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

Options

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

Answer

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

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

### Step 2: Set up the equation for \( g' \)
At a height \( h \) above the surface of the Earth, the distance from the center of the Earth becomes \( R + h \), where \( R \) is the radius of the Earth. Therefore, the new expression for gravity at height \( h \) is:
\[ g' = \frac{GM}{(R + h)^2} \]

### Step 3: Set \( g' \) to be one-fourth of \( g \)
According to the problem, we want to find \( h \) such that:
\[ g' = \frac{g}{4} \]
Substituting the expressions for \( g \) and \( g' \):
\[ \frac{GM}{(R + h)^2} = \frac{1}{4} \cdot \frac{GM}{R^2} \]

### Step 4: Simplify the equation
We can cancel \( GM \) from both sides:
\[ \frac{1}{(R + h)^2} = \frac{1}{4R^2} \]

### Step 5: Cross-multiply to eliminate the fractions
Cross-multiplying gives:
\[ 4R^2 = (R + h)^2 \]

### Step 6: Expand the right side
Expanding the right side:
\[ 4R^2 = R^2 + 2Rh + h^2 \]

### Step 7: Rearrange the equation
Rearranging gives:
\[ 4R^2 - R^2 = 2Rh + h^2 \]
\[ 3R^2 = 2Rh + h^2 \]

### Step 8: Rearrange to form a quadratic equation
Rearranging this into standard quadratic form:
\[ h^2 + 2Rh - 3R^2 = 0 \]

### Step 9: Solve the quadratic equation using the quadratic formula
Using the quadratic formula \( h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1 \), \( b = 2R \), and \( c = -3R^2 \):
\[ h = \frac{-2R \pm \sqrt{(2R)^2 - 4 \cdot 1 \cdot (-3R^2)}}{2 \cdot 1} \]
\[ h = \frac{-2R \pm \sqrt{4R^2 + 12R^2}}{2} \]
\[ h = \frac{-2R \pm \sqrt{16R^2}}{2} \]
\[ h = \frac{-2R \pm 4R}{2} \]

### Step 10: Calculate the possible values of \( h \)
Calculating the two possible values:
1. \( h = \frac{2R}{2} = R \)
2. \( h = \frac{-6R}{2} = -3R \) (not physically meaningful)

Thus, the height \( h \) at which \( g' = \frac{g}{4} \) is:
\[ h = R \]

### Final Answer
The height \( h \) at which the value of \( g \) becomes one-fourth of its original value is \( R \).

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