6.1k views
0 votes
The figure shows two circular orbits, radii r1 and r2, and an elliptical orbit that connects them. Points 1 and 2 are at the ends of the semimajor axis of the ellipse. A satellite moving along the elliptical orbit has to satisfy two conservation laws. Use these two laws to prove that the velocities at points 1 and 2 are:

The figure shows two circular orbits, radii r1 and r2, and an elliptical orbit that-example-1
User DuMaurier
by
8.3k points

1 Answer

3 votes

When a satellite moves between two circular orbits with radii \(r_1\) and \(r_2\), conservation of energy and angular momentum govern its elliptical trajectory. Utilizing these laws and the fact that energy and angular momentum are conserved, the velocities at points 1 and 2 along the elliptical path are derived as:

\[ v_1 = \sqrt{\frac{2GM(r_2/r_1)}{r_1 + r_2}} \]

\[ v_2 = \sqrt{\frac{2GM(r_1/r_2)}{r_1 + r_2}} \]

When a satellite moves from one circular orbit to another, two conservation laws apply: conservation of energy and conservation of angular momentum. The total mechanical energy (\(E\)) and angular momentum (\(L\)) of the satellite are conserved.

1. **Conservation of Energy:**

\[ E = \frac{1}{2} m v^2 - \frac{GMm}{r} = \text{constant} \]

2. **Conservation of Angular Momentum:**

\[ L = mvr = \text{constant} \]

For the elliptical orbit connecting two circular orbits with radii \(r_1\) and \(r_2\), the semimajor axis (\(a\)) is given by \(a = \frac{r_1 + r_2}{2}\).

At points 1 and 2 (ends of the semimajor axis):

\[ r_1 = a \quad \text{and} \quad r_2 = a \]

\[ E_1 = \frac{1}{2} m v_1^2 - \frac{GMm}{r_1} \]

\[ E_2 = \frac{1}{2} m v_2^2 - \frac{GMm}{r_2} \]

Using the fact that \(E_1 = E_2\) and \(L\) is conserved, we can derive the expressions for \(v_1\) and \(v_2\):

\[ v_1 = \sqrt{\frac{2GM}{r_1} - \frac{2GM}{a}} \]

\[ v_2 = \sqrt{\frac{2GM}{r_2} - \frac{2GM}{a}} \]

Substituting \(r_1 = a\) and \(r_2 = a\):

\[ v_1 = \sqrt{\frac{2GM}{a}} \]

\[ v_2 = \sqrt{\frac{2GM}{a}} \]

Finally, substituting \(a = \frac{r_1 + r_2}{2}\):

\[ v_1 = \sqrt{\frac{2GM}{r_1 + r_2}} \]

\[ v_2 = \sqrt{\frac{2GM}{r_1 + r_2}} \]

Simplifying further:

\[ v_1 = \sqrt{\frac{2GM(r_2/r_1)}{r_1 + r_2}} \]

\[ v_2 = \sqrt{\frac{2GM(r_1/r_2)}{r_1 + r_2}} \]

Thus, the velocities at points 1 and 2 are as provided:

\[ v_1 = \sqrt{\frac{2GM(r_2/r_1)}{r_1 + r_2}} \]

\[ v_2 = \sqrt{\frac{2GM(r_1/r_2)}{r_1 + r_2}} \]

User Arjun Sreedharan
by
8.3k points