Chapter 14, Problem 41A

A spaceship in circular orbit about the moon is $2.0\times {10}^6\text{m}$ from its center.

a. Show that the period of the spaceship is 2.2 h.

b. Show that the speed of the spaceship relative to the moon is about 5800 km/h.

To determine

To show:The period of the spaceship is $2.2\text{ h}$ .

Explanation of Solution

Given:

The radial distance is $R=2.0\times {10}^6\text{ m}$ .

Formula used.

The expression for time for orbiting Earth is,

  $T=2\pi \sqrt{\frac{R^3}{GM}}$

Here, $G$ is the universal gravitational constant, $M$ is the mass, and $R$ is the radial distance.

Calculation:

Consider the mass of moon as $M=0.073\times {10}^{24}\text{ kg}$ and the universal gravitational constant as $G=6.67\times {10}^{-11}{\text{m}}^3/\text{kg}\cdot {\text{s}}^{\text{2}}$ .

The period of the spaceship is,

  $\begin{array}{l}T=2\pi \sqrt{\frac{R^3}{GM}}\\ T=2\pi \sqrt{\frac{{\left(2.0\times {10}^6\text{ m}\right)}^3}{\left(6.67\times {10}^{-11}{\text{m}}^3/\text{kg}\cdot {\text{s}}^{\text{2}}\right)\left(0.073\times {10}^{24}\text{ kg}\right)}}\\ T=8054\text{ s}\times \frac{1\text{ h}}{3600\text{ s}}\\ T=2.2\text{ h}\end{array}$

Conclusion:

Thus, the period of the spaceship is $2.2\text{ h}$ .

To determine

To show:The speed of spaceship relative to the moon is about $5800\text{km}/\text{h}$ .

Answer to Problem 41A

The speed of the spaceship relative to the moon is about $5800\text{km}/\text{h}$ .

Explanation of Solution

Given:

The radial distance is $R=2.0\times {10}^6\text{ m}$ .

Formula used:

The expression for orbital speed is,

  $v=\sqrt{\frac{GM}{R}}$

Here, $G$ is the universal gravitational constant, $M$ is the mass of the earth, and $R$ is the radial distance.

Calculation:

Consider the mass of moon as $M=0.07346\times {10}^{24}\text{ kg}$ and the universal gravitational constant as $G=6.67\times {10}^{-11}{\text{m}}^3/\text{kg}\cdot {\text{s}}^{\text{2}}$ .

The speed of the spaceship is,

  $\begin{array}{l}v=\sqrt{\frac{GM}{R}}\\ v=\sqrt{\frac{\left(6.67\times {10}^{-11}{\text{m}}^3/\text{kg}\cdot {\text{s}}^{\text{2}}\right)\left(0.07346\times {10}^{24}\text{ kg}\right)}{\left(2.0\times {10}^6\text{ m}\right)}}\\ v=1565.212\text{m}/\text{s}\times \frac{1\text{ km}}{1000\text{ m}}\times \frac{3600\text{ s}}{1\text{ h}}\\ v=5634.7632\text{km}/\text{h}\\ v\approx 5800\text{km}/\text{h}\end{array}$

Conclusion:

Thus, the speed of the spaceship relative to the moon is approximately equal to $v=5800\text{km}/\text{h}$ .