The value of acceleration due to gravity will be 1% of its value at the surface of earth at a height of `(R_(e )=6400 km)`

Options

A. 6400 km
B. 577600 km
C. 2560 km
D. 6400 km

Answer

To find the height at which the acceleration due to gravity is 1% of its value at the surface of the Earth, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**: We know that the acceleration due to gravity at a height \( h \) is given by the formula:
\[ g_h = \frac{g}{(1 + \frac{h}{R})^2} \]
where \( g \) is the acceleration due to gravity at the surface of the Earth, \( R \) is the radius of the Earth, and \( g_h \) is the acceleration due to gravity at height \( h \).

2. **Set Up the Equation**: We want to find \( h \) such that \( g_h = \frac{g}{100} \). Thus, we can set up the equation:
\[ \frac{g}{100} = \frac{g}{(1 + \frac{h}{R})^2} \]

3. **Cancel \( g \)**: Since \( g \) is common on both sides, we can cancel it out:
\[ \frac{1}{100} = \frac{1}{(1 + \frac{h}{R})^2} \]

4. **Cross Multiply**: Rearranging gives:
\[ (1 + \frac{h}{R})^2 = 100 \]

5. **Take the Square Root**: Taking the square root of both sides:
\[ 1 + \frac{h}{R} = 10 \]

6. **Solve for \( h \)**: Rearranging gives:
\[ \frac{h}{R} = 10 - 1 = 9 \]
Therefore, we have:
\[ h = 9R \]

7. **Substituting the Value of \( R \)**: Given that the radius of the Earth \( R \) is approximately \( 6400 \) km, we substitute:
\[ h = 9 \times 6400 \text{ km} = 57600 \text{ km} \]

### Final Answer:
The height at which the acceleration due to gravity is 1% of its value at the surface of the Earth is \( 57600 \) km.

---