The value of 'g' reduces to half of its value at surface of earth at a height 'h', then :-

Options

A. `h=R`
B. `h=2 R`
C. `h=(sqrt(2)+1)R`
D. `h=(sqrt(2)-1)R`

Answer

To solve the problem where the value of 'g' reduces to half of its value at the surface of the Earth at a height 'h', we can follow these steps:

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

At a height \( h \) above the surface, the acceleration due to gravity \( g_h \) is given by:
\[ g_h = \frac{GM}{(R + h)^2} \]

### Step 2: Set up the equation based on the problem statement
According to the problem, at height \( h \), the gravity \( g_h \) is half of the gravity at the surface:
\[ g_h = \frac{g}{2} \]

Substituting the expressions for \( g \) and \( g_h \):
\[ \frac{GM}{(R + h)^2} = \frac{1}{2} \cdot \frac{GM}{R^2} \]

### Step 3: Simplify the equation
We can cancel \( GM \) from both sides (assuming \( GM \neq 0 \)):
\[ \frac{1}{(R + h)^2} = \frac{1}{2R^2} \]

Taking the reciprocal of both sides gives:
\[ (R + h)^2 = 2R^2 \]

### Step 4: Take the square root of both sides
Taking the square root of both sides results in:
\[ R + h = R\sqrt{2} \]

### Step 5: Solve for height \( h \)
Now, isolate \( h \):
\[ h = R\sqrt{2} - R \]
\[ h = R(\sqrt{2} - 1) \]

### Conclusion
Thus, the height \( h \) at which the value of 'g' reduces to half of its value at the surface of the Earth is:
\[ h = R(\sqrt{2} - 1) \]