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Currently in my academics I am studying about the Gravitation. In the chapter I came across a term called the Escape Velocity (It's the velocity of any celestial body which is required by an object to escape from body's gravitational field without any further propulsion). When I was going through the chapter I came to know that the escape velocity of black holes is greater than $c$ and that's why even light can't escape from it's gravitational field. So proceeding toward my question,

From the information I know,

$$v_{es}=\sqrt\frac{2GM}{R}=\sqrt{2gR}$$ where $v_{es}$ is the escape velocity, $G$ is the universal gravitation constant, $g$ is the acceleration due to gravityof the celestial body, $M$ is the mass of the celestial body and $R$ is the distance between the object and center of gravitation of celestial body.

So my question is that if Black Hole is having that enormous value of escape velocity they must be either having exceptionally large value value of $M$ (i.e their Mass) or they must be having very small value of $R$, which I myself didn't know how can it be defined for a black hole.

  • So does black holes have enormous mass which result in very large value of $v_{es}$

  • Also I want to how $R$ can affect the value of $v_{es}$ in case of black holes?

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possible duplicate of What is the escape velocity of a Black Hole? –  John Rennie Jun 9 at 8:54
@JohnRennie I hadn't asked for the escape velocity of black hole. I had actually debated on it's mass instead! –  Alpha Jun 9 at 10:08
That's not the definition of escape velocity; escape velocity is the velocity that if you had it, you'd escape without further propulsion, not the required velocity to escape. If you're willing to provide propulsion enough to drive yourself against gravity then you can escape from a planet at any speed. –  Eric Lippert Jun 9 at 13:57
@EricLippert I'm sorry actually my complete attention was over my question. I had made an edit now. I think it's perfect now. –  Alpha Jun 9 at 15:40
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1 Answer 1

up vote 7 down vote accepted

The radius of a non-rotating black hole is $$r_s = \frac{2GM}{c^2} \tag{1}$$ where $M$ is the mass, $G$ is Newton's constant, and $c$ is the speed of light. This is the distance from the center of the black hole to the event horizon. The event horizon is the surface that traps light and objects, it separates inside and outside the black hole. Anything that passes inside the event horizon can never escape, even light. This is why it is said that the escape velocity at the surface of a black hole is $c$.

It is also the case that any object with $r < r_s$ is a black hole. But since the mass should be roughly proportional to the volume, and the volume is proportional to $r^3$, anything heavy enough will form a black hole. Therefore from (1) the proper answer to your question is: because they are very massive, not because they are small.

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