At what height above the earth's surface does the value of g becomes 36% of the value at the surface of earth ?

Options

A. `(2R)/(5)`
B. `(2R)/(3)`
C. `(3R)/(7)`
D. `(R )/(3)`

Answer

To find the height above the Earth's surface where the value of gravitational acceleration (g) becomes 36% of its value at the surface, we can follow these steps:

### Step-by-Step Solution:

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

2. **Set up the equation for 36% of g**:
We want to find the height \( h \) where:
\[ g_h = 0.36g \]

3. **Substitute the expression for g_h**:
Substitute \( g_h \) in the equation:
\[ g \left( \frac{R}{R + h} \right)^2 = 0.36g \]
We can cancel \( g \) from both sides (assuming \( g \neq 0 \)):
\[ \left( \frac{R}{R + h} \right)^2 = 0.36 \]

4. **Take the square root of both sides**:
\[ \frac{R}{R + h} = \sqrt{0.36} = 0.6 \]

5. **Cross-multiply to solve for h**:
\[ R = 0.6(R + h) \]
Expanding the right side:
\[ R = 0.6R + 0.6h \]
Rearranging gives:
\[ R - 0.6R = 0.6h \]
\[ 0.4R = 0.6h \]

6. **Solve for h**:
\[ h = \frac{0.4R}{0.6} = \frac{2}{3}R \]

7. **Final Result**:
The height \( h \) above the Earth's surface where \( g \) becomes 36% of its value at the surface is:
\[ h = \frac{2}{3}R \]