Solution

From equation (8.8) we have

Since the Schwarzschild radius varies linearly with mass and has a value 3 km for a 1 M0 object, we can write an expression for RS for an object of any mass. It is

Real objects are not pointlike, but have some finite extent. An interpretation of Schwarzschild's result is that if an object is completely contained within its Schwarzschild radius, the singularity will occur.

We can understand the significance of this critical radius by recalling the discussion of gravitational redshift in Section 8.2. We saw that if a photon is emitted at a wavelength A1 at a distance r1 from a mass M, and is detected at r2, its wavelength A2 is given by

Remember, every object has its Schwarzschild radius. However, it can only be a black hole if it is contained within this radius. For example, the Sun is much larger than 3 km, so it is not a black hole.

The density of a 1 M0 black hole would be quite high, almost 1017 g/cm3. It is higher than the density of the nucleus of an atom. However, as we consider more massive black holes, the density goes down. This is because the radius is proportional to the mass, but the volume is proportional to the radius cubed (and therefore to the mass cubed). This means that the density will be proportional to 1/M2. Since we know the density for a 1 M0 black hole, we can write the density for any other mass black hole as p = (1 X 1017 g/cm3)(M/M0) "

By the time the mass reaches 108 M0, the density is only a few grams per centimeter cubed, just a few times the density of water.

We would expect the region just outside a black hole to be characterized by a large change in gravitational force over a small distance.

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