Chapter 12.2, Problem 12.85P

To determine

(a)

The gravitational force exerted on the spacecraft at given condition.

Answer to Problem 12.85P

The required value of the gravitational force is $1684N.$

Explanation of Solution

Given information:

$\begin{array}{l}{m}_s=500kg\\ h=4500km\\ r_m=1737km\\ m_m=0.01230m_e\end{array}$

Formula used:

$F=\frac{GMm}{r^2}$

Calculation:

We know that, $R_E=6.37\times {10}^{6}m$

$\Rightarrow r_E=R_E+h_E$;

$\Rightarrow \left(6.37\times {10}^6+4.5\times {10}^6\right)m$

$\Rightarrow 10.87\times {10}^{6}m$

And we also know that, $F=\frac{GMm}{r^2}$

And, $GM=g{R}^2$

Now, $F=g{R}^2\frac{m}{r^2}=W{\left(\frac{R}{r}\right)}^2$

Put the values in the above equation

$\Rightarrow F=\left(500kg\right)\left(9.81m/s^2\right)\left(\frac{6.37\times {10}^{6}m}{10.87\times {10}^{6}m}\right)$

$\Rightarrow F=1684N$

To determine

(b)

The radius of the orbit of the spacecraft about the moon at given condition.

Answer to Problem 12.85P

The required value of the radius is $2510km.$

Explanation of Solution

Given information:

$\begin{array}{l}{m}_s=500kg\\ h=4500km\\ r_m=1737km\\ m_m=0.01230m_e\end{array}$

Formula used:

$M=\frac{1}{G}{\left(\frac{2\pi }{\tau }\right)}^2{r}^3$

Calculation:

We have the relation, $M=\frac{1}{G}{\left(\frac{2\pi }{\tau }\right)}^2{r}^3$ from the problem 12.78

And, $\tau =\frac{2\pi r^{3/2}}{\sqrt{GM}}$

$\Rightarrow {\tau }_E={\tau }_M$;

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

(i)

Simplifies the above equation to find $r_M$

$\Rightarrow r_M={\left(\frac{M_M}{M_E}\right)}^{1/3}{r}_E={\left(0.01230\right)}^{1/3}\left(10.87\times {10}^{6}m\right)$

$\Rightarrow r_M=2.509\times {10}^{6}m$

$\Rightarrow r_M=2510km$

To determine

(c)

The acceleration of gravity at the surface of moon at given condition.

Answer to Problem 12.85P

The required value of the acceleration is $1.62m/s^2.$

Explanation of Solution

Given information:

$\begin{array}{l}{m}_s=500kg\\ h=4500km\\ r_m=1737km\\ m_m=0.01230m_e\end{array}$

Formula used:

$GM=g{R}^2$

Calculation:

We know that, $GM=g{R}^2$

Substitute into the equation (i)

$\Rightarrow \frac{2\pi r_E{}^{3/2}}{R_E\sqrt{g_E}}=\frac{2\pi r_M{}^{3/2}}{R_M\sqrt{g_M}}$

$\Rightarrow g_M={\left(\frac{R_E}{R_M}\right)}^2{\left(\frac{r_M}{r_E}\right)}^2{g}_E={\left(\frac{R_E}{R_M}\right)}^2{\left(\frac{M_M}{M_E}\right)}^2{g}_E$

$\Rightarrow g_M={\left(\frac{6370km}{1737km}\right)}^2\left(0.01230\right)\left(9.81m/s^2\right)$

$\Rightarrow g_{moon}=1.62m/s^2$