Question

The time period of a simple pendulum on the surface of the earth is $T$. The time period of the same pendulum at a height of 1280 km from the surface of the earth is

• A. $1.5T$
• B. $1.2T$
• C. $2T$
• D. $2.4T$

Hint:

When calculating the effect of height on gravity, remember that gravity decreases with the square of the distance from the center of the Earth.

Answer 1

The Correct Option is B

Step 1: The formula for the time period of a simple pendulum is given by: $$T = 2\pi \sqrt{\frac{l}{g}}$$ where $l$ is the length of the pendulum and $g$ is the acceleration due to gravity. Step 2: Gravity decreases with height. The formula for $g$ at a height $h$ above the earth’s surface is: $$g' = g \left( \frac{R}{R+h} \right)^2$$ where $R$ is the radius of the Earth. Step 3: The ratio of the time periods at the surface and at height $h$ is: $$\frac{T'}{T} = \sqrt{\frac{g}{g'}} = \sqrt{\left( \frac{R+h}{R} \right)^2}$$ Substitute the given values: $R = 6400 \, \text{km}$ and $h = 1280 \, \text{km}$: $$\frac{T'}{T} = \sqrt{\left( \frac{6400 + 1280}{6400} \right)^2} = \sqrt{\left( \frac{7680}{6400} \right)^2} = \sqrt{1.2^2} = 1.2$$ Thus, the time period at the height is $1.2T$.