At what altitude will the acceleration due to gravity be `25%` of that at the earth's surface (given radius of earth is `R`)?

Options

A. `R//4`
B. `R`
C. `3R//8`
D. `R//2`

Explanation

To solve the problem of finding the altitude at which the acceleration due to gravity is 25% of that at the Earth's surface, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship of gravity at different altitudes**:
The acceleration due to gravity at the Earth's surface is given by:
\[ g_0 = \frac{GM}{R^2} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

2. **Express gravity at an altitude \( h \)**:
At a height \( h \) above the Earth's surface, the acceleration due to gravity \( g \) is given by:
\[ g = \frac{GM}{(R + h)^2} \]

3. **Set up the equation for 25% of surface gravity**:
We want to find the altitude \( h \) where \( g \) is 25% of \( g_0 \):
\[ g = 0.25 g_0 \]
Substituting the expressions for \( g \) and \( g_0 \):
\[ \frac{GM}{(R + h)^2} = 0.25 \left(\frac{GM}{R^2}\right) \]

4. **Simplify the equation**:
Since \( GM \) appears on both sides, we can cancel it out:
\[ \frac{1}{(R + h)^2} = 0.25 \cdot \frac{1}{R^2} \]
This simplifies to:
\[ (R + h)^2 = 4R^2 \]

5. **Take the square root**:
Taking the square root of both sides gives:
\[ R + h = 2R \]

6. **Solve for \( h \)**:
Rearranging the equation to solve for \( h \):
\[ h = 2R - R = R \]

### Final Answer:
The altitude \( h \) at which the acceleration due to gravity is 25% of that at the Earth's surface is:
\[ h = R \]