Proving the gravitational force of a solid sphere using integration

[Sam Jelly]

Homework Statement

I am trying to calculate the gravitational force of mass m from a solid sphere with Radius R, mass M with uniform mass distribution. I am integrating the gravitaional force done by the thin circular plate. (I put mass m on top of every circular plate's center of mass). I know the solution to this is GmM/R^2, but my answer seems to be wrong. Is there any mistake? (mass m is on the surface of mass M)

Relevant Equations

dF = GmdM/r^2

This is my attempt at the solution. x from the equation dF = GmdM/x^2 represents the distance between the circular plate’s center of mass and mass m.

 

[Hill]

It appears that you assume that the force between the plate and the mass $m$ is the same as if the mass $dM$ of the plate were in the center of the plate. Why would it be so?

 

[Sam Jelly]

I assumed that mass dM was in the center of each plate because the mass is distributed uniformly therefore it will be at the center of mass.

 

[Orodruin]

I assumed that mass dM was in the center of each plate because the mass is distributed uniformly therefore it will be at the center of mass.

This is wrong.

Essentially you are making an assumption about the overall gravity of an object where you cannot replace it by a point mass in the CoM, but you want to show it for a sphere - for which you can.

 

[PeroK]

I assumed that mass dM was in the center of each plate because the mass is distributed uniformly therefore it will be at the center of mass.

If that were assumed to be true in general, there would be no point in proving it for the special case of a sphere!

 

[Orodruin]

As for a simple argument to see that it is not the case for a disc:

Consider the gravitational pull of a small part of the disc $dm$. As long as the volume it is contained in is small, the gravitational pull at the point of interest can be approximated by the point source formula
$$
d\vec g = \frac{G\, dm}{x^2+r^2} \vec e,
$$
where $\vec e$ is a unit vector pointing from the point of interest towards the mass element. Now, by symmetry, the final gravitational field $\vec g$ must point towards the center of the disc, but there are two effects that both come into play for any $r > 0$:

1. The denominator $x^2 + r^2 > x^2$ so the first factor is always smaller than it would be if the mass $dm$ was on the symmetry axis.
2. The component of the unit vector $\vec e$ will be smaller than 1

Both effects mean that the contribution of the mass $dm$ is smaller than it would be if you had put it at the center of mass. Since all mass elements (except for the one at $r = 0$) give smaller contributions to the gravitational field than they would if they were at the center of mass, the result of approximating the full mass of the disc to be at the center of mass must overestimate the actual gravitational field.