A planet moves around the sun. at a given point `P`, it is closest from the sun at a distance `d_(1)`, and has a speed `V_(1)`. At another point `Q`, when it is farthest from the sun at a distance `d_(2)`, its speed will be

Options

A. `(d_(1)^(2)V_(1))/(d_(2))`
B. `(d_(2)V_(1))/(d_(1))`
C. `(d_(1)V_(1))/(d_(2))`
D. `(d_(2)^(2)V_(1))/(d_(1)^(2))`

Explanation

To solve the problem, we can use the principle of conservation of angular momentum. The angular momentum of a planet moving in an elliptical orbit around the sun is conserved.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- At point P (closest to the sun), the distance from the sun is \( d_1 \) and the speed is \( V_1 \).
- At point Q (farthest from the sun), the distance from the sun is \( d_2 \) and we need to find the speed \( V_2 \).

2. **Conservation of Angular Momentum**:
- The angular momentum \( L \) of the planet at any point in its orbit is given by:
\[ L = m \cdot v \cdot r \]
- Here, \( m \) is the mass of the planet, \( v \) is its speed, and \( r \) is the distance from the sun.

3. **Setting Up the Equation**:
- At point P:
\[ L_P = m \cdot V_1 \cdot d_1 \]
- At point Q:
\[ L_Q = m \cdot V_2 \cdot d_2 \]
- Since angular momentum is conserved, we have:
\[ L_P = L_Q \]
- Therefore:
\[ m \cdot V_1 \cdot d_1 = m \cdot V_2 \cdot d_2 \]

4. **Cancelling the Mass**:
- Since the mass \( m \) is the same for both points, we can cancel it out:
\[ V_1 \cdot d_1 = V_2 \cdot d_2 \]

5. **Solving for \( V_2 \)**:
- Rearranging the equation gives:
\[ V_2 = \frac{V_1 \cdot d_1}{d_2} \]

### Final Answer:
The speed of the planet at point Q (farthest from the sun) is given by:
\[ V_2 = \frac{V_1 \cdot d_1}{d_2} \]