Chapter 20, Problem 8P

An Apollo command module orbited the Moon about 100 km above the surface. What was its orbital velocity? What was its orbital period? (Hint: Use the formula for circular velocity, Eq. 5-1a.) (Note: Relevant information can be found in Celestial Profile: The Moon.)

To determine

The orbital period of the Apollo command module.

Answer to Problem 8P

The orbital period of the Apollo command module is $1\text{h}\text{58}\text{min}$.

Explanation of Solution

The expression for the orbital period of the Apollo command module is given by,

    $T=\frac{2\pi \left(R+h\right)}{v_0}$        (I)

Here, $R$ is the radius of the planet, $h$ is the height of the planet, $v_0$ is the orbital velocity of the object and $T$ is the orbital period of the Apollo command module.

The expression for the orbital velocity of the object is,

    $v_0=\sqrt{g_p\left(R+h\right)}$        (II)

Here, $g_p$ is the gravitational acceleration at an altitude from the surface of a planet.

The expression for the gravitational acceleration at an altitude from the surface of a planet is,

    $g_p=\frac{GM}{{\left(R+h\right)}^2}$

Here, $G$ is the universal gravitational constant and $M$ is the mass of the planet.

Substitute $\frac{GM}{{\left(R+h\right)}^2}$ for $g_p$ in equation (II) to find $v_0$.

    $\begin{array}{c}{v}_0=\sqrt{\frac{GM}{{\left(R+h\right)}^2}\left(R+h\right)}\\ =\sqrt{\frac{GM}{\left(R+h\right)}}\end{array}$        (III)

Conclusion:

Substitute $6.67\times {10}^{-11}{\text{m}}^{\text{3}}/\text{kg}{\text{.s}}^{\text{2}}$ for $G$, $7.36\times {10}^{22}\text{kg}$ for $M$, $1737.4\text{km}$ for $R$ and $100\text{km}$ for $h$  in equation (III) to find $v_0$.

    $\begin{array}{c}{v}_0=\sqrt{\frac{\left(6.67\times {10}^{-11}{\text{m}}^{\text{3}}/\text{kg}{\text{.s}}^{\text{2}}\right)\left(7.36\times {10}^{22}\text{kg}\right)}{\left(1737.4\text{km}+100\text{km}\right)}}\\ =\sqrt{\frac{\left(6.67\times {10}^{-11}{\text{m}}^{\text{3}}/\text{kg}{\text{.s}}^{\text{2}}\right)\left(7.36\times {10}^{22}\text{kg}\right)}{\left(1837.4\text{km}\right)\left(\frac{1000\text{m}}{1\text{km}}\right)}}\\ =\sqrt{\frac{\left(6.67\times {10}^{-11}{\text{m}}^{\text{3}}/\text{kg}{\text{.s}}^{\text{2}}\right)\left(7.36\times {10}^{22}\text{kg}\right)}{\left(1837.4\times {10}^3\text{m}\right)}}\\ =1.63\times {10}^3\text{m}/\text{s}\end{array}$

Substitute $1.63\times {10}^3\text{m}/\text{s}$ for $v_0$, $1737.4\text{km}$ for $R$ and $100\text{km}$ for $h$  in equation (I) to find $T$.

    $\begin{array}{c}T=\frac{2\pi \left(1737.4\text{km}+100\text{km}\right)}{1.63\times {10}^3\text{m}/\text{s}}\\ =\frac{2\pi \left(1837.4\text{km}\left(\frac{1000\text{km}}{1\text{m}}\right)\right)}{1.63\times {10}^3\text{m}/\text{s}}\\ =\frac{11544724.68\text{m}}{1.63\times {10}^3\text{m}/\text{s}}\\ =7080\text{s}\end{array}$

Solve further as,

    $\begin{array}{c}T=7080\text{s}\left(\frac{1\text{hr}}{3600\text{s}}\right)\\ =1\text{h}\text{58}\text{min}\end{array}$

Therefore, the orbital period of the Apollo command module is $1\text{h}\text{58}\text{min}$.