Question

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. 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.

Accepted Answer

Step 1: Identify the key forces and energies involved in the problem. The boulder must overcome the gravitational pull of both the moon and the earth. This is a three-body system, so we need to consider the gravitational potential energy due to both the moon and the earth. Step 2: Write the total energy equation for the boulder. The total energy consists of the kinetic energy and the gravitational potential energy. The kinetic energy is given by $$ KE = \frac{1}{2}mv^2 $$, and the gravitational potential energy is $$ U = -\frac{GMm}{r} $$, where $$ G  is $$the gravitational constant, $$ M  is $$the mass of the celestial body, $$ m  is $$the mass of the boulder, and $$ r  is $$the distance from the center of the celestial body. Step 3: Set up the energy conservation equation. At the launch point on the moon, the boulder has an initial kinetic energy $$ KE_{initial} $$ and gravitational potential energy due to both the moon and the earth. At the earth's surface, the boulder will have zero velocity (minimum speed condition) and gravitational potential energy due to the earth. Write the equation: $$ KE_{initial} + U_{moon} + U_{earth} = U_{earth,final} . $$Step 4: Substitute the expressions for gravitational potential energy. For the moon, $$ U_{moon} = -\frac{G M_{moon} m}{R_{moon}} $$, where $$ M_{moon}  is $$the mass of the moon and $$ R_{moon}  is $$the radius of the moon. For the earth, $$ U_{earth} = -\frac{G M_{earth} m}{d} $$, where $$ M_{earth}  is $$the mass of the earth and $$ d  is $$the distance between the moon and the earth. At the earth's surface, $$ U_{earth,final} = -\frac{G M_{earth} m}{R_{earth}} $$, where $$ R_{earth}  is $$the radius of the earth. Step 5: Solve for the minimum initial velocity $$ v . $$Rearrange the energy conservation equation to isolate $$ v $$: $$ \frac{1}{2}mv^2 = \left( -\frac{G M_{earth} m}{R_{earth}} \right) - \left( -\frac{G M_{moon} m}{R_{moon}} - \frac{G M_{earth} m}{d} \right) . $$Simplify the terms and solve for $$ v . $$Note that the mass of the boulder $$ m $$ cancels out, leaving an expression for $$ v  in $$terms of $$ G $$, $$ M_{moon} $$, $$ M_{earth} $$, $$ R_{moon} $$, $$ R_{earth} $$, and $$ d .$$

Verified video answer for a similar problem:

Key Concepts

Gravitational Potential Energy

Kinematics of Projectile Motion

Three-Body Problem