# Does compression of matter increase gravity?

Black holes are thought to originate when a star burns out and collapses to a smaller physical size. This compresses it's matter, but would that increase it's gravity? And would it be enough to capture light? If that star's light could have been observed prior to it's collapse, how could it's burning out and collapse enable it to become a black hole and capture light?

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May we assume that you are at least familiar with the Newtonian formalism for gravity? The $G\frac{m_1 \, m_2}{r^2}$ bit? And if so, why do you doubt an increasing surface gravity? –  dmckee Jun 17 at 2:32

## 1 Answer

From afar, the answer is no. Let me put dmckee's comment here so as to make it more permanent: it gets to the nub of the problem:

May we assume that you are at least familiar with the Newtonian formalism for gravity? The $G\,\frac{M\,m}{r^2}$ bit? And if so, why do you doubt an increasing surface gravity?

Actually there is a little bit more to this: the inverse square law implies a Gauss Law/Poisson equation just as it does in the case of the electromagnetic field. Seehere for more details. I'd suggest you study this carefully so that you can reach the conclusion:

In a spherically symmetric mass distribution, the gravitational field at a distance $R$ from the centre of symmetry depends only on the mass contained within a spherical shell centred on the symmery centre

Analogous results also apply to the general relativistic solution: see the Schwarzschild solution for example. So if you stayed at a constant distance from the collapsing star, the gravitational field / local curvature would not change in the idealised spherically symmetric problem.

What would change, however, is that an event horizon would eventually "appear" [at a radius $r_s$

$$r_s = \frac{2\,G\,M}{c^2}$$

(the Schwarzschild Radius)](http://en.wikipedia.org/wiki/Schwarzschild_radius) when the star's matter reached a critical density. If other physics in the star keeps things such that the mass $M(r)$ enclosed in a spherical shell of radius $r$ fulfills $M(r) < r\,c^2/(2\,G)$ at all values of the radial co-ordinate $r$, then no horizon will form.

And, to reiterate, there would be no change in the gravity you perceive at some distance $r>r_2$. Mull over the commonly cited sentence, "if the Sun suddenly slumped into a black hole, the Earth's path through space would not change at all".

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