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In the case of the Black Hole (BH) the surface is a few Km from the center of mass whereas in the case of an active star with the same mass as the BH, the surface can be be many millions of Km distant from the center of mass. Therefore, because the inverse square law of distance, gravity in the case of a giant star is much more less on the surface of the star than in the case of the event horizon of a BH with the same mass.

However, for an external observer far away, a homongous mass giant star with a surface escape velocity equal the speed of light should theoretically appear as a dark star similar to a BH?

Qmechanic
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Markoul11
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2 Answers2

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In the case of the Black Hole (BH) the surface is a few Km from the center of mass ...

That is the event horizon of the black hole, not its surface. It is not clear that a black hole even has a surface, but if it does we certainly cannot see it.

... because the inverse square law of distance, gravity in the case of a giant star is much more less on the surface of the star than in the case of the event horizon of a BH with the same mass

Since a visible star has, by definition, a larger radius than the event horizon of a black hole of the same mass, then gravity at its surface would be less than gravity at the event horizon of a black hole with the same mass.

However, for an external observer far away, a homongous mass giant star with a surface escape velocity equal the speed of light should theoretically appear as a dark star similar to a BH?

If the surface escape velocity of the star is equal to the speed of light then it must collapse into a black hole, so what you then see is its event horizon, not its surface.

gandalf61
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It all comes down to the density of the mass-energy distribution. As long as the density is sufficiently low, so that for a static, non-rotating mass-energy distribution not all the mass is contained within the Schwarzschild radius $$ r_s= \frac{2GM}{c^2} $$ no black hole will form, where in the above expression $M$ means the total mass (including energy through $E = Mc^2)$, $G$ is Newton's constant, and $c$ is the speed of light in vacuum. Conversely, any amount of mass can be turned into a black hole if compressed into a radius smaller than $r_s$. The escape velocity from the surface of an object is determined solely from the spacetime geometry resulting from the corresponding mass-energy distribution, and does not directly depend on the amount of mass-energy involved.

paulina
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