Why does gravity cancel out for all points inside a sphere?

In summary: If you jump into a hollow spherical shell without any air inside, you will have a velocity relative to the shell due to the gravitational acceleration. However, the velocity will be very small because the mass of the shell is much greater than the mass of the object that is jumping in.
  • #36
Hi, was just reading this topic and was wondering if anyone could give me a link to a practice problem/solution for this type of question. I was trying to prove it through Gauss' law but get slightly confused on some of it.. and find it awful hard to follow when it isn't written in proper form (i.e /frac[r] (always find this hard to follow).

Would really like to see how it works out, i assume you do something similar to when you calculate a surface integral, with r varying between r < R.

Thanks,

Harry
 
Physics news on Phys.org
  • #37
also perhaps you could show it through treating it like a point mass( in EM a point charge).
thanks again.
 
  • #38
Another interesting GR feature of this effect is that even though the gravitational acceleration goes the zero everywhere inside such a sphere the gravitational time dilatation remains slowed to the same as on the surface. So you have what is essentially flat space-time inside and far removed from the mass, yet two observers with no relative motion between them them in these respective flat regions of space-time and yet their clocks will still have different relative rates.

This also means that the Gravitational constant (big G) associated with masses in the flat regions of space can 'apparently' differ if the difference in depth of field is not properly accounted for. Or conversely they can merely disagree on each others total mass which is the approach used by GR for good reason. Keeping the physical constants constant in this way keeps the laws of physics consistent in all cases, even though relativistically speaking you could get the same observational effects even if you assumed they varied.
 
  • #39
Hazzattack said:
... find it awful hard to follow when it isn't written in proper form (i.e /frac[r] (always find this hard to follow).

FYI, I have edited those posts (Post #'s 7 and 25) to make the equations readable. We have recently switched to new equation processor software, and some older posts do not display properly using the newer software.
 
Back
Top