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In the book "The Physics Of Stars" it is written that the gravitational binding energy stored in a sphere of radius $R$ and total mass $M$ is equal to $$-f\frac{GM^2}{R}$$

Where $f$ is some constant of proportionality

Also it is mentioned that a larger value of $f$ is obtained as density increases towards the center.

How can one prove both statements?

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    $\begingroup$ Dimensional analysis suffices for the first. $\endgroup$
    – G. Smith
    Commented Apr 19, 2021 at 20:20
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    $\begingroup$ @G. Smith Dimensional analysis doesn't specify the equation, it only demonstrates how this equation results energy as we expect it. I'm looking for a general derivation. $\endgroup$
    – sagittarius A
    Commented Apr 19, 2021 at 21:32
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    $\begingroup$ Dimensional analysis uniquely specifies this expression for the gravitational energy. $\endgroup$
    – G. Smith
    Commented Apr 19, 2021 at 22:23
  • $\begingroup$ @G. Smith. You could imagine adding an extra factor of M but divided by a product of some fundamental constant homogenous to a mass in principle. So I am not sure either dimensional analysis rigorously work in this example even though it would be a natural guess. $\endgroup$
    – StarBucK
    Commented Apr 19, 2021 at 23:39
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    $\begingroup$ @StarBucK No. The theory of Newtonian gravity has no fundamental constant with the dimensions of mass. $\endgroup$
    – G. Smith
    Commented Apr 20, 2021 at 0:13

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The binding energy can be calculated from the work needed to take shells of matter away from the star to infinity, until it's all gone (technically the negative of that).

If $\rho$ is the density, the radius of the star that's left is $r$ then the mass of the star that is left is $\frac{4}{3}\pi r^3\rho$, and the work needed to remove a shell of mass $4\pi r^2 \rho dr$ from $r$ to infinity is $$G( \frac{4}{3}\pi r^3\rho)\times (4 \pi r^2 \rho) \times \frac{1}{r} dr$$

integrate this from $0$ to $R$ and use $M=\frac{4}{3}\pi R^3\rho$ and you should get the result.

$f$ would have to be larger if the density increases towards the middle as follows:

Imagine an extreme case where almost no matter was at the edge of the star, but nearly all within $\frac{R}{2}$ - it's like having a star of the same mass, but half the radius, so $-f\frac{GM^2}{R}$ changes to $-f\frac{GM^2}{R/2}$ or $-2f\frac{GM^2}{R}$, so it's as if the $f$ has increased.

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  • $\begingroup$ For the first part, I have already tried this integral and I always end up at $\int r^4ρ^2 dr$ from 0 to R but going any more requires to specify the expression for density, is it possible to prove first statement in a way that does not require specific expression for density? I tried using integration by parts to get an integral which gives total mass, yet it was not successful $\endgroup$
    – sagittarius A
    Commented Apr 20, 2021 at 9:29
  • $\begingroup$ The answer has enough information for you to do it now. Hint: you won't need any difficult integrations $\endgroup$
    – John Hunter
    Commented Apr 20, 2021 at 12:32
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    $\begingroup$ @sagittariusA is it possible to prove first statement in a way that does not require specific expression for density? Yes. De-dimensionalize the integral by writing $r=Ru$ and $\rho(r)=\rho_\text{av}g(u)$ where $\rho_\text{av}=M/(\frac43\pi R^3)$. You’ll get $GM^2/R$ times a dimensionless integral over $u$ from $0$ to $1$ involving the dimensionless density profile $g(u)$. $\endgroup$
    – G. Smith
    Commented Apr 20, 2021 at 17:04

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