SOLUTIONS TO EXERCISE 21

(a) D = M/V, so 

      1.67 * 10^(-27) kg
  D = ------------------- = 1.67 * 10^18 kg/m^3
        (10^(-15) m)^3

This is roughly equal to the neutron star density computed in Exercise 19,
which was just under 10^18 kg/m^3.

(b) Using the formula for a black hole radius from Exercise 20,

r = GM/c^2 = (6.67 * 10^(-11)) * (2 * 10^30) / (3 * 10^8)^2 = 1.48 * 10^3 m.

This is a little less than one mile.
So the density of this black hole is M/(4/3 * pi * r^3)

 D = (2 * 10^30)/( 1.33 * 3.14 * (1480)^3 ) = 1.5 * 10^20 kg/m^3.

This density is about 150 times greater than that of a neutron star.

(c) Again, using the formula for a black hole radius from Exercise 20,

r = GM/c^2 = (6.67 * 10^(-11)) * (5 * 10^9 * 2 * 10^30) / (3 * 10^8)^2
  = 7.4 * 10^12 m.

This is about four billion miles, or roughly the distance from Pluto to the Sun.
So the density of this black hole is M/(4/3 * pi * r^3)

 D = (5 * 10^9 * 2 * 10^30) / ( 1.33 * 3.14 * (7.4 * 10^12)^3 )
   = 5.9 kg/m^3.

This is much lower than the density of a neutron star.  In fact, it is less than
1/150 the density of liquid water!  Apparently, a supermassive black hole can
be quite ethereal.