The value of 'g' at a certain height above the surface of the earth is 16% of its va lue on the surface. The height is (R = 6300 km)

Options

A. 10500 km
B. 12500 km
C. 3000 km
D. 9450 km

Answer

To solve the problem, we need to find the height \( h \) above the surface of the Earth where the acceleration due to gravity \( g' \) is 16% of the value of gravity \( g \) at the surface. The radius of the Earth \( R \) is given as 6300 km.

### Step-by-Step Solution:

1. **Understand the relationship between \( g' \) and \( g \)**:
We know that:
\[ g' = \frac{g}{(1 + \frac{h}{R})^2} \]
According to the problem, \( g' = 0.16g \) (since 16% is 0.16).

2. **Set up the equation**:
Substitute \( g' \) into the equation:
\[ 0.16g = \frac{g}{(1 + \frac{h}{R})^2} \]

3. **Cancel \( g \) from both sides**:
Since \( g \) is not zero, we can divide both sides by \( g \):
\[ 0.16 = \frac{1}{(1 + \frac{h}{R})^2} \]

4. **Cross-multiply**:
Rearranging gives:
\[ (1 + \frac{h}{R})^2 = \frac{1}{0.16} \]
Simplifying the right side:
\[ (1 + \frac{h}{R})^2 = 6.25 \]

5. **Take the square root of both sides**:
\[ 1 + \frac{h}{R} = \sqrt{6.25} = 2.5 \]

6. **Isolate \( \frac{h}{R} \)**:
\[ \frac{h}{R} = 2.5 - 1 = 1.5 \]

7. **Solve for \( h \)**:
Multiply both sides by \( R \):
\[ h = 1.5R \]
Given \( R = 6300 \) km:
\[ h = 1.5 \times 6300 = 9450 \text{ km} \]

### Final Answer:
The height \( h \) above the surface of the Earth is **9450 km**.