feared.]
Very nice, and the result is a thousand Earth’s weights now. I wonder if every atom inside Earth feels the gravity of every other atom at every moment. (I think not. Which is a heresy, so please don’t pay any attention to that.)
I guess the important thing to realise is that the size atoms is irrelevant to the problem. If we considered two atoms joined together to be a new “atom” then they would be twice as heavy, so the forces would be four times as strong, but there would be only half as many atoms, so there would be four times fewer pairs.
So the answer is just the integral as r and r’ range over the interior of the earth of G ρ(r) ρ(r’)/(r-r’)^2, where ρ(r) is the density. We can assume constant density, but I still can’t be bothered to do the integral.
The earth has mass 5.97*10^24 kg and radius 6.37*10^6 m, G = 6.674*10^-11 m^3 kg^-1 s^-2 and we want an answer in Newtons = m kg s^-2. So by dimensional analysis, the answer is about G M^2/r^2 = 5.86*10^25.
You estimate around 1 Earth weight on Earth’s surface.
That doesn’t seem right, though? Imagine a one dimensional version of the problem. If a stick of length 1 is divided into n atoms weighing 1/n each, then each pair of adjacent atoms is distance 1/n apart, so the force between them is 1. Since there are n such pairs, the total force grows at least linearly with n. And it gets even worse if some atoms are disproportionately closer to others (in molecules).
Cool insight. We’ll just pretend constant density of 3M/4r^3.
This kind of integral shows up all the time in E and M, so I’ll give it a shot to keep in practice.
You simplify it by using the law of cosines, to turn the vector subtraction 1/|r-r’|^2 into 1/(|r|^2+|r’|^2+2|r||r’|cos(θ)). And this looks like you still have to worry about integrating two things, but actually you can just call r’ due north during the integral over r without loss of generality.
So now we need to integrate 1/(r^2+|r’|^2+2r|r’|cos(θ)) r^2 sin(θ) dr dφ dθ. First take your free 2π from φ. Cosine is the derivative of sine, so substitution makes it obvious that the θ integral gives you a log of cosine. So now we integrate 2πr (ln(r^2+|r’|^2+2r|r’|) - ln(r^2+|r’|^2-2r|r’|)) / 2|r’| dr from 0 to R. Which mathematica says is some nasty inverse-tangent-containing thing.
Okay, maybe I don’t actually want to do this integral that much :P
EDIT: On second thoughts most of the following is bullshit. In particular, the answer clearly can’t depend logarithmically on R.
I had a long train journey today so I did the integral! And it’s more interesting than I expected because it diverges! I got the answer (GM^2/R^2)(9/4)(log(2)-43/12-log(0)). Of course I might have made a numerical mistake somewhere, in particular the number 43⁄12 looks a bit strange. But the interesting bit is the log(0). The divergence arises because we’ve modelled matter as a continuum, with parts of it getting arbitrarily close to other parts.
To get an exact answer we would have to look at how atoms are actually arranged in matter, but we can get a rough answer by replacing the 0 in log(0) by r_min/R, where r_min is the average distance between atoms. In most molecules the bond spacing is somewhere around 100 nm. So r_min ~ 10^-10, and R = 6.37*10^6 so log(r_min/R) ~ −38.7, which is more significant than the log(2)-43/12 = −2.89. So we can say that the total is about 38.7*9/4*GM^2/R^2 which is 87GM^2/R^2 or 5.1*10^27.
[But after working this out I suddenly got worried that some atoms get even closer than that. Maybe when a cosmic ray hits the earth it does so with such energy that it gets really really close to another nucleus, and then the gravitational force between them dominates the rest of the planet put together. Well the strongest cosmic ray on record is the Oh-My-God particle with mass 48J. So it would have produced a spacing of about h_barc/48, which is about 6.6\10^-28. But the mass of a proton is about 10^-27, so Gm^2/r^2 is about G, and this isn’t as significant as I feared.]
Very nice, and the result is a thousand Earth’s weights now. I wonder if every atom inside Earth feels the gravity of every other atom at every moment. (I think not. Which is a heresy, so please don’t pay any attention to that.)