So, I'm trying to calculate the "acceleration" (force / mass) on a spherical object of mass $M$ and radius $R$ due to its own gravity that holds it together. So, here is what I figured. The "acceleration" between two point particles separated by a distance $R$ in the sphere is given by $dF = \frac{Gdm}{r^2}$. So, I figued i can just integrate $dF$ over the sphere. First, to find $dm$:

$\frac{dm}{M} = \frac{dV}{V}$

$dm = \frac{M}{V}dV$

So, we have $F = \iiint_\text{sphere} \frac{GM}{Vr^2}dV$. In polar coordinates this is

$F = \int_{r = 0}^{R} \int_{\theta = 0}^{\pi} \int_{\phi = 0}^{2\pi} \frac{GM}{\frac{4}{3}\pi r^2 R^3} r^2 \sin\theta \; dr d\theta d\phi $. Simplyfying, we have

$F = \frac{3GM}{4\pi R^3} \int_0^R dr \int_0^\theta \sin\theta \; d\theta \int_0^{2\pi} d\phi$

Evaluating the integrals gives

$F = \frac{3GM}{4\pi R^3} \times R \times 2 \times 2\pi$,

which simplifes to

$F = \frac{3GM}{R^2}$

However, my lecturer gives this force as $\frac{GM}{R^2}$ without explanation. I already tried googling it, but all I can find is things about self gravitational potential energy, which I've gathered is something else. So, can someone see where I'm getting an extra factor of 3? And if I'm going about this entirely the wrong way, can someone point we in the right way of a correct derivation?

If anyone is wondering about the context, I want to calculate this force because I want to balance it with tidal forces.

  • $\begingroup$ it unclear to me what you're trying to calculate. Force is a vector quantity, and you seem not to care about tht at all. But if you'd just sum all internal forces as vectors, you'd get $\vec 0$, because of the Newton's Third Law. What exactly are you looking for? $\endgroup$ Commented Feb 26, 2020 at 11:41
  • $\begingroup$ Gravity doesn’t cause objects to accelerate themselves. Their net gravitational force on themselves is zero. You must have misunderstood what your lecturer did. It looks like they calculated the force of a large sphere on a small test mass just outside. $\endgroup$
    – G. Smith
    Commented Feb 26, 2020 at 17:12
  • $\begingroup$ I have indeed misunderstood. I will answer my own question. $\endgroup$ Commented Feb 26, 2020 at 17:15

1 Answer 1


I was looking for the surface gravity, which is indeed just given by

$\frac{GM}{R^2}$. My integration was nonsense as I did not consider the direction of the gravitational forces, only the magnitude.


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