Question

A satellite follows the elliptical orbit shown in FIGURE P12.77. The only force on the satellite is the gravitational attraction of the planet. The satellite’s speed at point a is 8000 m/s.
a. Does the satellite experience any torque about the center of the planet? Explain.
b. What is the satellite’s speed at point b?
c. What is the satellite’s speed at point c?

Short answer

a) Satellite will not experience any torque at the center of the planet.

b) Speed at point b is 2000 m/sec

c) Speed at point c is 4000m/sec

Step-by-step solution

Part(a) Step1: Given information

A satellite is orbiting the planet in elliptical orbit.

It is experiences the gravitational force of attraction.

The speed of the satellite at point a, is 8000 m/sec.

Part(a) Step2: Explanation

The torque is calculated as

torque = Force x perpendicular distance

As gravitation force is parallel to position of satellite, so it will not generate any Torque.

Part(b) Step 1: Given information

A satellite is orbiting the planet in elliptical orbit.

It is experiences the gravitational force of attraction.

The speed of the satellite at point a, is 8000 m/sec.

Part(b) Step 2: Explanation

From the given diagram we can find the r_a and r_b

$r_a=(30,000km)/2-9000km=6000km$

and

$r_b=(30,000km)/2+9000km=24,000km$

From the law of conservation of momentum, we get

L_a = L_b

L_a and L_b are angular momentum

$m{v}_b{r}_b=m{v}_a{r}_a.......................\left(1\right)$

Substitute values in equation (1) and solve for v_b

_{$v_b=\frac{(8000m/s)\times \left(6000km\right)}{24,000km}=2000m/s$}

Part(c) Step 1: Given information

A satellite is orbiting the planet in elliptical orbit.

It is experiences the gravitational force of attraction.

The speed of the satellite at point a, is 8000 m/sec.

Part(c) Step2: Explanation

First find the value of r_c from the figure

use the trigonometric ratio

$$\begin{gathered} r_c=d_c\sin \theta \\ =\sqrt{(12000{)}^2+(9000{)}^2}\times \left(\frac{12000}{15000}\right) \\ =12000\mathrm{km} \end{gathered}$$

From the momentum conservation on point b and c

$m{v}_b{r}_b=m{v}_c{r}_c$

Substitute values in equation (2) and solve

$$\begin{gathered} v_c=\frac{2000{\mathrm{ms}}^{-1}\times 24000\mathrm{km}}{12000\mathrm{km}} \\ =4000\mathrm{m}/\sec \end{gathered}$$