Question

Now we consider a black hole of the same mass as the Sun: Mbh 2x 1030 kg. (a) (2 marks) Show that if you are launching a rocket with velocity v upwards from a pl M, you can only escape the planet's gravity if you start from a radius r > 2GM/ t of mass Hint: Use Newtonian mechanics. What if your rocket is acutally a beam of light? If we forget about relativity for a minute, we can put c and see that for any given mass, there is a radius below which even light can't escape: 2GM horizon This radius is very small all the mass has to be packed into such a small a radius. If it is, light can't escape, and the 'planet' is a black hole. Note: This derivation is bogus: we got the right answer from a wrong analysis. We've used Amv2 for the kinetic energy of light: no good. We might expect a proper analysis to give a radius proportional to Phorizon- Solving the problem with general relativity gives us a lucky' result: the constant of is one. (b) 2 marks) What is the radius of a s black hole? The only thing we can tell about a black hole from outside is its mass (well, and spin and charge, which we'll ignore). We have no idea how matterlenergy is configured inside the black hole. To estimate the entropy of a black hole, we have to assume that it has the highest entropy possible Smax(m) given its mass. Otherwise it could violate the Second Law: collapsing a maximum entropy system with SSmar(m) into a black hole with SSmax(m) would decrease the entropy of the isolated system. (c) (5 marks) The lowest mass-energy thing that will fit in black hole of radius r is a photon with wavelength 2r. Find the number of such photons you can make with a mass M and show the entropy is approximately hc Stephen Hawking used a much more sophisticated argument in 1973 to show that the entropy of a black hole is exactly Sr-GM2 k, hc which is about five times bigger than your estimate. (d) (2 marks) Calculate the entropy of a one-solar-mass black hole

Answer 1

12GMm

2GM Tr

For light,

2GM

(b) For a solar mass-

$r_H\approx3\textup{ km}$

(c)

ic

$N=\frac{Mcr}{h}$

Put r (from part a) in N :-

2GM2

Thus,

2kBGM rN