The maximum and minimum distance of a comet form the sun are `8xx10^(12)m and 1.6xx10^(12)m`. If its velocity when nearest to the sun is `60m//s`, what will be its velocity in m/s when it is farthest

Options

A. `12`
B. `60`
C. `112`
D. `6`

Explanation

To solve the problem, we can use the principle of conservation of angular momentum. The angular momentum of a comet about the sun remains constant as it moves in its elliptical orbit.

### Step-by-Step Solution:

1. **Identify the distances and velocities:**
- Minimum distance (perihelion) from the sun, \( r_1 = 1.6 \times 10^{12} \, \text{m} \)
- Maximum distance (aphelion) from the sun, \( r_2 = 8.0 \times 10^{12} \, \text{m} \)
- Velocity at perihelion, \( v_1 = 60 \, \text{m/s} \)

2. **Use the conservation of angular momentum:**
The angular momentum \( L \) at perihelion and aphelion can be expressed as:
\[ L = m \cdot v_1 \cdot r_1 = m \cdot v_2 \cdot r_2 \]
where \( m \) is the mass of the comet, \( v_2 \) is the velocity at aphelion, and \( r_2 \) is the distance at aphelion.

3. **Since the mass \( m \) is constant, we can cancel it out:**
\[ v_1 \cdot r_1 = v_2 \cdot r_2 \]

4. **Rearranging the equation to find \( v_2 \):**
\[ v_2 = \frac{v_1 \cdot r_1}{r_2} \]

5. **Substituting the known values:**
\[ v_2 = \frac{60 \, \text{m/s} \cdot (1.6 \times 10^{12} \, \text{m})}{8.0 \times 10^{12} \, \text{m}} \]

6. **Calculating \( v_2 \):**
\[ v_2 = \frac{60 \cdot 1.6}{8.0} \, \text{m/s} \]
\[ v_2 = \frac{96}{8} \, \text{m/s} = 12 \, \text{m/s} \]

### Final Answer:
The velocity of the comet when it is farthest from the sun is \( 12 \, \text{m/s} \).