A body weighs 72 N on the surface of the earth. What is the gravitational force on it due to earth at a height equal to half the radius of the earth from the surface

Options

A. 16 N
B. 28 N
C. 32 N
D. 72 N

Explanation

To solve the problem, we need to determine the gravitational force acting on a body at a height equal to half the radius of the Earth from the surface. The weight of the body on the surface of the Earth is given as 72 N.

### Step-by-Step Solution:

1. **Understand the given information**:
- Weight of the body on the surface of the Earth (F) = 72 N.
- The height (h) at which we need to find the gravitational force is equal to half the radius of the Earth (R), so \( h = \frac{R}{2} \).

2. **Use the formula for gravitational force**:
- The gravitational force (F') at a height \( h \) above the surface of the Earth can be calculated using the formula:
\[ F' = F \cdot \left(\frac{R^2}{(R + h)^2}\right) \]
- Here, \( F \) is the weight of the body on the surface of the Earth, and \( R \) is the radius of the Earth.

3. **Substitute the value of \( h \)**:
- Since \( h = \frac{R}{2} \), we can substitute this into the formula:
\[ F' = F \cdot \left(\frac{R^2}{\left(R + \frac{R}{2}\right)^2}\right) \]
- Simplifying \( R + \frac{R}{2} \):
\[ R + \frac{R}{2} = \frac{2R + R}{2} = \frac{3R}{2} \]

4. **Rewrite the formula**:
- Now substituting back into the formula:
\[ F' = F \cdot \left(\frac{R^2}{\left(\frac{3R}{2}\right)^2}\right) \]
- This simplifies to:
\[ F' = F \cdot \left(\frac{R^2}{\frac{9R^2}{4}}\right) = F \cdot \left(\frac{4}{9}\right) \]

5. **Calculate the new gravitational force**:
- Now substituting \( F = 72 \, \text{N} \):
\[ F' = 72 \cdot \left(\frac{4}{9}\right) = \frac{72 \cdot 4}{9} = \frac{288}{9} = 32 \, \text{N} \]

6. **Conclusion**:
- The gravitational force on the body at a height equal to half the radius of the Earth from the surface is **32 N**.