13.7 Einstein's theory of gravity  (Page 4/19)

Calculating the schwarzschild radius

Calculate the Schwarzschild radius for both the Sun and Earth. Compare the density of the nucleus of an atom to the density required to compress Earth’s mass uniformly to its Schwarzschild radius. The density of a nucleus is about $2.3\times {10}^{17}{\text{kg/m}}^3$ .

Strategy

We use [link] for this calculation. We need only the masses of Earth and the Sun, which we obtain from the astronomical data given in Appendix D .

Solution

Substituting the mass of the Sun, we have

This is a diameter of only about 6 km. If we use the mass of Earth, we get $R_{\text{S}}=8.85\times {10}^{\mathrm{-3}}\text{m}$ . This is a diameter of less than 2 cm! If we pack Earth’s mass into a sphere with the radius $R_{\text{S}}=8.85\times {10}^{\mathrm{-3}}\text{m}$ , we get a density of

Significance

A neutron star    is the most compact object known—outside of a black hole itself. The neutron star is composed of neutrons, with the density of an atomic nucleus, and, like many black holes, is believed to be the remnant of a supernova—a star that explodes at the end of its lifetime. To create a black hole from Earth, we would have to compress it to a density thirteen orders of magnitude greater than that of a neutron star. This process would require unimaginable force. There is no known mechanism that could cause an Earth-sized object to become a black hole. For the Sun, you should be able to show that it would have to be compressed to a density only about 80 times that of a nucleus. (Note: Once the mass is compressed within its Schwarzschild radius, general relativity dictates that it will collapse to a singularity. These calculations merely show the density we must achieve to initiate that collapse.)

Got questions? Get instant answers now!