If radius of earth is R then the height ‘ h ’ at which value of ‘ g ’ becomes one-fourth is

Options

A. `R/4`
B. `(3R)/4`
C. `R`
D. `R/8`

Answer

To solve the problem of finding the height \( h \) at which the acceleration due to gravity \( g \) becomes one-fourth of its value at the surface of the Earth, we can follow these steps:

### Step 1: Understand the relationship of gravity at different heights
The acceleration due to gravity \( g \) at a distance \( r \) from the center of the Earth is given by the formula:
\[ g_h = \frac{G M}{r^2} \]
where \( G \) is the gravitational constant, and \( M \) is the mass of the Earth.

### Step 2: Define the variables
Let:
- \( R \) = radius of the Earth
- \( h \) = height above the Earth's surface
- \( g \) = acceleration due to gravity at the surface of the Earth, which is given by:
\[ g = \frac{G M}{R^2} \]

### Step 3: Write the expression for gravity at height \( h \)
At a height \( h \) above the surface, the distance from the center of the Earth is \( R + h \). Thus, the acceleration due to gravity at height \( h \) is:
\[ g_h = \frac{G M}{(R + h)^2} \]

### Step 4: Set up the equation for one-fourth gravity
We want to find \( h \) such that:
\[ g_h = \frac{1}{4} g \]
Substituting the expressions for \( g_h \) and \( g \):
\[ \frac{G M}{(R + h)^2} = \frac{1}{4} \cdot \frac{G M}{R^2} \]

### Step 5: Simplify the equation
Since \( G \) and \( M \) are common on both sides, we can cancel them out:
\[ \frac{1}{(R + h)^2} = \frac{1}{4 R^2} \]

### Step 6: Cross-multiply to solve for \( h \)
Cross-multiplying gives:
\[ 4 R^2 = (R + h)^2 \]

### Step 7: Expand the equation
Expanding the right side:
\[ 4 R^2 = R^2 + 2Rh + h^2 \]

### Step 8: Rearrange the equation
Rearranging gives:
\[ 0 = h^2 + 2Rh + R^2 - 4R^2 \]
\[ 0 = h^2 + 2Rh - 3R^2 \]

### Step 9: Factor the quadratic equation
Factoring the quadratic equation:
\[ (h - R)(h + 3R) = 0 \]

### Step 10: Solve for \( h \)
The solutions are:
\[ h - R = 0 \quad \Rightarrow \quad h = R \]
\[ h + 3R = 0 \quad \Rightarrow \quad h = -3R \quad (\text{not physically meaningful}) \]

Thus, the only valid solution is:
\[ h = R \]

### Final Answer
The height \( h \) at which the value of \( g \) becomes one-fourth is:
\[ \boxed{R} \]