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. `2.6xx10^6` m
B. `6.4xx10^6` m
C. `9.0xx10^6` m
D. `1.6xx10^6` m

Explanation

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

### Step 1: Understand the relationship between gravity at height and distance from the center of the Earth
The acceleration due to gravity \( g' \) at a height \( h \) above the Earth's surface can be expressed as:
\[ g' = \frac{GM}{(R + h)^2} \]
where:
- \( G \) is the gravitational constant,
- \( M \) is the mass of the Earth,
- \( R \) is the radius of the Earth,
- \( h \) is the height above the Earth's surface.

### Step 2: Relate \( g' \) to \( g \)
At the surface of the Earth, the acceleration due to gravity is given by:
\[ g = \frac{GM}{R^2} \]
We want to find the height \( h \) where \( g' = \frac{g}{2} \). Thus, we can set up the equation:
\[ \frac{GM}{(R + h)^2} = \frac{g}{2} \]

### Step 3: Substitute \( g \) into the equation
Substituting \( g \) in the equation gives:
\[ \frac{GM}{(R + h)^2} = \frac{1}{2} \cdot \frac{GM}{R^2} \]

### Step 4: Cancel \( GM \) from both sides
Since \( GM \) is present 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 simplify
Cross-multiplying gives:
\[ 2R^2 = (R + h)^2 \]

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

### Step 7: Rearrange the equation
Rearranging the equation gives:
\[ 0 = h^2 + 2Rh - R^2 \]

### Step 8: Solve the quadratic equation
This is a standard quadratic equation in \( h \):
\[ h^2 + 2Rh - R^2 = 0 \]
Using the quadratic formula \( h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = 2R, c = -R^2 \):
\[ h = \frac{-2R \pm \sqrt{(2R)^2 - 4 \cdot 1 \cdot (-R^2)}}{2 \cdot 1} \]
\[ h = \frac{-2R \pm \sqrt{4R^2 + 4R^2}}{2} \]
\[ h = \frac{-2R \pm \sqrt{8R^2}}{2} \]
\[ h = \frac{-2R \pm 2R\sqrt{2}}{2} \]
\[ h = -R + R\sqrt{2} \]
\[ h = R(\sqrt{2} - 1) \]

### Step 9: Calculate the height
If we take \( R \) as approximately \( 6.37 \times 10^6 \) meters (the radius of the Earth), then:
\[ h = 6.37 \times 10^6 \times (\sqrt{2} - 1) \]
Calculating \( \sqrt{2} - 1 \approx 0.414 \):
\[ h \approx 6.37 \times 10^6 \times 0.414 \approx 2.64 \times 10^6 \text{ meters} \]

### Final Answer
The height at which the acceleration due to gravity reduces to \( \frac{g}{2} \) is approximately \( 2.64 \times 10^6 \) meters.

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