Chapter 7, Problem 93A

a.

To determine

The period of Apollo 11 in minutes if it orbits the Moon at a height of $111\text{km}$ .

Answer to Problem 93A

  $119.16\min$

Explanation of Solution

Given:

The Moon’s radius is $1738\text{}\text{km}$ .

The mass of the Moon is $7.3\times {10}^{22}\text{}\text{kg}$ .

Formula used:

Period of a Planet orbiting the Moon can be calculated as follows:

  $T=2\pi \sqrt{\frac{r^3}{G{m}_M}}$

Where, $G$ is the gravitational constant, $r$ is the orbital radius, and $m_M$ is the mass of the Moon.

Calculation:

It is known that the value of gravitational constant is,

  $G=6.67\times {10}^{-11}\text{N}{\text{.m}}^{\text{2}}{\text{/kg}}^{\text{2}}$

Now, the orbital radius of Apollo 11 will be,

  $\begin{array}{c}r=R_M+h\\ =\left(1738+111\right)\text{km}\\ \text{=1849}\text{km}\end{array}$

Then using the given values of mass of the Moon and the orbital radius of the Apollo 11, the time period of Apollo 11 will be,

  $\begin{array}{c}T=2\pi \sqrt{\frac{{\left(1.849\times {10}^6\right)}^3}{\left(6.67\times {10}^{-11}\right)\left(7.3\times {10}^{22}\right)}}\\ =2\pi \left(1.139\times {10}^3\right)\\ =7.15\times {10}^3\text{s }\times \frac{1\min }{60\text{s}}\\ =119.16\min \end{array}$

Conclusion:

Thus, the time period of Apollo 11 is $119.16\min$ .

b.

To determine

The velocity with which Apollo 11 orbits the Moon.

Answer to Problem 93A

  $1.62\times {10}^3\text{m/s}$

Explanation of Solution

Given:

The Moon’s radius is $1738\text{}\text{km}$ .

The mass of the Moon is $7.3\times {10}^{22}\text{}\text{kg}$ .

Formula used:

The speed of an object in circular orbit is

  $v=\sqrt{\frac{G{m}_M}{r}}$

Where, $G$ is the gravitational constant, $r$ is the orbital radius, and $m_M$ is the mass of the Moon.

Calculation:

The value of gravitational constant is,

  $G=6.67\times {10}^{-11}\text{N}{\text{.m}}^{\text{2}}{\text{/kg}}^{\text{2}}$

Using the given values in above formula, the velocity of Apollo 11 will be

  $\begin{array}{c}v=\sqrt{\frac{G{m}_M}{r}}\\ =\sqrt{\frac{\left(6.67\times {10}^{-11}\right)\left(7.3\times {10}^{22}\right)}{\left(1.849\times {10}^6\right)}}\\ =1.62\times {10}^3\text{m/s}\end{array}$

Conclusion:

Thus, the orbital velocity of Apollo 11 is $1.62\times {10}^3\text{m/s}$ .