What is the gravitational force in m/s^2 on the surface of an infinitely long cylinder with radius r and density p?

I am pretty sure that it is finite as gravity is the inverse square of the distance, but I am unsure how to work it out.

I am hoping that this can be answered in the form of a formula containing only variables r and p and maybe a few constants if necessary.


1 Answer 1


Gauss' law for gravity says that $$\oint_{\partial S}{\vec{g}.d\vec{A}}=-4\pi Gm$$

If we use a cylindrical Gaussian surface of length $l$ and radius $R=$ radius of the infinite cylinder, the mass enclosed will be $\rho \pi R^2l$

The right hand side of the equation is $$g*2\pi Rl$$

Equating the two we get $$g*2\pi Rl=-4\pi G\rho \pi R^2l$$

So $$g=-2\pi G\rho R$$ at the surface of the cylinder.

  • $\begingroup$ Does the negative sign mean that grativy pushes away from the cylinder, or is it just a consequence of using scalars instead of vectors (meaning it doesn't really matter), or am I just misinterpreting something here? $\endgroup$
    – Harmless
    Commented Apr 12, 2017 at 10:15
  • $\begingroup$ @MarkAnonym: which way does the vector R point? $\endgroup$ Commented Apr 12, 2017 at 10:26
  • $\begingroup$ I think I have strayed a bit too far into mathematics I don't actually know - I assumed variables p and R would be scalars or at least positive but I'm not actually sure given the equation. The question I am asking is would the gravity still attract objects towards the cylinder despite the answer to the equation being negative? $\endgroup$
    – Harmless
    Commented Apr 12, 2017 at 11:28
  • $\begingroup$ @MarkAnonym: yes, $g$ is positive away from the cylinder, so $g < 0$ means things are attracted to it. On Earth $g\approx -9.8\,\mathrm{m}\mathrm{s}^{-2}$ for instance. Incidentally what you call $p$ is really $\rho$: the density of the cylinder. $\rho$ and $R$ are both scalars (another comment notwithstanding). $\endgroup$
    – user107153
    Commented Apr 12, 2017 at 12:18
  • $\begingroup$ Just to check, would gravity still work as an inverse square law? In other words, would it be possible for me to work out the gravity at any distance from the axis of the cylinder? $\endgroup$
    – Harmless
    Commented Apr 12, 2017 at 19:57

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