A rogue band of colonists on the moon declares war and prepares to use a catapult to launch large boulders at the earth. Assume that the boulders are launched from the point on the moon nearest the earth. For this problem you can ignore the rotation of

the two bodies and the orbiting of the moon.

a. What is the minimum speed with which a boulder must be launched to reach the earth?

Hint: The minimum speed is not the escape speed. You need to analyze a three-body system.

b. Ignoring air resistance, what is the impact speed on earth of a boulder launched at this minimum speed?

Gravitational constant $G=6.67\times {10}^{-11}{m}^3/kg{s}^2$

mass of the earth is role="math" localid="1649961110336" $m_e=5.98\times {10}^{24}kg$

mass of the moon is role="math" localid="1649961080261" $m_m=7.36\times {10}^{22}kg$

mass of the boulder is $m_b$

distance between earth and moon is role="math" localid="1649961173396" $r_{em}=3.84\times {10}^8$

distance reached by the boulder from the center of the moon is $r$

The minimum speed with which a boulder must be launched to reach the earth.

Expression for System's potential energy

$U_G=-G\left(\frac{m_e{m}_b}{r_{em}-r}+\frac{m_m{m}_b}{r}+\frac{m_e{m}_m}{r_{em}}\right)$

After finding the first derivative of $U_{G,}\left(\frac{\partial U_G}{\partial r}\right)=0$

Equation to find$r$becomes

$$\begin{gathered} \frac{r_{em}}{r}=1\pm \sqrt{\frac{m_e}{m_m}} \\ \Rightarrow \frac{3.84\times {10}^8}{r}=1\pm \sqrt{\frac{5.98\times {10}^{24}}{7.36\times {10}^{22}}} \\ r=3.836\times {10}^{7}m \end{gathered}$$Note: take the positive sign.

Equating the sum of initial and final potential and kinetic energies of the system

$U_i+K_i=U_f+K_f$

Since final velocity is zero, $K_f=0$

$\therefore K_i=-U_i+U_f$

Launch velocity$v$ is given by

role="math" localid="1649960963111" $$\begin{gathered} v=\sqrt{2G{m}_e\left(\frac{1}{r_{em}-R_m}-\frac{1}{r_{em}-r}\right)+2G{m}_m\left(\frac{1}{R_m}-\frac{1}{r}\right)} \\ v=\sqrt{2(6.67\times {10}^{-11})(5.98\times {10}^{24})\left(\frac{1}{3.84\times {10}^8-1.74\times {10}^6}-\frac{1}{3.84\times {10}^8-3.836\times {10}^7}\right)+2(6.67\times {10}^{-11})(7.36\times {10}^{22})\left(\frac{1}{1.74\times {10}^6}-\frac{1}{3.836\times {10}^7}\right)} \\ v=\sqrt{-2.21\times {10}^5+53.87\times {10}^5} \\ v=2.3\times {10}^{3}m/s=2.3km/s \end{gathered}$$

Impact speed on earth of a boulder launched at minimum speed $v_f$

$$\begin{gathered} v_f=\sqrt{2G\left[m_e\left(\frac{1}{R_e}-\frac{1}{r_{em}-r}\right)+m_m\left(\frac{1}{r_{em}-R_e}-\frac{1}{r}\right)\right]} \\ {v}_f=\sqrt{2\times 6.67\times {10}^{-11}\left[5.98\times {10}^{24}\left(\frac{1}{1.74\times {10}^6}-\frac{1}{3.84\times {10}^8-3.836\times {10}^7}\right)+7.36\times {10}^{22}\left(\frac{1}{3.84\times {10}^8-1.74\times {10}^6}-\frac{1}{3.836\times {10}^7}\right)\right]} \\ {v}_f=\sqrt{123.1\times {10}^6}=11\times {10}^{3}m/s=11km/s \end{gathered}$$

The impact speed on earth when boulder is launched at minimum speed$11km/s$