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 \( 1.2T \).