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I am trying to follow a derivation on this site to derive the minimum number of (hydrogen) atoms I need so that the gravitational force dominates over internal electrostatic forces.

The derivation begins by introducing the ratio of the two forces for two protons: $$\frac{1}{4 \pi \epsilon_0} \frac{e^2}{d^2} : G \frac{m^2}{d^2} = 1.24\cdot 10^{36}$$

It then considers a sphere of radius $R$, homogeneously filled with hydrogen, to which we add protons at the border. The gravitational energy per proton mass then is $$\frac{W}{m} = G \frac{M}{R}$$ This can be expressed in terms of the number of atoms, $\frac{W}{m} \sim N^{2/3}$.

So far, so good. The next argument is that gravity is not shielded, and long range. Sure, otherwise we wouldn't have used the formula above. And the electrostatic forces are shielded by the electrons in hydrogen, so they are short range only. I understand that.

But then, the final step is just the claim that

$$N > (1.24\cdot 10^{36})^{3/2}$$

and I don't see how this step can be made... I would have expected some discussion of the electrostatic energy per proton

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Your link to [this] site seems to be broken. Also there is a typo in the title. –  Willie Wong Aug 26 '11 at 12:58
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up vote 2 down vote accepted

Your equation $W/m\sim N^{2/3}$ gives the correct scaling, but to make this argument work you need to include the constant of proportionality: $$ {W\over m}\sim{GM\over R}={G(Nm)\over N^{1/3}r}={Gm\over r}N^{2/3}, $$ where $r=N^{-1/3}r$ is the typical distance between two protons.

Now $Gm/r$ is the gravitational potential energy for a single pair of particles, which is small compared to the electrostatic energy for a pair of particles by precisely the factor $\sim 10^{36}$ you gave. Because of shielding, we can assume that the electrical potential energy per particle is about the same as if there were just a single pair of particles. So if we compare gravitational to potential energies, we have $$ {(W/m)_{\rm grav}\over (W/m)_{\rm elec}}\sim (10^{-36})N^{2/3}. $$ So the two forms of energy are comparable when $N\sim (10^{36})^{3/2}=10^{54}$.

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Thanks, that clarifies it. –  Lagerbaer Aug 26 '11 at 17:45
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