Question

A satellite of mass \( \frac{M}{2} \) is revolving around Earth in a circular orbit at a height of \( \frac{R}{3} \) from the Earth's surface. The angular momentum of the satellite is \( M \sqrt{\frac{GM R}{x}} \). The value of \( x \) is:

Hint:

The angular momentum of a satellite depends on the mass, orbital velocity, and radius of the orbit. Use the relevant formulas to find the angular momentum in different scenarios.

Answer 1

Correct Answer: 3

The angular momentum \( L \) of a satellite in a circular orbit is given by:

\[ L = mvr, \]

Where:

• \( m \) is the mass of the satellite
• \( v \) is the orbital velocity
• \( r \) is the radius of the orbit (from the Earth's center)

Step 1: Orbital Velocity

The orbital velocity is given by:

\[ v = \sqrt{\frac{GM}{r}}, \]

Where \( G \) is the gravitational constant, and \( M \) is the Earth's mass.

Step 2: Orbit Radius

The satellite is at a height of \( \frac{R}{3} \) from the Earth's surface. So the total distance from the Earth's center is:

\[ r = R + \frac{R}{3} = \frac{4R}{3}. \]

Step 3: Substitute into Angular Momentum

Now substitute into the formula for \( L \):

\[ L = m \cdot v \cdot r = m \cdot \sqrt{\frac{GM}{r}} \cdot r. \]

Substitute \( r = \frac{4R}{3} \):

\[ L = m \cdot \sqrt{\frac{GM}{\frac{4R}{3}}} \cdot \frac{4R}{3} = m \cdot \sqrt{\frac{3GM}{4R}} \cdot \frac{4R}{3}. \]

Step 4: Use Given Satellite Mass

The satellite has mass \( m = \frac{M}{2} \), so we get:

\[ L = \frac{M}{2} \cdot \sqrt{\frac{3GM}{4R}} \cdot \frac{4R}{3}. \]

Step 5: Simplify

Simplify the constants:

\[ L = \frac{M}{2} \cdot \frac{4R}{3} \cdot \sqrt{\frac{3GM}{4R}}. \]

Combine terms under the square root and outside:

\[ L = M \cdot \sqrt{\frac{GM R}{3}}. \]

Step 6: Final Answer

The angular momentum is:

\[ L = M \cdot \sqrt{\frac{GMR}{3}}, \]

Therefore, the value of \( x = 3 \).