# A question about Black Holes and their Gravity

• Dean Whaley
In summary, as a star begins to die, and begins fusion of H, He, C, O... it eventually reaches Fe which cannot be fused to create energy. Then the stars own gravity overcomes it's outward radiation and the star implodes, and eventually a black hole is created.

#### Dean Whaley

So I understand that as a star begins to die, and begins fusion of H, He, C, O... it eventually reaches Fe which cannot be fused to create energy. Then the stars own gravity overcomes it's outward radiation and the star implodes, and eventually a black hole is created.

How can the star's gravitational force increase when it becomes more DENSE. I thought that gravitational force was only related to mass, and distance between centers shown by the formula for Universal Gravitation:
F = (Gm1m2)/d^2 . So as the star imploded I don't see how it's gravitational force became stronger, it didn't gain mass through this process, I would assume it lost mass as it fused various elements. It seems as though all that happened is that the star gained density.

Could someone please explain this to me, and correct any errors in my question?

Thanks!

It doesn't gain mass, it losses the pressure that is created by the fusion process; that's what changes the balance enough for the runaway implosion/explosion

But how does this implosion create something that has less mass, yet more gravitational force?

The separation from the effective point source is reduced.

Dean Whaley said:
But how does this implosion create something that has less mass, yet more gravitational force?
It has more gravitational force LOCALLY. If you were outside where the event horizon will form you would not see any difference.

If the star gets smaller, all distances in the star become smaller. As the sizes of stars at the end of their life vary extremely (typically more than a factor of 100), this is much more important than the mass changes.

Are you saying that the distance between the center of the star and an outside object has been reduced because the radius of the star is now almost 0 compared to when it was alive and well and its radius was somewhere around 400 million km? Therefore in the equation F= (Gm1m2/d^2), d could possibly be a lot smaller now if an object got close enough to the black hole. Whereas when the star was alive and well the minimum d could get to would be the radius of the star then the object would get sucked into the star and burn up if it hadnt already.

Are you saying that the distance between the center of the star and an outside object has been reduced because the radius of the star is now almost 0 compared to when it was alive and well and its radius was somewhere around 400 million km? Therefore in the equation F= (Gm1m2/d^2), d could possibly be a lot smaller now if an object got close enough to the black hole. Whereas when the star was alive and well the minimum d could get to would be the radius of the star then the object would get sucked into the star and burn up if it hadnt already.

Dean, think about it this way. Right now you weight X because you are standing on the surface of the Earth. Suppose you took EXACTLY the same mass and crunched it down to the size of a small mountain and you stood on it. Do you see how you would then weigh enormously more?

because I am now far closer to the center of the "earth" which has been compressed to the size of a mountain?

Dean Whaley said:
because I am now far closer to the center of the "earth" which has been compressed to the size of a mountain?
Well, what do you think?

I think that is correct because in F =(Gm1m2/d^2) d would be a lot smaller, which would produce a much larger gravitation force from this new small mountain planet

Right. That's what happens in the runaway implosion the local gravitational force is huge, causing further implosion, causing even bigger local gravitational force causing ...

Thanks a lot for your input, it makes sense now