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

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:

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 .

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

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:

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.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.

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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