The height above surface of earth where the value of gravitational acceleration is one fourth of that at surface, will be

Options

A. `R_(e)//4`
B. `R_(e)//2`
C. `3R_(e)//4`
D. `R_(e)`

Answer

To find the height above the surface of the Earth where the gravitational acceleration is one-fourth of that at the surface, we can use the formula for gravitational acceleration at a distance from the center of the Earth.

### Step-by-Step Solution:

1. **Understanding Gravitational Acceleration**:
The gravitational acceleration \( g \) 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. **Setting Up the Equation**:
We want to find the height \( h \) above the surface where the gravitational acceleration \( g' \) is:
\[ g' = \frac{g}{4} \]

3. **Using the Formula for Gravitational Acceleration at Height**:
The gravitational acceleration at a height \( h \) above the surface is given by:
\[ g' = \frac{GM}{(R + h)^2} \]

4. **Equating the Two Expressions for \( g' \)**:
We can set the two expressions for \( g' \) equal to each other:
\[ \frac{GM}{(R + h)^2} = \frac{g}{4} \]

5. **Substituting \( g \)**:
Substitute \( g = \frac{GM}{R^2} \) into the equation:
\[ \frac{GM}{(R + h)^2} = \frac{1}{4} \cdot \frac{GM}{R^2} \]

6. **Canceling \( GM \)**:
Since \( GM \) is common on both sides, we can cancel it out:
\[ \frac{1}{(R + h)^2} = \frac{1}{4R^2} \]

7. **Cross-Multiplying**:
Cross-multiplying gives:
\[ 4R^2 = (R + h)^2 \]

8. **Expanding the Right Side**:
Expanding the right side:
\[ 4R^2 = R^2 + 2Rh + h^2 \]

9. **Rearranging the Equation**:
Rearranging the equation leads to:
\[ 4R^2 - R^2 = 2Rh + h^2 \]
\[ 3R^2 = 2Rh + h^2 \]

10. **Assuming \( h \) is Small Compared to \( R \)**:
For small heights compared to the radius of the Earth, we can neglect \( h^2 \) in comparison to \( 2Rh \):
\[ 3R^2 \approx 2Rh \]

11. **Solving for \( h \)**:
Rearranging gives:
\[ h \approx \frac{3R^2}{2R} = \frac{3R}{2} \]

12. **Final Result**:
Thus, the height \( h \) above the surface of the Earth where the gravitational acceleration is one-fourth of that at the surface is:
\[ h = \frac{3R}{2} \]