Why the gravitational force come from the center of mass?

Question

Why we measure the force of gravity with the distance between an object and the radius of planet (that means the gravity force come from center of planet)?

And is that disagree with general relativity?

Can you be a little more specific about your question? Like, add some details as to why you feel it disagrees with GR or what exactly do you even mean by saying "gravitational force comes from the center of mass"? — , Jun 30 '17 at 13:01
See Newton's shell theorem & Birkhoff's theorem. Related: https://physics.stackexchange.com/q/21705/2451 and links therein. — Qmechanic, Jun 30 '17 at 13:11

score 1 · Answer 1 · answered Jun 30 '17 at 13:00

1

It does not come from the centre of mass , but for practical and theoritical reasons is assumed as such plus must celestial bodies being spherical objects seem like points for us thus this practice spawned to reduce complications of radii of body and its distance from point of observation.

And no it has nothing worth mentioning to do with general relativity .

answered Jun 30 '17 at 13:00

Bhavam Vidyarthi

35

so why we measure the force of gravity with the distance between the object and center of planet (radius of the planet) in the universal law of gravitation – Ahmed Emad Jun 30 '17 at 13:03
-1: If you mean "gravitational force would be the same if the body is replaced by a point particle of the same mass situated at the center of mass" by "gravitational force comes from the center of mass" then this is not assumed anywhere. In the cases of spherically symmetric mass distributions, it is proven and is thus used for such mass distributions. In the rest of the cases, at least for the theoretical purposes, it is neither used nor assumed. I don't know what engineers do for practical purposes but that is not a topic of discussion here anyway. – Jun 30 '17 at 13:04

score 1 · Answer 2 · answered Jun 30 '17 at 13:09

It can be shown using Gauß theorem applied to Coulomb (or Newton) $1/r^2$ law that the gravitational (electric) field generated by a uniform spherical distribution is equivalent to a gravitation (respectively electric) field generated by an equivalent point mass (or charge) situated in the centre of mass.

score 0 · Answer 3 · answered Jun 30 '17 at 13:12

Gravitational interaction occurs between all particles in every body. A moon orbits a planet because every particle in the moon gravitates to every particle in the planet, and the moon holds together because every particle in the moon gravitates to every other particle in the moon, etc.

For first order approximations, mathematically it works to assume that the mass (and thus the force) acts at a single point. This comes from the vector addition of all the gravitational forces between two bodies; the addition gives a single resultant vector (or pair of colinear vectors, considering the pair of objects) on a line between the centers of masses.

This is an approximation because it only works for rigid bodies, and no real object is perfectly rigid. When we stop idealizing the case and calculate the varying gravitational forces at various points, we see the tidal effects arise. In a non-rigid planet, its moon pulls harder on the near side than on the far side. This is to be expected because the individual particles on the near side are closer to the gravitating body than the particles on the far side, so the individual gravitational interaction is greater.

Why the gravitational force come from the center of mass?

3 Answers3