Question

The ratio of the time periods of a simple pendulum at heights $2R_E$ and $3R_E$ from the surface of the earth is ($R_E$ is radius of the earth)

• A. $1:2$
• B. $1:3$
• C. $3:4$
• D. $2:3$

Hint:

For a simple pendulum at altitude, $T \propto r$, where $r$ is the distance from the center of the earth.

Answer 1

The Correct Option is C

Step 1: Formula for Time Period and Gravity:
Time period of a simple pendulum: $T = 2\pi \sqrt{\frac{L}{g}}$. Acceleration due to gravity at height $h$: $g' = \frac{GM}{(R_E + h)^2}$. Thus, $T \propto \frac{1}{\sqrt{g'}} \propto \sqrt{(R_E + h)^2} \propto (R_E + h)$.
Step 2: Calculate for given heights:
Case 1: $h_1 = 2R_E$. Distance from center $r_1 = R_E + 2R_E = 3R_E$. $T_1 \propto 3R_E$. Case 2: $h_2 = 3R_E$. Distance from center $r_2 = R_E + 3R_E = 4R_E$. $T_2 \propto 4R_E$.
Step 3: Calculate Ratio:
\[ \frac{T_1}{T_2} = \frac{3R_E}{4R_E} = \frac{3}{4} \] Ratio is $3:4$.