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 Earth's surface 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. **Understanding the Problem**:
We need to find the height \( h \) above the Earth's surface where the acceleration due to gravity \( g' \) is \( \frac{g}{64} \), where \( g \) is the acceleration due to gravity at the Earth's surface.

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

3. **Setting Up the Equation**:
Since we know that \( g' = \frac{g}{64} \), we can substitute this into the formula:
\[ \frac{g}{64} = \frac{g}{(1 + \frac{h}{r})^2} \]

4. **Canceling \( g \)**:
We can cancel \( g \) from both sides of the equation (assuming \( g \neq 0 \)):
\[ \frac{1}{64} = \frac{1}{(1 + \frac{h}{r})^2} \]

5. **Cross-Multiplying**:
Cross-multiplying gives us:
\[ (1 + \frac{h}{r})^2 = 64 \]

6. **Taking the Square Root**:
Taking the square root of both sides, we get:
\[ 1 + \frac{h}{r} = 8 \]

7. **Solving for \( \frac{h}{r} \)**:
Rearranging the equation gives:
\[ \frac{h}{r} = 8 - 1 = 7 \]

8. **Finding \( h \)**:
Therefore, we can express \( h \) as:
\[ h = 7r \]

9. **Substituting the Radius of the Earth**:
The average radius of the Earth \( r \) is approximately \( 6400 \) kilometers. Thus:
\[ h = 7 \times 6400 \text{ km} = 44800 \text{ km} \]

10. **Final Result**:
Converting this into meters:
\[ h = 44800 \text{ km} = 44800 \times 1000 \text{ m} = 44800000 \text{ m} = 4.48 \times 10^7 \text{ m} \]
Therefore, the height above the Earth's surface where the acceleration due to gravity is \( \frac{1}{64} \) of its value at the surface is approximately \( 44800 \) km or \( 4.48 \times 10^7 \) m.