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, we need to find the height \( h \) at which the value of gravitational acceleration \( g \) reduces to half of its value at the surface of the Earth.

### Step-by-Step Solution:

1. **Understanding the Gravitational Acceleration**:
The gravitational acceleration at the surface of the Earth is given by:
\[ g_s = \frac{GM}{R^2} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

2. **Gravitational Acceleration at Height \( h \)**:
At a height \( h \) above the surface of the Earth, the gravitational acceleration \( g_h \) is given by:
\[ g_h = \frac{GM}{(R + h)^2} \]

3. **Setting Up the Equation**:
According to the problem, at height \( h \), the gravitational acceleration is half of that at the surface:
\[ g_h = \frac{g_s}{2} \]
Substituting the expressions for \( g_h \) and \( g_s \):
\[ \frac{GM}{(R + h)^2} = \frac{1}{2} \cdot \frac{GM}{R^2} \]

4. **Cancelling Common Terms**:
We can cancel \( GM \) from both sides (assuming \( GM \neq 0 \)):
\[ \frac{1}{(R + h)^2} = \frac{1}{2R^2} \]

5. **Cross-Multiplying**:
Cross-multiplying gives:
\[ 2R^2 = (R + h)^2 \]

6. **Expanding the Right Side**:
Expanding \( (R + h)^2 \):
\[ 2R^2 = R^2 + 2Rh + h^2 \]

7. **Rearranging the Equation**:
Rearranging gives:
\[ 0 = R^2 + 2Rh + h^2 - 2R^2 \]
\[ 0 = -R^2 + 2Rh + h^2 \]
\[ h^2 + 2Rh - R^2 = 0 \]

8. **Using the Quadratic Formula**:
This is a quadratic equation in \( h \). We can use the quadratic formula \( h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1 \), \( b = 2R \), and \( c = -R^2 \):
\[ h = \frac{-2R \pm \sqrt{(2R)^2 - 4 \cdot 1 \cdot (-R^2)}}{2 \cdot 1} \]
\[ h = \frac{-2R \pm \sqrt{4R^2 + 4R^2}}{2} \]
\[ h = \frac{-2R \pm \sqrt{8R^2}}{2} \]
\[ h = \frac{-2R \pm 2R\sqrt{2}}{2} \]
\[ h = -R + R\sqrt{2} \]
\[ h = R(\sqrt{2} - 1) \]

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