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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?

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    $\begingroup$ Any star with a surface escape velocity equal to the speed of light will immediately form a black hole, where immediately means within a few milliseconds. So your question cannot be answered because no such star can exist. $\endgroup$
    – John Rennie
    Commented Oct 29 at 9:29
  • $\begingroup$ Yes, I believe you are right. I don't think such an energetic star could be ever exist where the thermonuclear outwards explosive forces could compensate for the humongous gravitational inwards implosion forces of such massive star and maintain its enormous volume versus mass in equilibrium. However, assuming such star is possible would this be indistinguishable from a giant BH but without the star going supernovae? $\endgroup$
    – Markoul11
    Commented Oct 29 at 9:40
  • $\begingroup$ With the same logic, enormous volume size BHs millions of Km in diameter *similar to what was depicted is the movie "Interstellar", should not exist and cosmological BHs should be for example no more than 100km in diameter? $\endgroup$
    – Markoul11
    Commented Oct 29 at 9:50
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    $\begingroup$ You seem to equate "supernova remnant" with "black hole," but any object with "surface escape velocity equal the speed of light" is a black hole, no matter how it was created. $\endgroup$
    – Solomon Slow
    Commented Oct 29 at 10:56
  • $\begingroup$ @SolomonSlow OP uses supernova not supernova remnant; these are two distinct things with the latter being the outer layers of the former star racing outwards at a few percent of $c$ and not the leftover inner/core material (neutron star, white dwarf, BH). $\endgroup$
    – Kyle Kanos
    Commented Oct 29 at 11:59

2 Answers 2

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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.

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  • $\begingroup$ How do you see an event horizon? $\endgroup$
    – PM 2Ring
    Commented Oct 29 at 10:11
  • $\begingroup$ @PM2Ring Ok, to be precise, what you see is the absence of background stars inside the event horizon's disc and the distortion just outside the event horizon's disc caused by gravitational lensing. $\endgroup$
    – gandalf61
    Commented Oct 29 at 11:28
  • $\begingroup$ Ok, that's better. ;) We usually call that the black hole shadow, and the blackness extends to the photon sphere, as discussed in the answers to physics.stackexchange.com/q/810733/123208 $\endgroup$
    – PM 2Ring
    Commented Oct 29 at 11:32
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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.

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  • $\begingroup$ I am not asking for given mass what is the minimum size compression radius for which the mass will not further collapse in size forming a BH but I am asking what is absolute maximum radius of a BH given its mass value? In other words to all BHs when formed get their mass/volume compressed with the same compression factor? $\endgroup$
    – Markoul11
    Commented Oct 29 at 13:18

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