Consider a body which weighs `w` newton at the surface of the earth. Its weight at a height equals to half the radius of the earth, will be

Options

A. `(w)/(2)`
B. `(2w)/(3)`
C. `(4w)/(9)`
D. `(8w)/(27)`

Explanation

To solve the problem, we need to determine the weight of a body at a height equal to half the radius of the Earth. Let's break this down step by step.

### Step 1: Understanding Weight on the Surface of the Earth
The weight \( W \) of a body on the surface of the Earth is given by 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.

### Step 2: Gravitational Force Formula
The gravitational force (weight) can also be expressed using the formula:
\[ g = \frac{GM}{R^2} \]
where:
- \( G \) is the universal gravitational constant,
- \( M \) is the mass of the Earth,
- \( R \) is the radius of the Earth.

### Step 3: Weight at a Height
Now, we need to find the weight of the body at a height \( h \) above the surface of the Earth. The height given in the problem is half the radius of the Earth, so:
\[ h = \frac{R}{2} \]

The distance from the center of the Earth to the body at this height is:
\[ d = R + h = R + \frac{R}{2} = \frac{3R}{2} \]

### Step 4: Calculate the New Gravitational Acceleration
The new gravitational acceleration \( g' \) at this height can be calculated using the formula:
\[ g' = \frac{GM}{d^2} = \frac{GM}{\left(\frac{3R}{2}\right)^2} = \frac{GM}{\frac{9R^2}{4}} = \frac{4GM}{9R^2} \]

### Step 5: Weight at the Height
Now we can express the new weight \( W' \) of the body at this height:
\[ W' = mg' = m \left(\frac{4GM}{9R^2}\right) \]
Since we know that \( mg = W = \frac{GM}{R^2} \), we can substitute \( \frac{GM}{R^2} \) with \( W \):
\[ W' = m \left(\frac{4}{9} \cdot \frac{GM}{R^2}\right) = \frac{4}{9} W \]

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

### Final Answer
The weight at a height equal to half the radius of the Earth will be \( \frac{4}{9} W \).

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