Chapter 12, Problem 9C

To determine

(a)

The escape velocity on the earth.

Answer to Problem 9C

Escape velocity on earth is approximately $11,200\text{m/s}$.

Explanation of Solution

Given:

Mass of the earth = $M=5.972\times {10}^{24}\text{kg}$

Radius of earth = $R=6.4\times {10}^6\text{m}$

Universal gravitational constant = $G=6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}{\text{/kg}}^{\text{2}}$.

Formula used:

The escape velocity is given by, $v_e={\left(\frac{2GM}{R}\right)}^{1/2}$.

Calculation:

Substitute all given values in the above equation,

$\begin{array}{c}{v}_e={\left(\frac{2GM}{R}\right)}^{1/2}\\ v_e={\left(\frac{2(6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}{\text{/kg}}^{\text{2}})(5.972\times {10}^{24}\text{kg})}{6.4\times {10}^6\text{m}}\right)}^{1/2}\\ v_e={(}^{124478875}\text{m/s}\\ v_e=11,157\text{m/s}\\ \approx 11,200\text{m/s}\end{array}$.

Conclusion:

Escape velocity of object from the surface of earth is $11,200\text{m/s}$.

To determine

(b)

To prove:

The expression of Schwarzschild radius with the help of Newtonian mechanics is given by $R=\frac{2GM}{c^2}.$

Explanation of Solution

Given:

Mass of sphere = $M$

Radius of sphere = $R$

Speed of light = $c$

Mass of object = m.

Calculation:

By conservation of mechanical energy,

Loss of potential energy = gain in kinetic energy

$\begin{array}{l}\frac{GMm}{R}=\frac{1}{2}m{c}^2\\ R=\frac{GM}{c^2}\end{array}$.

Conclusion:

Thus, we have derived Schwarzschild radius using the concept of Newtonian mechanics.

To determine

(c)

The Schwarzschild radius of earth when escape velocity is the speed of light for any object. Also, compute the density of Earth and compare it with the current density of Earth.

Answer to Problem 9C

Schwarzschild radius of earth is $0.008852\text{m}$ and density of earth would be approximately $2.1\times {10}^{30}{\text{kg/m}}^{\text{3}}$ which is extremely high than the current density of Earth.

Explanation of Solution

Given:

Mass of the earth = $M=5.972\times {10}^{24}\text{kg}$

Current radius of earth = $R_e=6.4\times {10}^6\text{m}$

Universal gravitational constant = $G=6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}{\text{/kg}}^{\text{2}}$

Speed of light = $c=3\times {10}^8\text{m/s}$.

Formula used:

Schwarzschild radius is defined as,

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

Calculation:

Substitute all given values in the above equation,

$\begin{array}{l}{R}_s=\frac{(2)(6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}{\text{/kg}}^{\text{2}})(5.972\times {10}^{24}\text{kg})}{{(3\times {10}^8\text{m/s})}^2}\\ R_s=8.852\times {10}^{-3}\text{m}\\ R_s=0.008852\text{m}\end{array}$

Now, density is defined as

Density = $\rho =\frac{M}{v}$

Since, earth is in spherical so, volume of earth would be $\frac{4}{3}\pi R^3.$

So, density is defined as

$\begin{array}{l}\rho =\frac{M}{\frac{4}{3}\pi R^3}\\ \text{Since }M\text{ is constant so, }\\ \rho \propto \frac{1}{R^3}\\ \text{thus, }\frac{{\rho }_s}{{\rho }_e}={\left(\frac{R_e}{R_s}\right)}^3\end{array}$

Substituting the given values,

$\begin{array}{c}\frac{{\rho }_s}{5514{\text{kg/m}}^3}={\left(\frac{6.4\times {10}^6\text{m}}{8.852\times {10}^{-3}\text{m}}\right)}^3\\ {\rho }_s=2.1\times {10}^{30}{\text{kg/m}}^{\text{3}}\end{array}$

This is current density which is extremely very high to its initial value.

Conclusion:

Thus, Schwarzschild radius is $0.008852\text{m}$ and current density is $2.1\times {10}^{30}{\text{kg/m}}^{\text{3}}$ which is extremely very high.