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

Answer

To solve the problem, we will use the principle of conservation of angular momentum. The angular momentum of the comet remains constant as it moves in its elliptical orbit around the Sun.

### Step-by-Step Solution:

1. **Identify the Variables**:
- Minimum distance from the Sun (perihelion), \( r_1 = 1.6 \times 10^{12} \, \text{m} \)
- Maximum distance from the Sun (aphelion), \( r_2 = 8 \times 10^{12} \, \text{m} \)
- Velocity at perihelion, \( v_1 = 60 \, \text{m/s} \)
- Velocity at aphelion, \( v_2 \) (unknown)

2. **Apply Conservation of Angular Momentum**:
The angular momentum \( L \) at perihelion and aphelion can be expressed as:
\[ L = m \cdot r_1 \cdot v_1 = m \cdot r_2 \cdot v_2 \]
Since the mass \( m \) of the comet is the same at both points, we can cancel it out:
\[ r_1 \cdot v_1 = r_2 \cdot v_2 \]

3. **Rearranging the Equation**:
We can rearrange the equation to solve for \( v_2 \):
\[ v_2 = \frac{r_1 \cdot v_1}{r_2} \]

4. **Substituting the Known Values**:
Now, substitute the known values into the equation:
\[ v_2 = \frac{(1.6 \times 10^{12} \, \text{m}) \cdot (60 \, \text{m/s})}{8 \times 10^{12} \, \text{m}} \]

5. **Calculating**:
- First, calculate the numerator:
\[ 1.6 \times 10^{12} \cdot 60 = 96 \times 10^{12} \, \text{m} \cdot \text{m/s} \]
- Now, divide by the denominator:
\[ v_2 = \frac{96 \times 10^{12}}{8 \times 10^{12}} = \frac{96}{8} = 12 \, \text{m/s} \]

6. **Final Result**:
The velocity of the comet when it is farthest from the Sun is:
\[ v_2 = 12 \, \text{m/s} \]