A planet of mass m moves along an ellipse around the Sun so that its maximum and minimum distances from the Sun are equal to `r_1` and `r_2` respectively. Find the angular momentum M of this planet relative to the centre of the Sun.

Explanation

To find the angular momentum \( M \) of a planet of mass \( m \) moving in an elliptical orbit around the Sun, where the maximum and minimum distances from the Sun are \( r_1 \) and \( r_2 \) respectively, we can follow these steps:

### Step 1: Understand Angular Momentum Conservation
The angular momentum \( L \) of the planet about the Sun is conserved. At the maximum distance \( r_1 \) and minimum distance \( r_2 \), we can express the angular momentum as:
\[ L = m v_1 r_1 = m v_2 r_2 \]
where \( v_1 \) and \( v_2 \) are the speeds of the planet at distances \( r_1 \) and \( r_2 \) respectively.

### Step 2: Relate Speeds to Angular Momentum
From the conservation of angular momentum, we can express the speeds in terms of \( L \):
\[ v_1 = \frac{L}{m r_1} \quad \text{and} \quad v_2 = \frac{L}{m r_2} \]

### Step 3: Use Energy Conservation
The mechanical energy of the planet is also conserved. The total mechanical energy \( E \) at the two positions can be expressed as:
\[ E = \text{Kinetic Energy} + \text{Potential Energy} \]
At position 1:
\[ E = \frac{1}{2} m v_1^2 - \frac{GMm}{r_1} \]
At position 2:
\[ E = \frac{1}{2} m v_2^2 - \frac{GMm}{r_2} \]

### Step 4: Set Up the Energy Equations
Setting the energies equal gives:
\[ \frac{1}{2} m v_1^2 - \frac{GMm}{r_1} = \frac{1}{2} m v_2^2 - \frac{GMm}{r_2} \]

### Step 5: Substitute for \( v_1 \) and \( v_2 \)
Substituting \( v_1 \) and \( v_2 \) from Step 2 into the energy equation:
\[ \frac{1}{2} m \left(\frac{L}{m r_1}\right)^2 - \frac{GMm}{r_1} = \frac{1}{2} m \left(\frac{L}{m r_2}\right)^2 - \frac{GMm}{r_2} \]

### Step 6: Simplify the Equation
Cancelling \( m \) and rearranging gives:
\[ \frac{L^2}{2 r_1^2} - \frac{GM}{r_1} = \frac{L^2}{2 r_2^2} - \frac{GM}{r_2} \]

### Step 7: Rearranging and Solving for \( L^2 \)
Rearranging this equation leads to:
\[ L^2 \left(\frac{1}{2 r_1^2} - \frac{1}{2 r_2^2}\right) = GM \left(\frac{1}{r_1} - \frac{1}{r_2}\right) \]
This can be simplified to:
\[ L^2 = 2GM \left( \frac{r_1 r_2 (r_2 - r_1)}{r_1^2 r_2^2} \right) \]

### Step 8: Final Expression for Angular Momentum
Taking the square root gives:
\[ L = \sqrt{2GM \cdot \frac{r_1 r_2 (r_2 - r_1)}{r_1^2 r_2^2}} \]

### Final Answer
Thus, the angular momentum \( M \) of the planet relative to the center of the Sun is:
\[ M = \sqrt{2GM \cdot \frac{r_1 r_2 (r_2 - r_1)}{r_1^2 r_2^2}} \]