Would an Object Near a Pre-Blackhole Star Experience the Same Gravity as Post-Blackhole?

Question

My question was inspired by this question, which got me thinking. According to Newton's Law of Gravitation,

$$F = G\frac{m_1m_2}{r^2},$$

the gravity of an object is inversely proportional to the square distance between the objects, meaning that the closer the objects, or, rather, their centers of mass, get, the higher the gravitational force between them. If this is the case, why are black holes "special"? Seeing as a star is made of gas & plasma, would an object at what would become the event horizon after it becomes a black hole be "sucked in" and, assuming it isn't destroyed by heat or various pressures, not be able to get out of the gravitational pull?

If an object were extremely close to the center of gravity of a planet, whether it was solid or gas, would it be able to get out? The Schwartzchild radius for Earth, according to Wikipedia, is 8.87 millimeters. If someone were able to get that close to its center of gravity, would he/she be able to escape?

What about for smaller objects, which have a Schwarzchild radius measured in nanometers or smaller, which is the size of atoms & subatomic particles? I assume there is a limit where subatomic forces like the strong & weak forces take over, but what is that limit & why does it happen?

Answer 1

Here are several thought experiments (and what happens in each). I'll ignore relativistic effects like time distortion - not for your sake, but mine :)

• The earth collapses to a black hole beneath our feet. We fall with it, and end up inside a black hole, presumably dead.
• The earth collapses to a black hole beneath our feet, but we stay in the same place. In this case, we feel exactly the same gravitational pull (although a lot more fear).
• The earth remains the same size and shape, and we drill down to within its swarzchild radius. Say we're considering a particle at the very center of the earth. It will feel no gravitational pull at all, because each bit of the earth will be pulling it in a different direction. The pull from the mass at the north pole will be opposed by the pull from the mass at the south pole, and so on. Escape is easy from here.

Black holes are special not because the mass is large (many black holes have masses on the order of the mass of the sun), but because the radius is small. You can get close to the black hole, and every particle there will pull you in the same direction. The forces add up, instead of cancelling.

Answer 2

The answer is "maybe". Anomalous gravitational effects in black holes just exist beyond a region called "event horizon". After the collapse, only bodies moving very closely the "event horizont" can feel these bizzare gravitational effects. That's a good example about it: if the sun collapses to a black hole right now, gravitationaly, nothing would change for the planets, but if something is near enough (or inside) the "event horizon" of the "black hole-sun", then it will be subject to a colossal gravitational force.

If you want to know more about the "event horizon", you can study about "schwarzschild radius".

(I'm sorry about any English mistake)

Answer 3

Astrophysically no, because the star has to go through supernova to form a blackhole. Some of the mass will be pushed away by the explosion, so the black hole will have a smaller mass than the star before supernova.

If we ignore this and assume that mass stays the same then the object will feel the same gravity if it is at a safe distance, for the following reason. The (exterior) solution of Einstein's equations for a non-rotating neutral star is the Schwarzschild metric. After the gravitational collapse the metric will still be Schwarzschild. So if the object is far enough it should not feel any difference because geodesic equations will stay the same.

If the object is too close:

The object has to be inside the star to be very close to Schwarzshild radius. During gravitational collapse it will either be sucked in or pushed away by supernova. Either way it would feel a different gravity because the interior solution for a star is not schwarzschild.