Chapter 12.2, Problem 12.85P

A 500-kg spacecraft first is placed into a circular orbit about the earth at an altitude of 4500 km and then is transferred to a circular orbit about the moon. Knowing that the mass of the moon is 0.01230 times the mass of the earth and that the radius of the moon is 1737 km, determine (a) the gravitational force exerted on the spacecraft as it was orbiting the earth, (b) the required radius of the orbit of the spacecraft about the moon if the periodic times (see Prob. 12.83) of the two orbits are to be equal, (c) the acceleration of gravity at the surface of the moon.

To determine

Find the gravitational force exerted on the spacecraft as it was orbiting the earth.

Answer to Problem 12.85P

The gravitational force exerted on the spacecraft as it was orbiting the earth is $\underset{\_}{1,684\text{N}}$.

Explanation of Solution

Given information:

The mass $\left(m\right)$ of the spacecraft is 500 kg.

The altitude $\left(h_E\right)$ of circular orbit of spacecraft about the surface of the earth is 4,500 km.

The mass of the moon is 0.01230 times the mass of the earth $\left(M_M=0.01230M_E\right)$.

The radius $\left(R_M\right)$ of the moon is 1,737 km.

Calculation:

Write the general equation of weight (W).

$W=mg$

The radius $\left(R_E\right)$ of the earth is $6.37\times {10}^6\text{m}$.

Find the altitude $\left(r_E\right)$ of circular orbit of spacecraft about the surface of the earth using the Equation:

$r_E=R_E+h_E$

Substitute $6.37\times {10}^6\text{m}$ for $R_E$ and 4,500 km for $h_E$.

$\begin{array}{c}{r}_E=6.37\times {10}^6+\left(4,500\text{km}\times \frac{1,000\text{m}}{1\text{km}}\right)\\ =10.87\times {10}^6\text{m}\end{array}$

Consider the Newton’s law of universal gravitation:

$\begin{array}{c}F=G\frac{Mm}{r^2}\\ =GM\left(\frac{m}{r^2}\right)\end{array}$ (1)

Write the equation for mass of planet:

$GM=g{R}^2$ (2)

Substitute Equation (2) in Equation (1).

$\begin{array}{c}F=\left(g{R}^2\right)\left(\frac{m}{r^2}\right)\\ =\left(mg\right){\left(\frac{R_E}{r_E}\right)}^2\end{array}$

Substitute 500 kg for m, $9.81{\text{m/s}}^2$ for g, $6.37\times {10}^6\text{m}$ for $R_E$, $10.87\times {10}^6\text{m}$ for $r_E$.

$\begin{array}{c}F=\left(500\right)\left(9.81\right){\left(\frac{6.37\times {10}^6}{10.87\times {10}^6}\right)}^2\\ =1,684\text{N}\end{array}$

Thus, the gravitational force exerted on the spacecraft as it was orbiting the earth is $\underset{\_}{1,684\text{N}}$.

To determine

Find the required radius of the orbit of the spacecraft about the moon if the periodic times of the two orbits are to be equal.

Answer to Problem 12.85P

The required radius of the orbit of the spacecraft about the moon if the periodic times of the two orbits are to be equal is $\underset{\_}{2,510\text{km}}$.

Explanation of Solution

Calculation:

Write the equation of acceleration $\left(a_n\right)$ of satellites.

$a_n=\frac{v^2}{r}$ (3)

Write the Equation of attraction force between earth and satellites.

$F=m{a}_n$ (4)

Here, m is the mass of satellite.

Find the mass of earth (M):

Consider the Newton’s law of universal gravitation:

$F=G\frac{Mm}{r^2}$ (5)

Here, F is the attraction force, G is the universal constant, M is the mass of earth, and m is the mass of satellite.

Substitute Equation (3) in Equation (4).

$\begin{array}{l}F=G\frac{Mm}{r^2}\\ m{a}_n=G\frac{Mm}{r^2}\\ a_n=G\frac{M}{r^2}\end{array}$ (6)

Substitute Equation (1) in Equation (4).

$\begin{array}{c}\frac{v^2}{r}=G\frac{M}{r^2}\\ M=\frac{1}{G}r\left(v^2\right)\end{array}$ (7)

Write the equation of velocity of earth:

$v=\frac{2\pi r}{\Gamma }$ (8)

Substitute Equation (8) in Equation (7).

$\begin{array}{l}M=\frac{1}{G}r{\left(\frac{2\pi r}{\Gamma }\right)}^2\\ M=\frac{1}{G}\frac{4{\pi }^2{r}^3}{{\Gamma }^2}\\ {\Gamma }^2=\frac{4{\pi }^2{r}^3}{GM}\\ \Gamma =\frac{2\pi r_E^{\frac{3}{2}}}{\sqrt{G{M}_E}}\end{array}$

Find the radius of moon $\left(r_M\right)$.

The periodic time of two orbit is equal.

$\begin{array}{l}{\Gamma }_E={\Gamma }_M\\ \frac{2\pi r_E^{\frac{3}{2}}}{\sqrt{G{M}_E}}=\frac{2\pi r_M^{\frac{3}{2}}}{\sqrt{G{M}_M}}\\ r_M^{\frac{3}{2}}=\sqrt{\frac{M_M}{M_E}}\left(r_E^{\frac{3}{2}}\right)\\ r_M={\left(\frac{M_M}{M_E}\right)}^{\frac{1}{3}}\left(r_E\right)\end{array}$

Substitute $0.01230M_E$ for $M_M$.

$\begin{array}{l}{r}_M={\left(\frac{0.01230M_E}{M_E}\right)}^{\frac{1}{3}}\left(r_E\right)\\ r_M=0.23084r_E\end{array}$

Substitute $10.87\times {10}^6\text{m}$ for $r_E$.

$\begin{array}{c}{r}_M=0.23084\left(10.87\times {10}^6\right)\\ =2.509\times {10}^6\text{m}\times \frac{1\text{km}}{1,000\text{m}}\\ =2,509\text{km}\\ \approx \text{2,510}\text{km}\end{array}$

Thus, the required radius of the orbit of the spacecraft about the moon if the periodic times of the two orbits are to be equal is $\underset{\_}{2,510\text{km}}$.

To determine

Find the acceleration of gravity at the surface of the moon.

Answer to Problem 12.85P

The acceleration of gravity at the surface of the moon is $\underset{\_}{1.62{\text{m/s}}^2}$.

Explanation of Solution

Calculation:

The radius of the moon $\left(R_M\right)$ is $1.737\times {10}^6\text{m}$.

Find the acceleration of gravity at the surface of the moon  $\left(g_M\right)$ using Equation (9).

$\frac{2\pi r_E^{\frac{3}{2}}}{\sqrt{G{M}_E}}=\frac{2\pi r_M^{\frac{3}{2}}}{\sqrt{G{M}_M}}$

Substitute $g_E{R}_E^2$ for $G{M}_E$ and $g_M{R}_M^2$ for $G{M}_M$.

$\begin{array}{l}\frac{2\pi r_E^{\frac{3}{2}}}{\sqrt{g_E{R}_E^2}}=\frac{2\pi r_M^{\frac{3}{2}}}{\sqrt{g_M{R}_M^2}}\\ \frac{r_E^{\frac{3}{2}}}{R_E\sqrt{g_E}}=\frac{r_M^{\frac{3}{2}}}{R_M\sqrt{g_M}}\\ \frac{r_E^3}{R_E^2{g}_E}=\frac{r_M^3}{R_M^2{g}_M}\\ g_M={\left(\frac{R_E}{R_M}\right)}^2{\left(\frac{r_M}{r_E}\right)}^3{g}_E\end{array}$

Substitute $6.37\times {10}^6\text{m}$ for $R_E$, $1.737\times {10}^6\text{m}$ for $R_M$, $2.509\times {10}^6\text{m}$ for $r_M$, $10.87\times {10}^6\text{m}$ for $r_E$, and $9.81{\text{m/s}}^2$ for $g_E$.

$\begin{array}{c}{g}_M={\left(\frac{6.37\times {10}^6}{1.737\times {10}^6}\right)}^2{\left(\frac{2.509\times {10}^6}{10.87\times {10}^6}\right)}^3\left(9.81\right)\\ =1.62{\text{m/s}}^2\end{array}$

Thus, the acceleration of gravity at the surface of the moon is $\underset{\_}{1.62{\text{m/s}}^2}$.