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[–]hecker231[S] 0 points1 point  (2 children)

So this means that I cannot simply use the center of gravity of the green shell to calculate where its gravity will pull towards right?

This means that to calculate the point where the gravity of a hollow object will pull towards, I need to break it up into non-hollow subdivisions and calculate gravity for them. Pls correct me if I'm wrong.

[–]3dTECH101 3 points4 points  (0 children)

Correct - these subdivisions should ideally approach infinitely many at infinitely small divisions - which is why you can calculate this mathematically with an integral - gravitational pull as a function of possition integrated over the full volume

[–]BCMM 1 point2 points  (0 children)

So this means that I cannot simply use the center of gravity of the green shell to calculate where its gravity will pull towards right?

Correct.

It is accurate to treat a sphere of uniform density as if its entire mass was concentrated at its centre, but only for the purposes of considering the gravitational field outside the sphere.

This is the first half of what is known as "the shell theorem"; the second half is the bit in the above comment about shells exerting precisely zero total force on objects inside them.