# Gravity of Black hole far away.

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1. Sep 15, 2014

### avito009

Why the gravity around a black hole remains normal unless you get extremely close?

But first: How far from earth is there zero gravity?

Gravity declines as a function of the square of the distance from a mass. Earth's gravity is less and less as one goes further away, but is never zero. A quarter of a million miles from earth, gravity is still strong enough to hold the moon. Earth has very tiny affects on the orbits of Mars and Venus, millions of miles away.

The strength of the gravitational force between two objects with fixed masses decreases as the distance between them increases. It follows an inverse square law, with the force being proportional to 1/r2 (r being the distance). As the distance between two bodies increases the intensity of gravitational force decreases by a factor of 1/r2, where r = distance between two bodies.

Now the question is that why is a black hole's gravity not so strong far away when it is so strong near it?

A black hole is when a massive star dies, it leaves behind a small, dense remnant core. If the core's mass is more than about three times the mass of the Sun, the force of gravity overwhelms all other forces and produces a black hole.

But gravity near a black hole doesn't increases as 1/r2, but instead by the much greater: (1/r2) / (1-r/R), where R is the radius of the hole. But at a long distance from the black hole the force of gravity falls off as the inverse square of the distance, just as it does for normal objects.

Imagine a star 100 times as massive as our Sun and squashed down to -- well nothing. Infinite density but zero size -- this is a black hole. An object with the mass of 100 suns but with zero radius. So you end up with an object with so much mass and zero or at least negligible radius that all that mass is compressed into such a small area that's it's gravity is so great that not even light can escape it. (Mass of black hole is more so gravity is more)

So since gravity = GM/r2. R is the radius and so since R is very small the numerator would be greater because it has more mass and the denominator R is small so gravity is too much.

So the question I have is that why the gravity around a black hole remains normal unless you get extremely close?

2. Sep 15, 2014

### Bandersnatch

Where did you get that idea from? It's inverse square law all the way down.

And it gets so strong, because you can get so close to all that mass, without leaving any of it behind, as you'd do with a normal star.

3. Sep 15, 2014

### Chronos

avito009, you should read Romero's recent paper - http://arxiv.org/abs/1409.3318. He is an expert on the subject.