Question

What is the expression for the time period of a satellite orbiting Earth at a height \( h \) above the Earth's surface?

• A. \( T = 2\pi \sqrt{\frac{R^3}{GM}} \)
• B. \( T = 2\pi \sqrt{\frac{(R + h)^3}{GM}} \)
• C. \( T = \frac{1}{2\pi} \sqrt{\frac{GM}{(R + h)^3}} \)
• D. \( T = 2\pi \sqrt{\frac{(R^2 + h^2)}{GM}} \)

Hint:

When calculating the time period of a satellite, remember to include both the Earth's radius \( R \) and the satellite's height \( h \) above the surface. Use \( T = 2\pi \sqrt{\frac{(R + h)^3}{GM}} \) for accurate calculations.

Answer 1

The Correct Option is B

The time period of a satellite orbiting Earth at a height \( h \) above the Earth's surface is derived from Kepler's Third Law and the balance between the gravitational force acting on the satellite and the centripetal force required for its orbital motion.
Step 1: Gravitational Force Provides the Centripetal Force. The gravitational force \( F_{\text{gravity}} \) between the satellite and Earth is given by Newton's law of gravitation: \[ F_{\text{gravity}} = \frac{GMm}{(R + h)^2}, \] where: \( G \) is the gravitational constant, \( M \) is the mass of the Earth, \( m \) is the mass of the satellite, \( R + h \) is the distance from the center of Earth to the satellite (the radius of Earth plus the height of the satellite above the surface). This gravitational force provides the necessary centripetal force for the satellite's circular motion: \[ F_{\text{centripetal}} = \frac{mv^2}{R + h}, \] where \( v \) is the orbital velocity of the satellite.
Step 2: Equating Gravitational and Centripetal Forces. For the satellite to maintain its orbit, the gravitational force must be equal to the centripetal force. Thus, we can equate the two forces: \[ \frac{GMm}{(R + h)^2} = \frac{mv^2}{R + h}. \] We can cancel the mass \( m \) of the satellite from both sides, and simplify the equation: \[ v^2 = \frac{GM}{R + h}. \]
Step 3: Relating Orbital Velocity to Time Period. The orbital velocity \( v \) is related to the time period \( T \) of the satellite by the formula: \[ v = \frac{2\pi(R + h)}{T}. \] Now, substitute the expression for \( v^2 \) from Step 2 into this equation: \[ \left(\frac{2\pi(R + h)}{T}\right)^2 = \frac{GM}{R + h}. \] Simplifying the equation: \[ \frac{4\pi^2(R + h)^2}{T^2} = \frac{GM}{R + h}. \]
Step 4: Solving for Time Period \( T \). Rearranging the equation to solve for \( T^2 \): \[ T^2 = \frac{4\pi^2(R + h)^3}{GM}. \] Finally, taking the square root of both sides: \[ T = 2\pi \sqrt{\frac{(R + h)^3}{GM}}. \]
Conclusion: The expression for the time period \( T \) of a satellite orbiting Earth at a height \( h \) above the Earth's surface is: \[ T = 2\pi \sqrt{\frac{(R + h)^3}{GM}}. \] Thus, the correct answer is \( \mathbf{(2)} \).