A body weighs W newton at the surface of the earth. Its weight at a height equal to half the radius of the earth, will be

Options

A. W
B. 2W
C. `(4W)/(9)`
D. `(W)/(4)`

Answer

To solve the problem of finding the weight of a body at a height equal to half the radius of the Earth, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Weight at the Surface**:
The weight of the body at the surface of the Earth is given as \( W \) Newton. The weight can be expressed using the formula:
\[ W = mg \]
where \( m \) is the mass of the body and \( g \) is the acceleration due to gravity at the surface of the Earth.

2. **Finding the Mass of the Body**:
From the equation \( W = mg \), we can rearrange it to find the mass \( m \):
\[ m = \frac{W}{g} \]

3. **Calculating the New Gravitational Acceleration at Height**:
We need to find the gravitational acceleration \( g' \) at a height \( h \) equal to half the radius of the Earth. The formula for gravitational acceleration at a height \( h \) above the surface is:
\[ g' = \frac{GM}{(R + h)^2} \]
where \( G \) is the universal gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth. Here, \( h = \frac{R}{2} \).

4. **Substituting the Height**:
Substituting \( h = \frac{R}{2} \) into the equation gives:
\[ g' = \frac{GM}{\left(R + \frac{R}{2}\right)^2} = \frac{GM}{\left(\frac{3R}{2}\right)^2} = \frac{GM}{\frac{9R^2}{4}} = \frac{4GM}{9R^2} \]

5. **Relating \( g' \) to \( g \)**:
We know that the gravitational acceleration at the surface of the Earth is given by:
\[ g = \frac{GM}{R^2} \]
Thus, we can express \( g' \) in terms of \( g \):
\[ g' = \frac{4}{9}g \]

6. **Finding the Weight at Height**:
The weight of the body at the height \( h \) is given by:
\[ W' = mg' = m \left(\frac{4}{9}g\right) \]
Substituting \( m = \frac{W}{g} \) into this equation:
\[ W' = \left(\frac{W}{g}\right) \left(\frac{4}{9}g\right) \]
The \( g \) terms cancel out:
\[ W' = \frac{4}{9}W \]

### Final Answer:
The weight of the body at a height equal to half the radius of the Earth is:
\[ W' = \frac{4}{9}W \]