Solution to Exercise 3

1. The radius is R = 2MG / c2. The product 2G / c2 is

2 G / c 2 = 2 x 6.67 x 10 -8 cm3 g-1 s-2 / (3 x 1010 cm/s x 3 x 1010 cm/s)
= 1.482 x 10 -28 cm / g

Therefore, R = ( 1.482 x 10 -28 cm/g ) x M. Thus, plugging in the mass M in grams gives the radius in centimeters. The results are summarized in the following table.


Radius for Black Holes of a Particular Mass
Object Mass Black Hole Radius
Earth 5.98 x 10 27 g 0.9 cm
Sun 1.989 x 10 33 g 2.9 km
5 Solar Mass Star 9.945 x 10 33 15 km
Galactic Core 10 9 Solar Masses 3 x 10 9 km


2. The radius of the Sun is 6.96 x 1010 cm = 6.96 x 105 km. Taking this as a characteristic size for a normal star in a binary containing a black hole of 5 solar masses, we find from the above table that the ratio of the radii for the normal star and the black hole is

R(star) / R(black hole) ~ 6.96 x 105 km / 15 km ~ 46,400.

The accretion ring in the Hubble image of NGC 4261 is about 200 LY ~ 1.9 x 10 15 km in radius. The billion solar mass black hole has a radius of about 3 x 10 9 km. Thus, the accretion ring is about

1.9 x 1015 / 3 x 109 = 6.3 x 105

times larger than the black hole. Since the black hole is almost a million times smaller than the accretion ring, it could not be seen in the Hubble image, even if it were not obscured.