The period of a simple pendulum on the surface of earth is T. At an attitude of half of the radius of earth from the surface, its period will be

Options

A. `sqrt((3)/(2))T`
B. `(3T)/(2)`
C. `(2T)/(3)`
D. `sqrt((2)/(3))` T

Answer

To solve the problem, we need to find the period of a simple pendulum at an altitude of half the radius of the Earth from the surface. Let's break down the solution step by step.

### Step 1: Understand the formula for the period of a simple pendulum
The period \( T \) of a simple pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity.

### Step 2: Determine the acceleration due to gravity at the surface of the Earth
At the surface of the Earth, the acceleration due to gravity is denoted as \( g \).

### Step 3: Calculate the acceleration due to gravity at the altitude
At an altitude of \( h = \frac{R}{2} \) (where \( R \) is the radius of the Earth), the distance from the center of the Earth becomes \( R + h = R + \frac{R}{2} = \frac{3R}{2} \).

The formula for the acceleration due to gravity at a distance \( r \) from the center of the Earth is:
\[ g' = \frac{GM}{r^2} \]
At the altitude of \( \frac{R}{2} \):
\[ g' = \frac{GM}{\left(\frac{3R}{2}\right)^2} = \frac{GM}{\frac{9R^2}{4}} = \frac{4GM}{9R^2} = \frac{4}{9}g \]

### Step 4: Substitute \( g' \) into the period formula for the pendulum at altitude
Now we can find the period \( T' \) at this altitude:
\[ T' = 2\pi \sqrt{\frac{L}{g'}} \]
Substituting \( g' \):
\[ T' = 2\pi \sqrt{\frac{L}{\frac{4}{9}g}} = 2\pi \sqrt{\frac{9L}{4g}} = 2\pi \cdot \frac{3}{2} \sqrt{\frac{L}{g}} = 3 \cdot 2\pi \sqrt{\frac{L}{g}} = 3T \]

### Step 5: Final result
Thus, the period of the simple pendulum at an altitude of half the radius of the Earth from the surface is:
\[ T' = \frac{3}{2}T \]

### Summary
The period of the pendulum at the altitude of half the radius of the Earth is \( \frac{3}{2}T \).

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