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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 '14 at 8:54
@JohnRennie I hadn't asked for the escape velocity of black hole. I had actually debated on it's mass instead! – Gamma Jun 9 '14 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 '14 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. – Gamma Jun 9 '14 at 15:40
up vote 4 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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because they are very massive, not because they are small This doesn't make much sense. The necessary condition is that a certain mass is concentrated within a region of space with a certain size: both $r$ and $m$ matter, and in fact what matters is $r/m$. A main sequence star that forms a black hole doesn't gain mass. The difference is the decreased size. – Ben Crowell Nov 6 '14 at 22:50

It's not just the mass that matters or just the radius. It's $r/m$. If $r/m$ is less than $2G/c^2$, then you have a black hole.

Black holes can theoretically have big masses or small masses. It's theoretically possible to have a black hole with a mass of 1 kg -- we just don't know of any processes in nature that would compress a 1 kg mass enough to make it a black hole.

When a star evolves off the main sequence and becomes a black hole, the reason for the change is mainly the charge in radius. The star actually loses a little mass because it blows some off when it goes through its death throes.

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$R$ in the formula for escape velocity is simply distance from the center of the object, not necessarily the radius of the object. If the object is something with a well-defined and nearly-spherical surface, like a planet, then we can let R be the radius of that surface, and speak about the escape velocity from the surface, but you can also talk about the escape velocity from an arbitrary distance.

In the case of a black hole, we define the radius of the black hole (the Schwarzschild radius) as the radius inside which the escape velocity is greater than the speed of light.

Now, there is a kicker, which is that the escape velocity formula is only valid if the entire mass $M$ is contained inside the sphere with radius $R$. For any given $M$ there is a Schwarzschild radius, which can be obtained by taking the escape velocity formula, setting $v_{es} = c$ to get $R_s = {2GM\over c^2}$ as others have pointed out. If an object is contained within its own Schwarzschild radius, it is a black hole. This implies that the object is very massive, or very small. Or perhaps both, depending on your definitions. As a concrete example, the Earth would have to be squeezed down to a sphere of 8.9mm radius to form a black hole.

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