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?
$\begingroup$
$\endgroup$
1
-
1$\begingroup$ Gravitational field/force is not just based on mass, even in Newtonian gravity. $\endgroup$– ProfRobApr 24, 2016 at 19:57
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
5
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.
answered Apr 24, 2016 at 19:54
-
1$\begingroup$ Good answer. I think the it should be, "force would be $10^{10} = 10,000,000,000$ times greater". And the difference in force would be $[(3000 + 2)/3000]^2 \approx 0.001$. $\endgroup$ Apr 24, 2016 at 19:59
-
$\begingroup$ @DilithiumMatrix: I've corrected the calculations; I was thinking of Larry Niven's Neutron Star, and lost a bunch of zeros ... $\endgroup$ Apr 24, 2016 at 21:50
-
$\begingroup$ Thanks for the great answer. So if I understand correctly, the gravitational "range" of a star doesn't become larger as it collapses into a black hole, it is just that the area within the star's former volume becomes accessible to objects to then experience the higher gravitational forces closer to the core. Is it safe to assume that if we could observe the very most interior of an active star undergoing fusion, we would only observe blackness because the light generated from the reaction would not be able to escape the strongest gravitational forces at the core? $\endgroup$– DeclanApr 25, 2016 at 1:47
-
1$\begingroup$ @Declan: the core of a star is not a black hole, and so it isn't black. It has a higher density towards the center, which is why the core can support fusion: gravity provides the pressure required to sustain the reaction. The photons created here don't make it to the surface, but the heat does travel outwards via diffusion. $\endgroup$ Apr 25, 2016 at 2:27
-
$\begingroup$ Thanks @peter dierh. You confirmed the point I was trying to get at which is that the photons at the core don't make it to the surface (which might lead to the presumption that from an outside observer's perspective they would be prevented from visually seeing the core if the outer layers of the star could be shed away purely for observational purposes). $\endgroup$– DeclanApr 25, 2016 at 2:42