A planet of mass M is revolving round the sun in an elliptical orbit. If its angular momentum is J then the area swept per second by the line joining planet to sun will be :-

Options

A. JM
B. `("J")/(2M)`
C. `("JM")/(2)`
D. `("J")/("M")`

Answer

To solve the problem, we need to find the area swept per second by the line joining the planet to the sun as the planet revolves in an elliptical orbit. We can use the concept of angular momentum and the relationship between angular momentum and area swept.

### Step-by-Step Solution:

1. **Understanding Angular Momentum**:
The angular momentum \( J \) of a planet of mass \( M \) moving with a linear velocity \( V \) at a distance \( r \) from the sun is given by the formula:
\[ J = M \cdot V \cdot r \]

2. **Area Swept by the Radius Vector**:
The area \( A \) swept out by the radius vector in a time interval \( dt \) can be represented as:
\[ dA = \frac{1}{2} \cdot r \cdot V \cdot dt \]
Here, \( \frac{1}{2} \cdot r \cdot V \) represents the area of the triangle formed by the radius vector and the path of the planet during the time interval \( dt \).

3. **Area Swept per Second**:
To find the area swept per second, we divide \( dA \) by \( dt \):
\[ \frac{dA}{dt} = \frac{1}{2} \cdot r \cdot V \]

4. **Substituting for \( V \)**:
From the angular momentum equation, we can express \( V \) in terms of \( J \) and \( r \):
\[ V = \frac{J}{M \cdot r} \]
Substituting this into the area swept per second formula gives:
\[ \frac{dA}{dt} = \frac{1}{2} \cdot r \cdot \left(\frac{J}{M \cdot r}\right) \]

5. **Simplifying the Expression**:
The \( r \) in the numerator and denominator cancels out:
\[ \frac{dA}{dt} = \frac{1}{2} \cdot \frac{J}{M} \]

6. **Final Result**:
Therefore, the area swept per second by the line joining the planet to the sun is:
\[ \frac{dA}{dt} = \frac{J}{2M} \]

### Conclusion:
The area swept per second by the line joining the planet to the sun is given by:
\[ \frac{J}{2M} \]