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 determination

Given: The orbital radius \( r \) is defined as \( r = R + \frac{R}{3} \), which simplifies to \( r = \frac{4R}{3} \).

Step 2: Calculation of orbital velocity

For a satellite in circular orbit, the gravitational force provides the necessary centripetal force: \( \frac{GMm}{r^2} = \frac{mv^2}{r} \). Solving for velocity \( v \) yields: \( v = \sqrt{\frac{GM}{r}} \).

Step 3: Satellite's angular momentum formula

The angular momentum \( L \) of the satellite is calculated using the formula: \( L = mvr \).

Step 4: Inputting given parameters

The mass of the satellite is \( m = \frac{M}{2} \). The orbital radius is \( r = \frac{4R}{3} \). The orbital speed is \( v = \sqrt{\frac{GM}{\frac{4R}{3}}} = \sqrt{\frac{3GM}{4R}} \).

Step 5: Substitution into angular momentum equation

Substituting the values from Step 4 into the angular momentum formula: \( L = \frac{M}{2} \cdot \frac{4R}{3} \cdot \sqrt{\frac{3GM}{4R}} \). This simplifies to: \( L = \frac{2MR}{3} \sqrt{\frac{3GM}{4R}} \).

Step 6: Simplification of the square root term

The square root term can be simplified as: \( \sqrt{\frac{3GM}{4R}} = \sqrt{\frac{3}{4}} \sqrt{\frac{GM}{R}} = \frac{\sqrt{3}}{2} \sqrt{\frac{GM}{R}} \). Substituting this back into the expression for \( L \): \( L = \frac{2MR}{3} \times \frac{\sqrt{3}}{2} \sqrt{\frac{GM}{R}} \). Further simplification leads to: \( L = \frac{M\sqrt{3}}{3} \sqrt{GMR} \), and \( L = \frac{M}{\sqrt{3}} \sqrt{GMR} \).

Step 7: Given expression for angular momentum

An alternative form of the angular momentum is provided: \( L = M \sqrt{\frac{GMR}{x}} \).

Step 8: Equating the two expressions for angular momentum

By equating the derived expression for \( L \) with the given expression: \( \frac{M}{\sqrt{3}} \sqrt{GMR} = M \sqrt{\frac{GMR}{x}} \). Since \( M \) and \( \sqrt{GMR} \) are non-zero, they can be cancelled from both sides, leaving: \( \frac{1}{\sqrt{3}} = \frac{1}{\sqrt{x}} \).

Step 9: Solving for x

Squaring both sides of the equation \( \frac{1}{\sqrt{3}} = \frac{1}{\sqrt{x}} \) results in: \( \frac{1}{3} = \frac{1}{x} \), which implies \( x = 3 \).

Final Answer:

\[ \boxed{x = 3} \]

Key Concepts:

• The gravitational force serves as the centripetal force in orbital motion.
• The orbital velocity for a circular orbit is given by \( v = \sqrt{\frac{GM}{r}} \).
• The angular momentum of a satellite is defined as \( L = mvr \).
• Careful simplification is required when combining constants within square roots.