A satellite of mass m orbits around the Earth of mas M in an elliptical orbit of semi - major and semi - minor axes 2a and a respectively. The angular momentum of the satellite about the centre of the Earth is

Options

A. `pimsqrt((GMa)/(4))`
B. `pim sqrt((GMa)/(4))`
C. `msqrt((GM a)/(8))`
D. `msqrt((GM a)/(2))`

Answer

To find the angular momentum of a satellite of mass \( m \) orbiting the Earth of mass \( M \) in an elliptical orbit with semi-major axis \( 2a \) and semi-minor axis \( a \), we can follow these steps:

### Step 1: Calculate the Area of the Ellipse
The area \( A \) of an ellipse is given by the formula:
\[ A = \pi \times \text{semi-major axis} \times \text{semi-minor axis} \]
Substituting the values:
\[ A = \pi \times (2a) \times a = 2\pi a^2 \]

### Step 2: Determine the Time Period of the Satellite
According to Kepler's third law, the time period \( T \) of a satellite in an elliptical orbit is given by:
\[ T = 2\pi \sqrt{\frac{a^3}{GM}} \]
where \( a \) is the semi-major axis. Here, the semi-major axis is \( 2a \), so we substitute:
\[ T = 2\pi \sqrt{\frac{(2a)^3}{GM}} = 2\pi \sqrt{\frac{8a^3}{GM}} = 2\pi \cdot 2\sqrt{\frac{a^3}{GM}} = 4\pi \sqrt{\frac{a^3}{GM}} \]

### Step 3: Use Kepler's Second Law
Kepler's second law states that the rate at which area is swept out by the line connecting the satellite and the Earth is constant:
\[ \frac{dA}{dt} = \frac{L}{2m} \]
Where \( L \) is the angular momentum. We can rearrange this to find \( L \):
\[ L = 2m \frac{dA}{dt} \]

### Step 4: Calculate \( \frac{dA}{dt} \)
The average rate of area swept out in one complete revolution (time period \( T \)) is:
\[ \frac{dA}{dt} = \frac{A}{T} = \frac{2\pi a^2}{4\pi \sqrt{\frac{a^3}{GM}}} = \frac{2a^2}{4\sqrt{\frac{a^3}{GM}}} = \frac{a^2}{2\sqrt{\frac{a^3}{GM}}} \]

### Step 5: Substitute \( \frac{dA}{dt} \) into the Angular Momentum Equation
Now substituting \( \frac{dA}{dt} \) into the equation for \( L \):
\[ L = 2m \cdot \frac{a^2}{2\sqrt{\frac{a^3}{GM}}} = m \cdot \frac{a^2}{\sqrt{\frac{a^3}{GM}}} \]
This simplifies to:
\[ L = m \cdot a^2 \cdot \sqrt{\frac{GM}{a^3}} = m \cdot \sqrt{GMa^{-1}} \]

### Step 6: Final Expression for Angular Momentum
Thus, the angular momentum \( L \) can be expressed as:
\[ L = m \cdot \sqrt{\frac{GM}{2}} \cdot a \]

### Conclusion
The final expression for the angular momentum of the satellite about the center of the Earth is:
\[ L = m \cdot \sqrt{GMa} \cdot \frac{1}{2} \]