A comet revolves around the sun in an eliptical orbit. When it is closest to the sun at a distance d, its corresponding kinetic energy is `k_(0)`. If it is farthest from the sun at distance 3d then the corresponding kinetic energy will be

Options

A. `(k_(0))/(9)`
B. `(8k_(0))/(9)`
C. `(k_(0))/(4)`
D. `(4k_(0))/(9)`

Answer

To solve the problem, we need to analyze the motion of the comet in its elliptical orbit around the Sun. We will use the principles of conservation of angular momentum and kinetic energy.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- The comet is in an elliptical orbit around the Sun. The closest point to the Sun is at distance \(d\) (perihelion), and the farthest point is at distance \(3d\) (aphelion).
- At the closest point, the kinetic energy is given as \(K_0\).

2. **Using Kepler's Second Law**:
- Kepler's Second Law states that a line segment joining a planet (or comet) and the Sun sweeps out equal areas during equal intervals of time. This implies that the product of the distance from the Sun and the velocity of the comet remains constant.
- Mathematically, this can be expressed as:
\[ r_1 v_1 = r_2 v_2 \]
- Where \(r_1 = d\) (closest distance) and \(v_1\) is the velocity at that point, and \(r_2 = 3d\) (farthest distance) and \(v_2\) is the velocity at that point.

3. **Setting Up the Equation**:
- From Kepler's law:
\[ d \cdot v_1 = 3d \cdot v_2 \]
- Simplifying gives:
\[ v_1 = 3v_2 \]

4. **Relating Kinetic Energies**:
- The kinetic energy \(K\) is given by the formula:
\[ K = \frac{1}{2} mv^2 \]
- At the closest point, the kinetic energy is:
\[ K_0 = \frac{1}{2} m v_1^2 \]
- At the farthest point, the kinetic energy \(K_1\) is:
\[ K_1 = \frac{1}{2} m v_2^2 \]

5. **Substituting for \(v_2\)**:
- Since \(v_1 = 3v_2\), we can express \(v_2\) in terms of \(v_1\):
\[ v_2 = \frac{v_1}{3} \]
- Now substituting \(v_2\) into the expression for \(K_1\):
\[ K_1 = \frac{1}{2} m \left(\frac{v_1}{3}\right)^2 = \frac{1}{2} m \frac{v_1^2}{9} = \frac{1}{9} \left(\frac{1}{2} m v_1^2\right) = \frac{K_0}{9} \]

6. **Final Result**:
- Therefore, the kinetic energy when the comet is at the farthest distance \(3d\) is:
\[ K_1 = \frac{K_0}{9} \]

### Answer:
The corresponding kinetic energy when the comet is farthest from the Sun at a distance \(3d\) will be \(\frac{K_0}{9}\).