# Gravity in the center of a hollow neutron star [duplicate]

Imagine a hollow 100 metre diameter (for example) sphere made of incredible dense material (ie neutron star dust etc) but is self supporting (ie the central cavity).

Assuming that the sphere skin is reletively thick so that the whole object is exceptionally heavy (eg the mass of planet earth.

What would someone who was stationed dead center of the sphere feel? IE would they be crushed by the potential gravity or ripped apart and smeared on the inside other sphere cavity (due to the gravity of the surrounding material?

• Explain, how the neutron star condition is different from the gravity in the center of the Earth? – Val Jul 2 '13 at 15:24
• possible duplicate of Would you be weightless at the center of the Earth? – John Rennie Jul 2 '13 at 15:27
• Isn't this pretty much a duplicate of: physics.stackexchange.com/q/68519 – Abhimanyu Pallavi Sudhir Jul 2 '13 at 15:40
• this is not duplicate of the Earth gravity. It is a result of human researching black holes and quantum gravity without ever looked at the basic physics, what happens under his foot. New generation is much more ambicious than their fathers, who had no quantum black holes to draw all the focus from the basics. – Val Jul 2 '13 at 15:46
• I think that this shouldn't be treated as a duplicate because the OP seems to think that the result should be dependent on how dense the hollow shell is. They have specifically used the words "made of incredible dense material". So I believe there is understanding to be gained by asking this question, hence why I have given my answer indicating the result is independent of density. – Will Jul 2 '13 at 16:00

Gravity is gravity is gravity. Whether treated using Newtonian gravity or GR, the result is the same. An observer anywhere within the hollow of the sphere feels no force. This result is independent of the density of the hollow sphere (assuming only that the density is the same on every "shell" of distance $r$ and the sphere isn't rotating) - it's a result of symmetry.

To elaborate:

(1) Using Newtonian gravity, this can easily be seen via the "Newton's shell theorem". The observer inside the hollow experiences no gravitational force.

(2) Using GR, this can easily be seen via "Birkhoff's theorem". The observer inside the hollow is in flat (Minkowski) space-time - there is no curvature and therefore experiences no "gravitational force".

• Minor suggestions to answer (v4): (i) Replace the words Gauss shell theorem with Newton's shell theorem. (ii) Correct spelling of Birkhoff. – Qmechanic Jul 2 '13 at 16:28

The answer is that the person in the hollow shell would feel no gravity at all (assuming the shell is the only significant source of gravity). The person can be anywhere inside the shell, and this will still be true. This can be understood as a consequence of Gauss's law for gravity.

$$\oint \mathbf{g} \cdot d \mathbf{A} = -4\pi GM$$

where the integral is a surface integral over gravity, $G$ is the gravitational constant, and $M$ is the mass enclosed by the surface.

It's difficult to explain why this in a mathematical way, unless you're familiar with vector fields and flux.

An easier to understand explanation is that all the gravity cancels out. If the person inside the sphere is really close to one side (left side for example), then the mass on the left side pulls more strongly because it's closer. But if the person is on the left side, then there's far more mass pulling from the right side, even if each amount of mass pulls more weakly. The gravity from both sides cancels out perfectly, regardless of where you are in the sphere. The result is that the person inside feels nothing.