Problem with black holes? I understand that black holes are formed when supermassive stars (>10 times the size of our Sun) die. When they die, their cores continue to shrink. I have read that eventually the core reaches a "critical density" where it collapses and becomes a point of immense gravity. What I do not understand is that why is the gravity of the black hole greater than the gravity of the original star when the mass of the entire system did not change at all. In my perspective, the gravity of the black hole must have not changed because Newton's law of gravitation states that the gravitational force is directly proportional to the masses of two objects and inversely proportional to the square of the radius between them. Of course, Newtonian mechanics may not apply to complex objects such as black holes but, if anyone could clear up my misconception, that would be great.
 A: I don't think you do have any misconceptions other than not thinking about where the mass is. When the star is a massive main-sequence star the mass is distributed in a sphere that is many solar radii in radius. Newton's shell theorem says that the gravity exerted at a given radius from the centre of that (symmetric) sphere is equivalent to that exerted as if all the mass interior to that radius was situated at the centre. Any mass exterior to this point has no net influence. Thus if you get close to the centre of the star, only a small fraction of the total mass is interior to this radius.
Once the black hole has collapsed then you can get very close to the centre of the mass distribution but there is still a huge amount of mass (the entire mass of the black hole) that is still interior to any radius you choose. It is this mass interior to any radius which is responsible for the very strong gravitational field at small radii.
If you were to observe the black hole from a radius that was bigger than the original size of the star before collapse, then actually its gravitational field at this point could be weaker than for the original star, because only a fraction of the stellar mass ends up inside the black hole. The rest is probably/possibly blown into space during a supernova explosion.
A: Indeed, Newtonian mechanics does not apply here, but it can still be used to help your understanding. As a body grows smaller, its surface gravity increases. This is just a direct application of the $\frac{1}{r^2}$ law.
As the pressure that was keeping the star large dies away, it shrinks in volume and its surface gravity increases. If the gravity at a point is greater than a certain value, it will prevent light from escaping, and this only depends on the surface gravity, not the total energy of the gravitational field. So a black hole doesn't have "more" gravity than the star, but its surface gravity is greater - so much so that light cannot get away.
This would be true of any object if it were made small enough. Indeed, Heaviside predicted "dark stars" long before general relativity was conceived, based entirely on Newtonian gravity, because of this principle of very dense things having large surface gravity. He did have to make the assumption that light was affected by gravity, which not known at the time, and relativity does indeed show this to be true, as well as explaining more fully the collapse process, and why black holes are so strange.
EDIT: I should point out that this is more of a comment that grew out of hand, but does provide some explanation.
A: First a quick point: There is a change in the mass of such a system, because when a star dies it looses much of it's mass during the explosion. Black holes formed during such processes are generally not larger then a few solar masses.
So the mass is actually smaller, but the gravitational field of a black hole is much larger because the object is so much smaller as well. If we assume the star is spherical, we can discuss this explicitly. The gravitational acceleration at the surface of a spherical object with mass $M$ contained within the sphere of radius $R$ is
$$g=\frac{GM}{R^2}$$
Take a solar mass object (2 $\times$ 10$^{30}$ kg). If the object is Sun-sized ($R=7\times 10^8$ m), the acceleration will be
$$g_S=272~m/s^2$$
Conversely, the typical size of a black hole is given by the Schwarzchild radius $R_S=2GM/c^2$, so the gravitational acceleration at the surface of a black hole will be
$$g_{BH}=\frac{GM}{(2GM/c^2)^2}\approx 1.52\times 10^{13}~m/s^2$$
(with a Schwarzschild radius of 2960 m). Now of course this isn't what the acceleration really is - we need General Relativity for a real calculation. But it gives you a sense of why a black hole is such an extreme object.
A: Why is the gravity of the black hole greater than the gravity of the original star?
This question assumes that the mass of a BH is contained in an "infinitely" dense singularity at the center of its Schwarzschild envelope.  But this assumption doesn't really make any sense.  
One could be equally correct in saying that the original collection of particles that formed the BH were transformed into $E = mc^2$ when the dynamics of particle momentum exceeded relativistic limits; it's assumed that this transformation would take place coincidently with the formation of the Schwarzschild envelope...where particle velocity approaches the speed of light.  
One could further assume that this radiant ($E = mc^2$) energy is manifested at or near the Schwarzschild boundary.  These assumptions would not change the "nature" of a black-hole, except that it would eliminate the need for a singularity and infinite gravity.
