# Why can center of mass be used in calculating gravity?

Why can gravitational forces be based on the center of mass. Due to the fact that gravity is related to the square of the distance should not the gravitational sum of every particle exceed the force when calculated by the center of mass.

• – DilithiumMatrix Nov 20 '15 at 15:36
• That is only true in the case of a radially symmetric sphere, and you being outside of the sphere. In the general case, if you are far from the body relative to the extent of the body, the approximation will not be too bad. – Jon Custer Nov 20 '15 at 15:37

Well that depends. As always, you should be very careful with such reasoning as

Due to the fact that gravity is related to the square of the distance should not the gravitational sum of every particle exceed the force when calculated by the center of mass.

because this is a problematic statement. In general your (Newtonian) gravitational potential is given by the Poisson equation: $$\vec\nabla^2\phi=4\pi G\rho(\vec x)$$ Which can be solved via the green function for the Laplacian: $$\phi=\int{\frac{G\rho(\vec x')}{|\vec x - \vec x'|}\mathrm{d}^3x'}+\mathrm{C}$$

and this potential is related to the force of gravity by: $$\vec a=-\vec\nabla\phi$$ You cannot say in general that this results in an inverse square law. In case of a spherically symmetric mass distribution you can attempt solving it explicitly (in every case when the mass distribution is homogenous, haven't tried it out otherwise) and find the inverse square law. As stated initially in this paragraph, this must not be. For example the earth's gravitational field is expanded in spherical harmonics for needs of expanded accuracy.

EDIT: Okay, too make that clearer: Yes, the lowest order approximation always gives you an inverse square dependence of the force of gravity. The point I mean here is that you cannot always say "The gravitational force is a square distance force from the center of mass" because that is only an approximation (albeit in most applications a good one - not always sufficient, though)

• But the leading, long range term will be $1/r$. (As the lowest multipole term dominates in the far field and all mass distributions have a monopole term because there are no negative masses). So while not an exact statement the $1/r$ behaviour is a good approximation for large distances. – Sebastian Riese Nov 20 '15 at 16:38
• Of couse that is true, but that was not really the point of my answer. I have clarified that. Still, my point is only that the inverse square law using the centers of mass is no exact application in general. – failtrolol Nov 20 '15 at 16:48
• Although artificial, there are also cases of infinitely long mass distributions. For such, not even far away from the mass distribution the gravitational field will behave like $1/r^2$. – andrehgomes Mar 14 '18 at 21:10