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Home » Space/Science

A black hole's escape velocity. June 10, 2022 6:39 pm ER

You will recall that in the previous thread, I cribbed (from Wikipedia) the formula for the escape velocity from a body’s surface

v= sqrt (2GM/r) where
v = escape velocity
G = universal gravitational constant
M = mass of the body to be escaped from
r = distance from the center of the mass

So suppose you have say, a ten solar mass (M) star with a radius (r) of say, 10 million kilometers.

The formula above allows you to calculate the escape velocity at its surface, a distance r from its center. Spherical objects tend to act gravitationally as if ALL their mass were concentrated at the center.

Now let us assume this star collapses into a black hole. a tiny blob with a radius (r) of say, just 10 kilometers. The mass (M) of the black hole is still ten solar masses but its radius is a million times smaller. So why is the black hole’s gravity so much stronger than the original star’s? Actually, it isn’t!. The gravity field at the original distance of “10 million kilometers” is exactly the same. The black hole isn’t any heavier than the original star. So why do black holes have such a fearsome reputation as infinitely deep gravity wells? G and M have not changed, so why is v so huge?

The reason is that r is much smaller now, and when you divide by a smaller number, the quotient gets bigger. The gravitational attraction is the same, but now you can get much closer to it.

So black holes don’t really suck all that much, you can just get close enough to be sucked in a lot easier.

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