Assuming earth to be a sphere of radius 6400km, calculate the height above the earth's surface at which the value of acceleration due to gravity reduces to half its value on the erath's surface.

Explanation

To solve the problem of finding the height above the Earth's surface at which the acceleration due to gravity reduces to half its value on the Earth's surface, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the formula for gravitational acceleration**:
The acceleration due to gravity at the surface of the Earth (g) 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.

2. **Gravitational acceleration at height \( h \)**:
The acceleration due to gravity at a height \( h \) above the Earth's surface is given by:
\[ g' = \frac{GM}{(R + h)^2} \]

3. **Set up the equation**:
According to the problem, we want to find the height \( h \) where \( g' = \frac{g}{2} \). Therefore, we can write:
\[ \frac{GM}{(R + h)^2} = \frac{1}{2} \cdot \frac{GM}{R^2} \]

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

5. **Cross-multiply to solve for \( R + h \)**:
Cross-multiplying gives:
\[ 2R^2 = (R + h)^2 \]

6. **Expand the right side**:
Expanding \( (R + h)^2 \) gives:
\[ 2R^2 = R^2 + 2Rh + h^2 \]

7. **Rearrange the equation**:
Rearranging the equation leads to:
\[ R^2 = 2Rh + h^2 \]

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

9. **Use the quadratic formula**:
To solve for \( h \), we can use the quadratic formula:
\[ h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = 2R \), and \( c = -R^2 \):
\[ h = \frac{-2R \pm \sqrt{(2R)^2 - 4 \cdot 1 \cdot (-R^2)}}{2 \cdot 1} \]
\[ h = \frac{-2R \pm \sqrt{4R^2 + 4R^2}}{2} \]
\[ h = \frac{-2R \pm \sqrt{8R^2}}{2} \]
\[ h = \frac{-2R \pm 2R\sqrt{2}}{2} \]
\[ h = -R + R\sqrt{2} \]

10. **Substituting the value of R**:
Given \( R = 6400 \) km, we substitute:
\[ h = -6400 + 6400\sqrt{2} \]
\[ h = 6400(\sqrt{2} - 1) \]

11. **Calculate the numerical value**:
Approximating \( \sqrt{2} \approx 1.414 \):
\[ h \approx 6400(1.414 - 1) \approx 6400 \times 0.414 \approx 2649.6 \text{ km} \]

### Final Answer:
The height above the Earth's surface at which the value of acceleration due to gravity reduces to half its value on the Earth's surface is approximately **2649.6 km**.