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 \( L \) 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: Understand the Geometry of the Elliptical Orbit
The semi-major axis \( a \) is the longest radius of the ellipse, while the semi-minor axis \( b \) is the shortest radius. In this case, the semi-major axis is given as \( 2a \) and the semi-minor axis as \( a \).

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

### Step 3: Use Kepler's Second Law
Kepler's Second Law states that the area swept out by the line connecting the satellite to the center of the Earth in a given time is constant. This 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 4: Determine the Time Period of the Satellite
The time period \( T \) of a satellite in orbit is given by:
\[ T = 2\pi \sqrt{\frac{R^3}{GM}} \]
where \( R \) is the semi-major axis. In this case, \( R = 2a \), so:
\[ T = 2\pi \sqrt{\frac{(2a)^3}{GM}} = 2\pi \sqrt{\frac{8a^3}{GM}} = 4\pi \sqrt{\frac{2a^3}{GM}} \]

### Step 5: Relate Area and Time Period
From Kepler's Second Law, we can express the area swept in one complete revolution as:
\[ \frac{dA}{dt} = \frac{2\pi a^2}{T} \]
Substituting \( T \):
\[ \frac{dA}{dt} = \frac{2\pi a^2}{4\pi \sqrt{\frac{2a^3}{GM}}} = \frac{a^2}{2\sqrt{\frac{2a^3}{GM}}} = \frac{a^2 \sqrt{GM}}{2\sqrt{2} a^{3/2}} = \frac{\sqrt{GM} a^{1/2}}{2\sqrt{2}} \]

### Step 6: Solve for Angular Momentum \( L \)
Now substituting \( \frac{dA}{dt} \) back into the equation from Kepler's Second Law:
\[ \frac{L}{2m} = \frac{\sqrt{GM} a^{1/2}}{2\sqrt{2}} \]
Thus,
\[ L = 2m \cdot \frac{\sqrt{GM} a^{1/2}}{2\sqrt{2}} = m \cdot \frac{\sqrt{GM} a^{1/2}}{\sqrt{2}} \]

### Final Result
The angular momentum \( L \) of the satellite about the center of the Earth is:
\[ L = m \sqrt{\frac{GM}{2}} a^{1/2} \]