The height above the surface of the earth where acceleration due to gravity is 1/64 of its value at surface of the earth is approximately.

Options

A. `45 xx 10^(6)` m
B. `54 xx 10^(6)` m
C. `102 xx 10^(6)` m
D. `72 xx 10^(6)` m

Explanation

To solve the problem of finding the height above the surface of the Earth where the acceleration due to gravity is 1/64 of its value at the surface, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between gravity at the surface and at height**:
The acceleration due to gravity at the surface of the Earth is given by:
\[ g = \frac{GM}{R^2} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

2. **Set up the equation for gravity at height \( H \)**:
The acceleration due to gravity at a height \( H \) above the surface is given by:
\[ g' = \frac{GM}{(R + H)^2} \]
We want \( g' \) to be \( \frac{g}{64} \).

3. **Substitute \( g \) into the equation**:
We can substitute \( g \) into the equation:
\[ \frac{g}{64} = \frac{GM}{(R + H)^2} \]
This leads to:
\[ \frac{GM}{R^2 \cdot 64} = \frac{GM}{(R + H)^2} \]

4. **Cancel \( GM \) from both sides**:
Since \( GM \) appears on both sides, we can cancel it out:
\[ \frac{1}{64} = \frac{1}{(R + H)^2/R^2} \]
Rearranging gives:
\[ \frac{(R + H)^2}{R^2} = 64 \]

5. **Take the square root of both sides**:
Taking the square root gives:
\[ \frac{R + H}{R} = 8 \]

6. **Solve for \( R + H \)**:
Multiplying both sides by \( R \):
\[ R + H = 8R \]
Rearranging gives:
\[ H = 8R - R = 7R \]

7. **Substitute the value of \( R \)**:
The radius of the Earth \( R \) is approximately \( 6.4 \times 10^6 \) meters. Therefore:
\[ H = 7 \times 6.4 \times 10^6 \text{ m} \]

8. **Calculate \( H \)**:
\[ H = 44.8 \times 10^6 \text{ m} \]

### Final Answer:
The height above the surface of the Earth where the acceleration due to gravity is \( \frac{1}{64} \) of its value at the surface is approximately \( 44.8 \times 10^6 \) meters.