Chapter 14, Problem 17A

To determine

To find: The escape speed on the moon.

Answer to Problem 17A

The escape speed on the moon is $2.4\text{km}/\text{s}$ .

Explanation of Solution

Introduction:

Escape speed is the minimum inertial speed that will cause an object to move upward forever.

To find the escape speed, apply energy conservation.

  $U_i+K_i=U_f+K_f$

Where,

  $U_i$ is the initial potential energy.

  $U_f$ is the final potential energy.

  $K_i$ is the initial kinetic energy.

  $K_f$ is the final kinetic energy.

At infinity potential energy is zero. To escape, set both terms on the right to zero.

The potential energy of an object of mass m at the surface is

  $U=-\frac{GmM}{R}$

Substituting the values in energy equivalence equation.

  $\begin{array}{c}–\frac{GmM}{R}+K_{escape}=0\\ \frac{1}{2}m{v}_{escape}{}^2=\frac{GmM}{R}\\ v_{escape}=\sqrt{\frac{2GM}{R}}\end{array}$

Where,

G is the gravitational constant.

M is the mass of an object. (Moon in this case)

R is the radius of Moon.

Substituting the corresponding values,

  $\begin{array}{c}{v}_{escape}=\sqrt{\frac{2(6.67384\times {10}^{-11}\text{N}\cdot {\text{m}}^2\cdot {\text{kg}}^{-2})(7.34767\times {10}^{22})}{1737400}}\\ =2375.89\text{m}/\text{s}\\ =2.375\text{km}/\text{s}\\ \approx 2.4\text{km}/\text{s}\end{array}$

Conclusion:

Therefore, the escape speed is $2.4\text{km}/\text{s}$