# Is the point mass assumption valid for non-spherical objects?

In case of a shell or a solid sphere we can assume it to act as a point mass to calculate the potential and field for any object as long as it lies outside the shell/sphere.

I have learnt the proof for each of these geometries separately and it does not seem general for all objects.

Can this assumption be made for non-spherical objects too? Can we assume it to be a point mass at the center given certain conditions?

• From far away everything looks spherically symmetric: this is a manifestation of the fundamental principle :) – nwolijin Feb 3 at 9:18

Can this assumption be made for non-spherical objects too?

No.

Some examples:

• Deviations of the gravitational acceleration from what is expected, both in magnitude and direction, are one of the tools people use to find buried mineral deposits.
• Many Earth satellites take advantage of the fact that the Earth is not spherical. It instead has a notable equatorial bulge. This equatorial bulge, amongst other things, makes the orbital plane of a non-equatorial satellite precess. A satellite placed in just the right inclination orbit will see the plane precess by 360° per year, making the satellite stay in sync with the Sun. These sun-synchronous orbits are very important for remote sensing.
• The Moon has several large mass concentrations (mascons) on the near side of the Moon. These mascons have a significant impact on low lunar orbits. Two lunar orbiting satellites released by the Apollo program into circular orbits had their eccentricity grow to such extent that the satellites eventually impacted the lunar surface.

Can we assume it to be a point mass at the center given certain conditions?

Yes. The departures from spherical behavior fall off as the inverse cube of distance, or even higher powers. As spherical gravity falls off as the inverse square of distance, at sufficiently large distances those non-spherical can essentially be ignored.

Yes, the condition is to be far enough, and that is the basis for the point-mass assumption.

Whatever is the shape of a body, its gravitational potential at a large distance will be better and better described as equivalent to the potential of a point-like charge. It is a simple result, based on the analysis of the asymptotic behavior of the potential $$\phi({\bf r})= \int_{body} \frac{\rho({\bf r'})}{|{\bf r}-{\bf r'}|} {\mathrm d}^3{\bf r'}.$$ If the point $${\bf r}$$ is far from the body characterized by a mass density $$\rho({\bf r})$$ the asymptotic behavior is dominated by the first term of the asymptotic expansion $$\frac{1}{|{\bf r}-{\bf r'}|}=\frac{1}{(r^2+r'^2-2{\bf r}\cdot{\bf r'})^{\frac12}} =\frac{1}{r(1+r'^2-2{\bf r}\cdot{\bf r'})^{\frac12}}\simeq \frac{1}{r},$$ Then, the large distance behavior of the potential is $$\phi({\bf r})= \frac{\int_{body} \rho({\bf r'}) {\mathrm d}^3{\bf r'}}{r}.$$ Therefore the body acts as a source of gravitational potential as if its whole mass would be concentrated in a point. Of course, at finite distances, the other terms of the expansion will play a role and deviations from the point-like approximation (multiples) will become more and more important by approaching the body unless, for special geometries like a homogeneous spherical distribution, all the other terms will vanish.

In short, if there are the same forces acting on all points (for example there is no more gravitational force acting on one corner than another), then you can suppose that it is point like. For example, a body with non-zero angular momentum has different forces acting on different points. Also, if the size of the object is negligible, then you can assume pointyness.