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Since the black hole is formed by a star of high mass after supernovae it should have less mass than original star. Then how the gravity of black hole is much stronger than a star that formed it? Since F∝Mass of object.

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marked as duplicate by Qmechanic Apr 25 '18 at 6:42

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Because $F$ is in fact proportional to $M/r^2$ (thinking of gravity in a Newtonian way). The mass that remains has been compressed into a small radius. Once a certain mass is inside it's Schwarzschild radius then it is a black hole.

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Let's consider a simpler situation: a star and a black hole with the same mass. The gravitational force exerted by these objects will be the same when observed from the same distance. That is, the force felt by an observer 100 million miles away from the star will be the same as the force felt 100 million miles away from the black hole.

The difference comes about because a black hole is much smaller than a star, so an observer can get much closer to the black hole while still feeling the gravitational pull from all of the mass, even inside the event horizon. If you get that close to a star, you will be inside it, and that means the gravitational force is actually reduced because the part of the start behind you (that is, further from the center than you) is pulling you away from the center, reducing the force felt.

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  • $\begingroup$ ... the gravitational force is actually reduced because the part of the star behind you (that is, further from the center than you) is pulling you away from the center, reducing the force felt. I think this is not correct. If the mass of the star is distributed with spherical symmetry, then the mass further from the centre than you exerts no resultant force on you (Newton's Shell Theorem). $\endgroup$ – sammy gerbil Aug 1 '18 at 13:18

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