Does the mass of a star change as it collapses into a black hole?

Question

I know (I think!) that when a really big star collapses on itself it creates a black hole. My question: When a star collapses, is the mass equal to the mass of the star when it's not a black hole? Or does it change while collapsing? This question came to me and my friend while studying Newton's law: $$F=G \\frac{m_1 \\cdot m_2}{r^2}$$ If the mass of the star doesn't change, then it can't have enough force to "eat" light (unless it has that force in the first place). Does the force change because of the density?

Answer 1

The masses are not equal - the black hole is less (initially). When a large star reaches the end of its life, it ejects its outer layers in a supernova explosion. The remaining core collapses, as you say, into a singular point of, theoretically, infinite density creating a gravity well from which nothing can escape, not even a ray of light -hence, a black hole. I said "initially" because material such as gas, dust, and even entire stars that get too close to the black hole (the point of no return is the event horizon) will be "eaten" by it, increasing its mass.

Answer 2

Long story short, the key issue here is that r^2 in the denominator. If the radius of a collapsing star were to become one-tenth of its original radius, with no change in mass, then the gravity at the surface would be 100 times the gravity at the original surface. But in becoming a black hole, a star must decrease its radius far more than that.

According to Wikipedia's excellent article on Escape Velocity, the speed an object needs to break free from the surface of an object is equal to Sqaure root of (GM/r), so it depends both on the mass and the radius. To find the conditions necessary for a black hole, we equate the this to the speed of light: Sqrt(GM/r) = c. So GM/r = c^2, and thus r = GM/c^2. (At least, this is how you'd do it using classical, Newtonian physics).

See also: Schwartzchild radius.