BLACK HOLE CALCULATIONS

Astrophysical theory suggests that a burned-out star whose mass is at least three solar masses will collapse under its own gravity to form a black hole. If it does, what is the radius of its event horizon?

Question

BLACK HOLE CALCULATIONS

Astrophysical theory suggests that a burned-out star whose mass is at least three solar masses will collapse under its own gravity to form a black hole. If it does, what is the radius of its event horizon?

Accepted Answer

IDENTIFY, SET UP, and EXECUTE:

The radius in question is the Schwarzschild radius. We use Eq. (13.30) ([latex]R_{\mathrm{S}}=\frac{2 G M}{c^{2}}[/latex]) with a value of M equal to three solar masses, or [latex]M=3\left(1.99 imes 10^{30} \mathrm{~kg} ight)=6.0 imes 10^{30} \mathrm{~kg}[/latex]:

[latex]\begin{aligned}R_{\mathrm{S}} &=\frac{2 G M}{c^{2}}=\frac{2\left(6.67 imes 10^{-11}\mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} ight)\left(6.0 imes 10^{30}\mathrm{~kg} ight)}{\left(3.00 imes 10^{8} \mathrm{~m} / \mathrm{s} ight)^{2}} \&=8.9 imes 10^{3} \mathrm{~m}=8.9 \mathrm{~km}=5.5 \mathrm{mi}\end{aligned}[/latex]

EVALUATE: The average density of such an object is

[latex] ho=\frac{M}{\frac{4}{3} \pi R^{3}}=\frac{6.0 imes 10^{30} \mathrm{~kg}}{\frac{4}{3} \pi\left(8.9 imes 10^{3} \mathrm{~m} ight)^{3}}=2.0 imes 10^{18} \mathrm{~kg} / \mathrm{m}^{3}[/latex]

This is about [latex]10^{15}[/latex] times as great as the density of familiar matter on earth and is comparable to the densities of atomic nuclei. In fact, once the body collapses to a radius of [latex]R_S[/latex], nothing can prevent it from collapsing further. All of the mass ends up being crushed down to a single point called a singularity at the center of the event horizon. This point has zero volume and so has infinite density.

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