A point charge `–q` revolves around a fixed charge `+Q` in elliptical orbit. The minimum ar.d maximum distance of q from Q are `r_1` and `r_2`, respectively. The mass of revolving particle is `Q gt q` and assume no gravitational effects. Find the velocity of q at positions when it is at `r_1` and `r_2` distance from Q.

Explanation

To find the velocities of the point charge `-q` at the minimum distance `r1` and maximum distance `r2` from the fixed charge `+Q`, we can use the principles of conservation of angular momentum and conservation of mechanical energy.

### Step-by-Step Solution:

1. **Define Angular Momentum Conservation:**
Since there are no external torques acting on the system, the angular momentum of the charge `-q` about the charge `+Q` is conserved.
\[ L = m v_1 r_1 = m v_2 r_2 \]
Here, \( v_1 \) is the velocity at distance \( r_1 \) and \( v_2 \) is the velocity at distance \( r_2 \).

2. **Rearranging Angular Momentum Equation:**
From the angular momentum equation, we can express \( v_1 \) in terms of \( v_2 \):
\[ v_1 = \frac{v_2 r_2}{r_1} \]

3. **Define Energy Conservation:**
The total mechanical energy of the system is conserved. The mechanical energy at distance \( r_1 \) is equal to the mechanical energy at distance \( r_2 \):
\[ \frac{1}{2} m v_1^2 - \frac{k Q q}{r_1} = \frac{1}{2} m v_2^2 - \frac{k Q q}{r_2} \]

4. **Substituting for \( v_1 \):**
Substitute \( v_1 \) from step 2 into the energy conservation equation:
\[ \frac{1}{2} m \left(\frac{v_2 r_2}{r_1}\right)^2 - \frac{k Q q}{r_1} = \frac{1}{2} m v_2^2 - \frac{k Q q}{r_2} \]

5. **Simplifying the Energy Equation:**
Expanding the left-hand side:
\[ \frac{1}{2} m \frac{v_2^2 r_2^2}{r_1^2} - \frac{k Q q}{r_1} = \frac{1}{2} m v_2^2 - \frac{k Q q}{r_2} \]
Rearranging gives:
\[ \frac{1}{2} m \frac{v_2^2 r_2^2}{r_1^2} - \frac{1}{2} m v_2^2 = -\frac{k Q q}{r_2} + \frac{k Q q}{r_1} \]

6. **Factoring Out \( v_2^2 \):**
Factoring out \( v_2^2 \) from the left side:
\[ \frac{1}{2} m v_2^2 \left(\frac{r_2^2}{r_1^2} - 1\right) = \frac{k Q q}{r_1} - \frac{k Q q}{r_2} \]

7. **Solving for \( v_2^2 \):**
Rearranging gives:
\[ v_2^2 = \frac{2 k Q q \left(\frac{1}{r_1} - \frac{1}{r_2}\right)}{m \left(\frac{r_2^2}{r_1^2} - 1\right)} \]

8. **Finding \( v_1 \):**
Now substituting \( v_2 \) back to find \( v_1 \):
\[ v_1 = \frac{v_2 r_2}{r_1} \]

### Final Expressions:
- The velocity at minimum distance \( r_1 \):
\[ v_1 = \frac{r_2}{r_1} \sqrt{\frac{2 k Q q \left(\frac{1}{r_1} - \frac{1}{r_2}\right)}{m \left(\frac{r_2^2}{r_1^2} - 1\right)}} \]

- The velocity at maximum distance \( r_2 \):
\[ v_2 = \sqrt{\frac{2 k Q q \left(\frac{1}{r_1} - \frac{1}{r_2}\right)}{m \left(\frac{r_2^2}{r_1^2} - 1\right)}} \]