# B Is the inverse square law exact near a spherical body?

#### fizzy

I'm forking this off another thread where I brought it up but it was getting OT.

"Inverse square law is exact everywhere outside of spherically symmetric objects"
It is good enough for a first approximation but it is certainly not exact.

Consider a test mass one radius from a spherical body. Work out the contributions form two points diametrically opposed on the surface, once is distant by r , the other by 3r. Compare that to the same masses placed a the centre of mass.

1/r^2+1/(3r)^2 = 2/(2r)^2 ??

A similar inequality and of the same sign will apply for all pairs of points symmetrically placed either side of the plane through the c.o.m. and perpendicular to the line joining the test mass and the centre. Summing all such pairs to represent the whole shows the inequality holds for the whole body.

The inaccuracy reduces as the separation becomes much larger than the diameter of the sphere.

mfb replied:
It is exact. You can't limit the calculation to two points, you have to integrate over the whole shell. Do it and see what you get.
Alternatively: It is a direct consequence of Gauss' law.

This is a well-known result from classical mechanics and not the topic of this thread.
Well I was not considering just two points. I extended the two point conceptually to integrate over the whole body. There may be a flaw in my logic, so maybe someone can point it out.

Because of the non linear nature , the nearer point will have attraction greater than the centre of mass proxy point and that difference will greater than deficit of the more distant symmetrically placed sibling point.

For all such pairs of points there is an inequality of the same sign. There will be exactly the same number of such points in either hemisphere so whichever way we integrate we end up with an inequality w.r.t the point mass at the centre .

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#### Ibix

Google "shell theorem". Newton himself proved you wrong. The maths is quite messy if you aren't familiar with Gauss' Theorem; it's a one-liner if you are.

#### fizzy

Thanks Ibix, can't see what was wrong with my logic but doing the integral seems clear.

#### Ibix

You haven't laid it out completely, but you appear to be summing $F=GMm/r^2$ for each elementary point on the sphere. This is wrong - force is a vector, so you need to sum $\vec F=(GMm/r^2)\vec{\hat r}$. Assuming your test mass is on the z axis, the x and y components will clearly cancel by symmetry, so you should only be summing the z components, which will, in general, have a trigonometric factor you appear not to have accounted for. Since that's always less than one your inequality won't hold all around the sphere.

#### mfb

Mentor
Original discussion

Let's put our observer above the north pole to establish a coordinate system. The points at the north pole and south pole lead to a larger attraction than an equal mass at the center, but two opposite points at the equator will lead to a smaller attraction: They have a larger separation and the smaller absolute forces don't point in the same direction, so we lose something there as well.
If you integrate over the whole sphere you get a force exactly equal to a point mass in the center.
It is good enough for a first approximation but it is certainly not exact.
Don't make statements like that if you are not really certain.

#### fizzy

yes, effectively there will be a difference in angle between the back of front elements of each pair that I had not accounted for. That could be it.

#### Dale

Mentor
Summing all such pairs to represent the whole shows the inequality holds for the whole body.
Instead of summing you need to integrate. For a body of finite density the mass at a point is 0, and thus the error that you describe sums to 0. So instead you need to consider differential volumes $dV$ with mass $dm=\rho dV$. When you do so you will find that the inverse square law is indeed exact outside of a spherically symmetric object for Newtonian gravity.

I extended the two point conceptually to integrate over the whole body. There may be a flaw in my logic, so maybe someone can point it out.
You don’t integrate points, you integrate differential volumes.

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