To solve the problem, we need to find the change in kinetic energy of the comet when it moves from its closest point to the sun to its farthest point. We will use the principle of conservation of angular momentum and the formula for kinetic energy.
### Step-by-Step Solution:
1. **Identify Given Values:**
- Mass of the comet, \( m = 10^8 \, \text{kg} \)
- Distance when closest to the sun, \( r_a = 2.5 \times 10^{11} \, \text{m} \)
- Speed when closest to the sun, \( v = 2 \times 10^4 \, \text{m/s} \)
- Distance when farthest from the sun, \( r_b = 5 \times 10^{10} \, \text{m} \)
2. **Use Conservation of Angular Momentum:**
- Angular momentum at point A (closest to the sun):
\[
L_a = m \cdot v \cdot r_a
\]
- Angular momentum at point B (farthest from the sun):
\[
L_b = m \cdot v_0 \cdot r_b
\]
- Since angular momentum is conserved, \( L_a = L_b \):
\[
m \cdot v \cdot r_a = m \cdot v_0 \cdot r_b
\]
- We can cancel \( m \) from both sides:
\[
v \cdot r_a = v_0 \cdot r_b
\]
- Rearranging for \( v_0 \):
\[
v_0 = v \cdot \frac{r_a}{r_b}
\]
3. **Substitute Values:**
- Substitute \( v = 2 \times 10^4 \, \text{m/s} \), \( r_a = 2.5 \times 10^{11} \, \text{m} \), and \( r_b = 5 \times 10^{10} \, \text{m} \):
\[
v_0 = 2 \times 10^4 \cdot \frac{2.5 \times 10^{11}}{5 \times 10^{10}}
\]
- Simplifying:
\[
v_0 = 2 \times 10^4 \cdot 5 = 10 \times 10^4 = 1 \times 10^5 \, \text{m/s}
\]
4. **Calculate Kinetic Energies:**
- Kinetic energy at point A:
\[
KE_a = \frac{1}{2} m v^2 = \frac{1}{2} \cdot 10^8 \cdot (2 \times 10^4)^2
\]
\[
KE_a = \frac{1}{2} \cdot 10^8 \cdot 4 \times 10^8 = 2 \times 10^{16} \, \text{J}
\]
- Kinetic energy at point B:
\[
KE_b = \frac{1}{2} m v_0^2 = \frac{1}{2} \cdot 10^8 \cdot (1 \times 10^5)^2
\]
\[
KE_b = \frac{1}{2} \cdot 10^8 \cdot 1 \times 10^{10} = 0.5 \times 10^{18} \, \text{J} = 5 \times 10^{17} \, \text{J}
\]
5. **Calculate Change in Kinetic Energy:**
- Change in kinetic energy:
\[
\Delta KE = KE_b - KE_a = 5 \times 10^{17} - 2 \times 10^{16}
\]
\[
\Delta KE = 5 \times 10^{17} - 0.2 \times 10^{17} = 4.8 \times 10^{17} \, \text{J}
\]
### Final Answer:
The change in kinetic energy when the comet is farthest from the sun is \( 4.8 \times 10^{17} \, \text{J} \).
