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

Step 1: Orbital radius

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

Step 2: Orbital velocity using gravitational and centripetal force balance

For circular motion, gravitational force provides the centripetal force: \[ \frac{GMm}{r^2} = \frac{mv^2}{r} \] Simplifying, we get: \[ v = \sqrt{\frac{GM}{r}} \]

Step 3: Angular momentum of the satellite

The angular momentum \( L \) of the satellite is: \[ L = mvr \]

Step 4: Substituting given values

Satellite mass: \[ m = \frac{M}{2} \] Orbital radius: \[ r = \frac{4R}{3} \] Orbital speed: \[ v = \sqrt{\frac{GM}{\frac{4R}{3}}} = \sqrt{\frac{3GM}{4R}} \]

Step 5: Substitute into the formula for angular momentum

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

Step 6: Simplifying the square root

\[ \sqrt{\frac{3GM}{4R}} = \sqrt{\frac{3}{4}} \sqrt{\frac{GM}{R}} = \frac{\sqrt{3}}{2} \sqrt{\frac{GM}{R}} \] Therefore, \[ L = \frac{2MR}{3} \times \frac{\sqrt{3}}{2} \sqrt{\frac{GM}{R}} \] \[ L = \frac{M\sqrt{3}}{3} \sqrt{GMR} \] \[ L = \frac{M}{\sqrt{3}} \sqrt{GMR} \]

Step 7: Alternative given form of angular momentum

It is given that: \[ L = M \sqrt{\frac{GMR}{x}} \]

Step 8: Equating both expressions

\[ \frac{M}{\sqrt{3}} \sqrt{GMR} = M \sqrt{\frac{GMR}{x}} \] Cancel \( M \) and \( \sqrt{GMR} \) (since both are non-zero): \[ \frac{1}{\sqrt{3}} = \frac{1}{\sqrt{x}} \]

Step 9: Squaring both sides

\[ \frac{1}{3} = \frac{1}{x} \Rightarrow x = 3 \]

Final Answer:

\[ \boxed{x = 3} \]

Concept :

• Gravitational force acts as the centripetal force for orbital motion.
• Orbital velocity for a circular orbit: \( v = \sqrt{\frac{GM}{r}} \).
• Angular momentum of a satellite: \( L = mvr \).
• Simplify carefully when combining constants under square roots.

Answer 2

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 \( x = 3 \).