A satellite of mass m orbits the earth in an elliptical orbit having aphelion distance `r_(a)` and perihelion distance `r_(p)`. The period of the orbit is T. The semi-major and semi-minor axes of the ellipse are `(r_(a) + r_(p))/(2)` and `sqrt(r_(p)r_(a))` respectively. The angular momentum of the satellite is: (A) `(2m pi (r_(B)+r_(p)) sqrt(r_(a) r_(p)))/T` (B) `(m pi (r_(a)+r_(p)) sqrt(r_(a) r_(p)))/(2T)` (C) `(m pi (r_(a)+r_(p)) sqrt(r_(a)r_(p)))/(4T)` (D) `(mpi (r_(a)+r_(p)) sqrt(r_(a)r_(p)))/T`

Options

A. `(2m pi (r_(B)+r_(p)) sqrt(r_(a) r_(p)))/T`
B. `(m pi (r_(a)+r_(p)) sqrt(r_(a) r_(p)))/(2T)`
C. `(m pi (r_(a)+r_(p)) sqrt(r_(a)r_(p)))/(4T)`
D. `(mpi (r_(a)+r_(p)) sqrt(r_(a)r_(p)))/T`

Answer

To find the angular momentum of a satellite orbiting the Earth in an elliptical orbit, we can follow these steps:

### Step 1: Understand the parameters of the elliptical orbit
The satellite has:
- Aphelion distance \( r_a \)
- Perihelion distance \( r_p \)
- Period of the orbit \( T \)
- Semi-major axis \( a = \frac{r_a + r_p}{2} \)
- Semi-minor axis \( b = \sqrt{r_a r_p} \)

### Step 2: Use Kepler's Second Law
Kepler's Second Law states that the area swept out by the satellite in a given time is constant. The relationship can be expressed as:
\[ \frac{dA}{dt} = \frac{L}{2m} \]
where \( L \) is the angular momentum and \( m \) is the mass of the satellite.

### Step 3: Calculate the area of the ellipse
The total area \( A \) of the elliptical orbit is given by:
\[ A = \pi a b \]
Substituting the values of \( a \) and \( b \):
\[ A = \pi \left(\frac{r_a + r_p}{2}\right) \left(\sqrt{r_a r_p}\right) \]

### Step 4: Relate area to angular momentum
The total area covered in time \( T \) is:
\[ A = \frac{L}{2m} \cdot T \]
Setting the two expressions for area equal gives:
\[ \frac{L}{2m} \cdot T = \pi \left(\frac{r_a + r_p}{2}\right) \left(\sqrt{r_a r_p}\right) \]

### Step 5: Solve for angular momentum \( L \)
Rearranging the equation to solve for \( L \):
\[ L = \frac{2m \pi \left(\frac{r_a + r_p}{2}\right) \left(\sqrt{r_a r_p}\right)}{T} \]
This simplifies to:
\[ L = \frac{m \pi (r_a + r_p) \sqrt{r_a r_p}}{T} \]

### Conclusion
Thus, the angular momentum of the satellite is:
\[ L = \frac{m \pi (r_a + r_p) \sqrt{r_a r_p}}{T} \]
This corresponds to option (D).