The altitude at which the weight of a body is only 64% of its weight on the surface of the earth is (Radius of the earth is 6400 km)

Options

A. 1600 m
B. 16 m
C. 160 m
D. 1600 km

Answer

To find the altitude at which the weight of a body is only 64% of its weight on the surface of the Earth, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the problem**: We need to find the altitude \( h \) where the weight of a body is 64% of its weight on the surface of the Earth. The weight of a body is given by the formula \( W = mg \), where \( g \) is the acceleration due to gravity.

2. **Use the formula for gravity at height**: The acceleration due to gravity at a height \( h \) above the Earth's surface is given by:
\[ g_h = \frac{g}{(1 + \frac{h}{R})^2} \]
where \( R \) is the radius of the Earth.

3. **Set up the equation**: We know that at height \( h \), the weight is 64% of the weight at the surface:
\[ g_h = 0.64g \]

4. **Substitute into the gravity formula**:
\[ \frac{g}{(1 + \frac{h}{R})^2} = 0.64g \]
We can cancel \( g \) from both sides (assuming \( g \neq 0 \)):
\[ \frac{1}{(1 + \frac{h}{R})^2} = 0.64 \]

5. **Take the reciprocal**:
\[ (1 + \frac{h}{R})^2 = \frac{1}{0.64} \]
Simplifying \( \frac{1}{0.64} \):
\[ \frac{1}{0.64} = \frac{100}{64} = \frac{25}{16} \]

6. **Take the square root**:
\[ 1 + \frac{h}{R} = \frac{5}{4} \]

7. **Solve for \( \frac{h}{R} \)**:
\[ \frac{h}{R} = \frac{5}{4} - 1 = \frac{1}{4} \]

8. **Multiply by the radius of the Earth**:
\[ h = R \cdot \frac{1}{4} \]
Given \( R = 6400 \, \text{km} \):
\[ h = 6400 \cdot \frac{1}{4} = 1600 \, \text{km} \]

### Final Answer:
The altitude at which the weight of a body is only 64% of its weight on the surface of the Earth is **1600 km**.