The minimum and maximum distances of a planet revolving around sun are r and R . If the minimum speed of planet of its trajectory is `v_(0)` , its maximum speed will be

Options

A. `(v_(0)R)/r`
B. `(v_(0)r)/R`
C. `(v_(0)R^(2))/(r^(2))`
D. `(v_(0)r^(2))/(R^(2))`

Answer

To solve the problem, we need to apply the principle of conservation of angular momentum, as the planet is revolving around the sun in an elliptical orbit. Here are the steps to find the maximum speed of the planet:

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a planet revolving around the sun with minimum distance \( r \) (perigee) and maximum distance \( R \) (apogee).
- The minimum speed at the distance \( r \) is given as \( v_0 \).
- We need to find the maximum speed \( v_{max} \) at the distance \( R \).

2. **Using Conservation of Angular Momentum**:
- The angular momentum \( L \) of the planet at any point in its orbit is given by:
\[ L = m \cdot v \cdot r \]
- At the minimum distance \( r \) (where speed is \( v_0 \)):
\[ L_{min} = m \cdot v_0 \cdot r \]
- At the maximum distance \( R \) (where speed is \( v_{max} \)):
\[ L_{max} = m \cdot v_{max} \cdot R \]

3. **Setting Angular Momenta Equal**:
- Since angular momentum is conserved, we can set the two expressions equal:
\[ m \cdot v_0 \cdot r = m \cdot v_{max} \cdot R \]
- The mass \( m \) cancels out from both sides:
\[ v_0 \cdot r = v_{max} \cdot R \]

4. **Solving for Maximum Speed**:
- Rearranging the equation to solve for \( v_{max} \):
\[ v_{max} = v_0 \cdot \frac{r}{R} \]

5. **Final Expression**:
- Thus, the maximum speed \( v_{max} \) of the planet at its maximum distance \( R \) is:
\[ v_{max} = v_0 \cdot \frac{R}{r} \]

### Conclusion:
The maximum speed of the planet when it is at the maximum distance \( R \) is given by:
\[ v_{max} = v_0 \cdot \frac{R}{r} \]