Question

Now we consider a black hole of the same mass as the Sun: Mbh 2 x 1050 k (a) (2 marks) Show that if you are launching a rocket with velocity v upwards from a planet of mass M, you can only escape the planet's gravity if you start from a radius r > 2GM/v2 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 v -c and see that for any given mass, there is a radius below which even light can't escape 2GM /horizon- c2 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 zmv- for the kinetic energy of light: no good. We might expect a proper analysis to give a radius proportional to rhorizon. Solving the problem with general relativity gives us a 'lucky' result: the constant of proportionality is one (b) (2 marks) What is the radius of a one-solar-mass black hole'? The only thing we can tell about a black hole from outside is its mass (well, and spin and charge, which we'l ignore). We have no idea how matter/energy 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 S - Smax (m) into a black hole with S < Smax(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 15GM2 k. hc bh Stephen Hawking used a much more sophisticated argument in 1973 to show that the entropy of a black hole is exactly bh which is about five times bigger than your estimate.

Answer 1

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