Chapter 33, Problem 18P

Solution

a.

To show: The Schwarzschild radius for a star with a mass, equal to the Sun is 2.95km

Given:

A star whose mass is equal to the Sun

Formula used:

  $R_s=\frac{2GM}{c^2}$

Here, R_s is Schwarzschild radius, G is gravitational constant , M is mass and c is speed of light Calculation:

Mass of Sun is equal to $1.99\times {10}^{30}$ kg

G is equal to $\text{6}{\text{.67×10}}^{\text{-11}}\text{N}\cdot {\text{kg}}^{\text{-2}}{\text{m}}^{\text{2}}$

Speed of light c is equal to $3\times {10}^8\text{m}/\text{s}$

From $R_s=\frac{2GM}{c^2}$

  $R_s=\frac{2\times 6.67\times {10}^{-11}\times 1.99\times {10}^{30}}{3\times {10}^8\times 3\times {10}^8}$

  $\begin{array}{c}{R}_s=\frac{26.55\times {10}^{19}}{9\times {10}^{16}}\\ =2.95\times {10}^3\text{m}\\ \text{=2}\text{.95}\text{km}\end{array}$

b.

To Show: The Schwarzschild radius for a star with a mass, equal to the Earth is 8.9mm.

Given:

A star whose mass is equal to the Earth

Formula used:

  $R_s=\frac{2GM}{c^2}$

Here R_s is Schwarzschild radius, G is gravitational constant , M is mass and c is speed of light Calculation:

Mass of Earth is equal to $5.97\times {10}^{24}\text{kg}$ kg G is equal to $6.67\times {10}^{-11}\text{N}\cdot {\text{kg}}^{\text{-2}}{\text{m}}^{\text{2}}$

Speed of light c is equal to $3\times {10}^8\text{m}/\text{s}$

From $R_s=\frac{2GM}{c^2}$

  $R_s=\frac{2\times 6.67\times {10}^{-11}\times 5.97\times {10}^{24}}{3\times {10}^8\times 3\times {10}^8}$

  $\begin{array}{c}{R}_s=\frac{79.64\times {10}^{13}}{9\times {10}^{16}}\\ =8.85\times {10}^{-3}\text{m}\\ \approx 8.9\text{mm}\end{array}$