Question

A satellite is revolving around the Earth at some height $(h)$. If $R$ is the radius of orbit, then the time period of satellite is proportional to

• A. $(R+h)^3$
• B. $(R+h)^{3/2}$
• C. $(R+h)^{5/2}$
• D. $(R+h)^6$

Hint:

Kepler’s Third Law: $T^2 \propto r^3$ So, $T \propto r^{3/2}$ — very important for satellites and planets.

Answer 1

The Correct Option is B

Concept: From Kepler’s Third Law of Planetary Motion, the time period of a satellite is related to the radius of its orbit as: \[ T^2 \propto r^3 \] where $r = R + h$ (distance from the center of Earth).

Step 1: Write Kepler’s law.
\[ T^2 \propto r^3 \]

Step 2: Take square root on both sides.
\[ T \propto r^{3/2} \]

Step 3: Substitute $r = R + h$.
\[ T \propto (R + h)^{3/2} \]

Step 4: Final conclusion.
Thus, the time period is proportional to: \[ (R + h)^{3/2} \]