Question

A 55,000 kg space capsule is in a 28,000-km-diameter circular orbit around the moon. A brief but intense firing of its engine in the forward direction suddenly decreases its speed by 50%. This
causes the space capsule to go into an elliptical orbit. What are the space capsule’s (a) maximum and (b) minimum distances from the center of the moon in its new orbit?
Hint: You will need to use two conservation laws.

Short answer

a) Maximum distance from center is 14000 km

b) Minimum distance from center is 2000 km

Step-by-step solution

Part(a) Step1: Given information

Mass of the capsule, m_Capsule=55,000 kg
Diameter of the orbit, D=28000km=28000 x 10³m
Decreases in speed, is 50%

Part(a) Step2: Explanation

Given that the velocity decreases to half.

By conservation of angular momentum, at that instance, the capsule is at maximum distance from the moon.

So if we can say in the new orbit, the maximum distance from the center of the moon is equal to the radius of the orbit

So,

$$\begin{gathered} x_{\max }=\frac{D}{2} \\ =\frac{28000km}{2} \\ =14000\mathrm{km} \end{gathered}$$

Max distance is 14000 km.

Part(b) Step1: Given information

Mass of the capsule, m_Capsule=55,000 kg
Diameter of the orbit, D=28000km=28000 x 10³m
Decreases in speed, is 50%

Part(b)Step2: Explanation

The velocity of the satellite in orbit can be calculated by
$v=\sqrt{\frac{GM}{r}}$

Where

M = mass of the moon,

G = gravitational constant and

r = orbital radius.

Substitute values and calculate velocity as

$$\begin{gathered} v=\sqrt{\frac{GM}{r}} \\ =\sqrt{\frac{(6.67\times {10}^{-11}N{m}^2/k{g}^2)\times (7.34\times {10}^{22}kg)}{\left(\frac{2800\times {10}^3}{2}m\right)}} \\ =591.35\mathrm{m}/\mathrm{s} \end{gathered}$$

The velocity of the satellite at apogee

$$\begin{gathered} v_1=\frac{v}{2}=0.5\times 591.35m/s \\ =295.7\mathrm{m}/\mathrm{s} \end{gathered}$$

Now use momentum conservation

$$\begin{gathered} x_{\max }{v}_1=x_{\min }{v}_2 \\ {v}_2=\frac{x_{\max }{v}_1}{x_{\min }} \end{gathered}$$

Now use energy conservation

$$\begin{gathered} \frac{1}{2}m{v}_1^2-\frac{GmM}{x_{\max }}=\frac{1}{2}m{v}_2^2-\frac{GmM}{x_{\min }} \\ \frac{1}{2}{v}_1^2-\frac{GM}{x_{\max }}=\frac{1}{2}{v}_2^2-\frac{GM}{x_{\min }} \\ \frac{1}{2}\times \frac{1}{4}\frac{GM}{x_{\max }}-\frac{GM}{x_{\max }}=\frac{1}{2}\times \frac{1}{4}\frac{GM}{x_{\max }}\frac{x_{\max }^2}{x_{\min }^2}-\frac{GM}{x_{\min }} \\ \frac{1}{8x_{\max }}-\frac{1}{x_{\max }}=\frac{x_{\max }}{8x_{\min }^2}-\frac{1}{x_{\min }} \\ {x}_{\min }=\frac{x_{\max }}{7} \\ {x}_{\min }=\frac{14000\mathrm{km}}{7}=2000km \end{gathered}$$

Minimum distance is 2000 km.