If `g` is acceleration due to gravity on the surface of the earth, having radius `R`, the height at which the acceleration due to gravity reduces to `g//2` is

Options

A. `R//2`
B. `sqrt(2)R`
C. `R/(sqrt(2))`
D. `(sqrt(2)-1)R`

Explanation

To solve the problem of finding the height at which the acceleration due to gravity reduces to \( \frac{g}{2} \), we can follow these steps:

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

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

### Step 3: Set \( g' \) equal to \( \frac{g}{2} \)
We want to find the height \( h \) where the gravitational acceleration is half of that at the surface:
\[ \frac{GM}{(R + h)^2} = \frac{1}{2} \cdot \frac{GM}{R^2} \]

### Step 4: Cancel out \( GM \) from both sides
Since \( GM \) appears on both sides of the equation, we can cancel it out:
\[ \frac{1}{(R + h)^2} = \frac{1}{2R^2} \]

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

### Step 6: Expand the right-hand side
Expanding \( (R + h)^2 \) results in:
\[ 2R^2 = R^2 + 2Rh + h^2 \]

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

### Step 8: Rearranging to form a quadratic equation
This can be rearranged to form a quadratic equation:
\[ h^2 + 2Rh - R^2 = 0 \]

### Step 9: Use the quadratic formula to solve for \( h \)
Using the quadratic formula \( h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = 2R, c = -R^2 \):
\[ h = \frac{-2R \pm \sqrt{(2R)^2 - 4 \cdot 1 \cdot (-R^2)}}{2 \cdot 1} \]
\[ h = \frac{-2R \pm \sqrt{4R^2 + 4R^2}}{2} \]
\[ h = \frac{-2R \pm \sqrt{8R^2}}{2} \]
\[ h = \frac{-2R \pm 2R\sqrt{2}}{2} \]
\[ h = -R + R\sqrt{2} \]

### Step 10: Simplify the expression for height
Thus, the height \( h \) can be simplified to:
\[ h = R(\sqrt{2} - 1) \]

### Final Answer
The height at which the acceleration due to gravity reduces to \( \frac{g}{2} \) is:
\[ h = R(\sqrt{2} - 1) \]