Acceleration the due to gravity become `[(g)/(2)]` at a height equal to

Options

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

Explanation

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

### Step 1: Understand the formula for acceleration due to gravity at height \( h \)
The acceleration due to gravity \( g' \) at a height \( h \) above the Earth's surface is given by the formula:
\[ g' = g \left(1 - \frac{2h}{R}\right) \]
where \( R \) is the radius of the Earth.

### Step 2: Set up the equation
We need to find the height \( h \) where \( g' = \frac{g}{2} \). Therefore, we set up the equation:
\[ \frac{g}{2} = g \left(1 - \frac{2h}{R}\right) \]

### Step 3: Simplify the equation
We can divide both sides of the equation by \( g \) (assuming \( g \neq 0 \)):
\[ \frac{1}{2} = 1 - \frac{2h}{R} \]

### Step 4: Rearrange the equation
Now, we can rearrange the equation to isolate \( \frac{2h}{R} \):
\[ \frac{2h}{R} = 1 - \frac{1}{2} \]
\[ \frac{2h}{R} = \frac{1}{2} \]

### Step 5: Solve for \( h \)
Now, we can solve for \( h \):
\[ 2h = \frac{R}{2} \]
\[ h = \frac{R}{4} \]

### Conclusion
Thus, the height \( h \) at which the acceleration due to gravity becomes \( \frac{g}{2} \) is:
\[ h = \frac{R}{4} \]

### Final Answer
The height at which the acceleration due to gravity becomes \( \frac{g}{2} \) is \( \frac{R}{4} \).

---