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Is the event horizon of a black hole the actual physical surface of the object?

a) Yes
b) No
c) Only for stellar-mass black holes
d) Depends on the black hole's rotation

1 Answer

4 votes

Final answer:

The event horizon is not the physical surface of a black hole but a boundary from which no light can escape. It is determined by the Schwarzschild radius, which depends on the black hole's mass.

Step-by-step explanation:

The event horizon of a black hole is not the actual physical surface of the object, so the correct answer to the question is b) No. The event horizon is defined as the boundary around a black hole from which nothing, not even light, can escape, and it marks where the escape velocity equals the speed of light.

The true physical surface of a black hole, if it can be said to have one, is thought to be the singularity—a point of infinite density and zero volume at the black hole's center. The Schwarzschild radius (Rs) is the calculation used to determine the size of the event horizon, given by Rs = 2GM/c2, where G is the universal gravitational constant, M is the mass of the body, and c is the speed of light.

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