Answer to Question #151640 in Mechanics | Relativity for Ina

Question

1) You are the head engineer for a 3,000 kg rocket that is nearing the moon at a speed of 20,000 kilometers per hour (5,556 m/s). It needs to quickly slow to 10,000 kilometers per hour (2,778 m/s) to begin a controlled descent to the surface. If the rocket descends too fast it will crash and too slowly it will run out of fuel. The rocket has a new braking system consisting of several 6.00 kg metal spheres on the front of the rocket that can be fired straight ahead. Each has an explosive charge that fires it at a speed of 120,000 meters per second with respect to the rocket. You need to send the rocket’s computer system a message telling it how many spheres to fire. How many spheres would need to be fired for a safe landing?

Answer 1

Explanations & Calculations

• The firing of the balls retard the speed of the rocket in successive steps.
• Since each ball is fired with respect to the moving rocket, speed relative to the given fixed frame (earth) should be calculated.
• Consider algebraically, $\small V_0=5556ms^{-1}\,\,\,\,v=120000ms^{-1}$ and the rocket's speed after a ball is fired $\small =V'$ Similarly $\small 3000kg=M,\,\,\,6kg=m$
• The speed of a ball relative to the earth is

$\qquad\qquad \begin{aligned} \small V_{b,E}&= \small V+V_{R,E}\\ &= \small \overrightarrow{v }+\overrightarrow{V}\\ &=\small \overrightarrow{V+v} \end{aligned}$

• As the firing of each ball is conserved in the linear momentum, this could be used to develop a relationship between the speeds mentioned above.

$\qquad\qquad \begin{aligned} \small MV&=\small (M-m)V'+m(V+v)\\ \small V'&=\small V-\frac{m}{(M-m)}\times v \end{aligned}$

• Considering the successive firings, (applying conservation of linear momentum for each firing) this could be extended as follows

$\qquad\qquad \begin{aligned} \small V_1'&=\small V_0-\frac{m}{(M-m)}\times v\\ \small V_2'&=\small V_1-\frac{m}{(M-m-m)}\times v\\ \small V_3'&=\small V_2-\frac{m}{(M-m-m-m)}\times v\\ \cdots\\ \cdots\\ \small V_n'&=\small V_{n-1}-\frac{m}{(M-mn)}\times v \end{aligned}$

• Adding together yeilds,

$\qquad\qquad \begin{aligned} \small V_n'&=\small V_0-mv \Sigma_1^n\frac{1}{(M-nm)} \end{aligned}$

• Substituting values

$\qquad\qquad \begin{aligned} \small 2778&=\small 5556-(6\times120000)\times \Sigma_1^n\frac{1}{(3000-6n)} \end{aligned}$

• This gives $\small \bold{n=11}$
• It is necessary to fire 11 balls.