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 find the maximum speed of a planet revolving around the Sun, given its minimum speed \( v_0 \) at a distance \( r \) (perihelion) and maximum distance \( R \) (aphelion), we can use the principle of conservation of angular momentum.

### Step-by-Step Solution:

1. **Understanding Angular Momentum**:
The angular momentum \( L \) of a planet moving in an elliptical orbit is conserved. It can be expressed as:
\[ L = m \cdot v \cdot r \]
where \( m \) is the mass of the planet, \( v \) is its speed, and \( r \) is the distance from the Sun.

2. **Setting Up the Equation**:
At perihelion (minimum distance \( r \)), the speed is \( v_{\text{min}} = v_0 \).
At aphelion (maximum distance \( R \)), the speed is \( v_{\text{max}} \).
Therefore, we can write:
\[ L_{\text{perihelion}} = L_{\text{aphelion}} \]
This gives us:
\[ m \cdot v_0 \cdot r = m \cdot v_{\text{max}} \cdot R \]

3. **Canceling the Mass**:
Since the mass \( m \) is the same on both sides, we can cancel it out:
\[ v_0 \cdot r = v_{\text{max}} \cdot R \]

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

5. **Substituting the Given Values**:
Since \( v_0 \) is given as the minimum speed, we can directly substitute:
\[ v_{\text{max}} = v_0 \cdot \frac{r}{R} \]

### Final Answer:
Thus, the maximum speed of the planet is:
\[ v_{\text{max}} = v_0 \cdot \frac{r}{R} \]