In the cases of black holes that form from supernova and collapse of a massive star, I understand that in most of these cases, the star loses significant amounts of mass from the explosion. Presumably, after this point as the remaining mass becomes more dense as it further collapses unto itself, it eventually becomes a black hole with gravitational force greater than that of its parent star. But, if gravity is based on mass, how can the black hole have greater gravitational force than the star from which it is formed?
Answer
For a given mass the gravitational attraction remains the same -- but only if you are far away.
For example, the surface gravity of Sol, our sun, is $274$ $ m/s^2$, about 28 times the surface gravity of Terra, which is $9.8$ $ m/s^2$.
But as the material is compacted, the surface gravity increases: this is because the effective mass can be treated as concentrated at a point in Newtonian gravitation: $F=GMm/R^2$, and here $M$ is the constant mass of the (remanant) of the star, while $m$ is the observer, and $R$ is the distance from the surface to the center of the star.
As the star becomes smaller, the distance between the surface and the center shrinks. For Sol the effective radius would shrink from $5\times 10^5$ $km$ to about $3$ $km$, the Schwarzchild radius. This puts you $100,000$ times closer, so the gravitational force would be $10^{10}$ times greater, and would vary measurable from your feet to your head.
So it all depends upon your distance. The earth would receive the same pull as always, minus the supernova and the missing mass, of course.
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