Possible Duplicate:
What is the escape velocity of a Black Hole?

This is gravitational speed for earth: $$v=\sqrt {g_{e}r}.$$ What is gravitational speed for a black hole? I want an approximate speed.

  • 3
    $\begingroup$ If, by gravitational speed, you mean escape velocity, and if, by black hole, you mean the event horizon of a black hole, then the answer requires only the understanding of the meaning of the terms; no calculation required. $\endgroup$ – Alfred Centauri Jun 30 '12 at 12:17
  • $\begingroup$ escape velocity for earth is $v=\sqrt {2g.r}$ this is not my question $\endgroup$ – gravity Jun 30 '12 at 14:44
  • 5
    $\begingroup$ Dear gravity, could you please explain what is your question then, more precisely, what is meant by "gravitational speed"? This is surely not a standard term of any sort, is it? Otherwise your question is analogous to the following puzzle: the potato voltage for a strawberry is 7 apples. What is the the potato voltage for a dark chocolate? $\endgroup$ – Luboš Motl Jun 30 '12 at 15:36

In your expression:

$$ v=\sqrt {g_{e}r} $$

presumably $g_{e}$ is the acceleration due to gravity $GM_e/r^2$, in which case to make the physics clearer let me rewrite your equation for $v$ as:

$$ v = \sqrt{\frac{GM_e}{r}} $$

where $G$ is the gravitational constant, $M_e$ is the mass of the Earth and $r$ is the distance from the centre of the Earth. This tells you the velocity of an object falling towards the earth with an initial speed of zero at infinity. I'm guessing that your question asks what the expression for $v$ looks like for an object falling into a black hole, and if so the answer is:

$$v = \left( 1 - \frac{2M}{r} \right) \sqrt {\frac{2M}{r}}$$

where $M$ is the black hole mass in units of length: $M = GM_{bh}/c^2$. We generally write the mass this way to eliminate all those factors of $G$ and $c$ from the algebra. The proof of this is straightforward, but I won't clutter up this answer by giving it here unless someone really wants me to.

The expression for $v$ is actually pretty similar to the non-relativistic one, but if you look closely there are some very weird things about this equation. For example, the event horizon is at $r = 2M$ and if we plug this value into the equation for $v$ we get:

$$v = \left( 1 - \frac{2M}{2M} \right) \sqrt {\frac{2M}{r}} = zero!$$

The equation tells us that the velocity is zero at the event horizon. Surely something is wrong? Well no, it isn't (and don't call me Shirley) the velocity really is zero. The reason for this s that the $r$ used in this equation is the Scharzschild radius, i.e. the radial distance as measured by an observer far from the black hole, and the velocity is the velocity measured by the distant observer. But as our object accelerates towards the event horizon we see it's time slowing down and as a result it's velocity appears to reduce. This is the well known phenomenon that an object falling towards a black hole takes an infinite time to reach the event horizon.

The obvious next question is to ask what speed the object measures itself passing the event horizon. This is a little more complicated than you might think, because the object has no way of telling when it passes the event horizon. The best it can do is calculate where the event horizon should be, and measure the speed it passes that point. Again I'll just quote the result (proof available on request): the object passes the event horizon at the speed of light.

This is sort of obvious if you consider that the external observer sees the object freeze at the event horizon. We know that time slows for a moving object, so at what speed does the time slow to nothing. That happens at the speed of light. So if we see time slowing to a stop for the object that implies it's moving at the speed of light.


I think that the speed or escape velocity from a black hole (the speed necessary to escape the gravitational pull (overcome the gravitational force) of a celestial body) can be written like this

$v=\sqrt{\frac{2GM}{R}}$, where

$G$ is the gravitational constant

$M$ is the mass of the celestial body


$R$ is the radius of the celestial body

This might not be the exact speed but an approximate speed in the case of a black hole.


Not the answer you're looking for? Browse other questions tagged or ask your own question.