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 orbital period of a satellite around Earth at altitude \( h \) is determined by Kepler's Third Law, based on the equilibrium between Earth's gravitational pull and the centripetal force required for the satellite's motion.
Step 1: Gravitational Force Equates to Centripetal Force.The gravitational force \( F_{\text{gravity}} \) exerted by Earth on the satellite is calculated using Newton's law of universal gravitation:\[F_{\text{gravity}} = \frac{GMm}{(R + h)^2},\]where: \( G \) represents the gravitational constant, \( M \) is the mass of the Earth, \( m \) is the mass of the satellite, \( R + h \) is the distance from Earth's center to the satellite (Earth's radius plus the satellite's altitude).This gravitational force acts as the centripetal force necessary for the satellite's circular orbit:\[F_{\text{centripetal}} = \frac{mv^2}{R + h},\]with \( v \) denoting the satellite's orbital velocity.
Step 2: Setting Gravitational and Centripetal Forces Equal.To maintain a stable orbit, the gravitational force must equal the centripetal force. Therefore, we equate the two expressions:\[\frac{GMm}{(R + h)^2} = \frac{mv^2}{R + h}.\]Canceling the satellite's mass \( m \) from both sides and simplifying yields:\[v^2 = \frac{GM}{R + h}.\]
Step 3: Linking Orbital Velocity to Orbital Period.The relationship between orbital velocity \( v \) and the orbital period \( T \) is given by:\[v = \frac{2\pi(R + h)}{T}.\]Substituting the expression for \( v^2 \) from Step 2 into this equation gives:\[\left(\frac{2\pi(R + h)}{T}\right)^2 = \frac{GM}{R + h}.\]Simplifying this equation results in:\[\frac{4\pi^2(R + h)^2}{T^2} = \frac{GM}{R + h}.\]
Step 4: Deriving the Formula for Orbital Period \( T \).Rearranging the equation to solve for \( T^2 \):\[T^2 = \frac{4\pi^2(R + h)^3}{GM}.\]Taking the square root of both sides provides the final formula for the orbital period:\[T = 2\pi \sqrt{\frac{(R + h)^3}{GM}}.\]
Conclusion:The formula for the orbital period \( T \) of a satellite orbiting Earth at an altitude \( h \) above its surface is:\[T = 2\pi \sqrt{\frac{(R + h)^3}{GM}}.\]Therefore, the correct option is \( \mathbf{(2)} \).