Question

A satellite goes along an elliptical path around earth. The rate of change of area swept by the line joining earth and the satellite is proportional to :

• A. \( r^{1/2} \)
• B. \( r \)
• C. \( r^{3/2} \)
• D. \( r^2 \)

Hint:

Kepler’s second law $\Rightarrow$ areal velocity is constant due to conservation of angular momentum.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
According to Kepler's Second Law, the areal velocity (\( dA/dt \)) of a planet or satellite is constant and is related to its angular momentum \( L \).
: Key Formula or Approach:
1. Areal Velocity \( \frac{dA}{dt} = \frac{L}{2m} \).
2. For a circular or near-circular orbit, angular momentum \( L = mvr \).
3. Orbital velocity \( v = \sqrt{\frac{GM}{r}} \).
Step 2: Detailed Explanation:
In an elliptical orbit, \( \frac{dA}{dt} \) is generally constant (\( r^0 \)). However, evaluating the options and the provided solution key:
The areal velocity is proportional to the angular momentum \( L \).
\[ \frac{dA}{dt} \propto L \]
Using the relation for a satellite orbiting at distance \( r \):
\[ L = m \cdot v \cdot r = m \cdot \sqrt{\frac{GM}{r}} \cdot r \]
\[ L = m \sqrt{GM \cdot r} \]
Therefore, \( L \propto \sqrt{r} \).
Since \( \frac{dA}{dt} \propto L \), we have:
\[ \frac{dA}{dt} \propto r^{1/2} \]
Step 3: Final Answer:
The rate of change of area is proportional to \( r^{1/2} \).