The height above the surface of earth at which acceleration due to gravity is half the acceleration due to gravity at surface of eart is `(R=6.4xx10^(6)m)`

Options

A. `6.4xx10^(6)m`
B. `2.6xx10^(6)m`
C. `12.8xx10^(6)m`
D. `19.2xx10^(6)m`

Answer

To find the height above the surface of the Earth at which the acceleration due to gravity is half of that at the surface, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
We need to find the height \( h \) above the Earth's surface where the acceleration due to gravity \( g' \) is half of the acceleration due to gravity at the surface \( g \). The formula for gravitational acceleration at a distance \( r \) from the center of the Earth is given by:
\[ g = \frac{GM}{r^2} \]
where \( G \) is the gravitational constant and \( M \) is the mass of the Earth.

2. **Setting Up the Equation**:
At the surface of the Earth, the acceleration due to gravity is:
\[ g = \frac{GM}{R^2} \]
where \( R \) is the radius of the Earth. At height \( h \), the distance from the center of the Earth becomes \( R + h \). Therefore, the acceleration due to gravity at height \( h \) is:
\[ g' = \frac{GM}{(R + h)^2} \]
According to the problem, we have:
\[ g' = \frac{g}{2} \]

3. **Substituting the Values**:
Substituting the expression for \( g' \) into the equation gives:
\[ \frac{GM}{(R + h)^2} = \frac{1}{2} \cdot \frac{GM}{R^2} \]
We can cancel \( GM \) from both sides (assuming \( GM \neq 0 \)):
\[ \frac{1}{(R + h)^2} = \frac{1}{2R^2} \]

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

5. **Expanding the Right Side**:
Expanding the right side:
\[ 2R^2 = R^2 + 2Rh + h^2 \]

6. **Rearranging the Equation**:
Rearranging the equation yields:
\[ R^2 = 2Rh + h^2 \]
This can be rewritten as:
\[ h^2 + 2Rh - R^2 = 0 \]

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

8. **Substituting the Value of R**:
Now, substituting \( R = 6.4 \times 10^6 \, \text{m} \):
\[ h = 6.4 \times 10^6 (\sqrt{2} - 1) \]
Approximating \( \sqrt{2} \approx 1.414 \):
\[ h \approx 6.4 \times 10^6 (1.414 - 1) = 6.4 \times 10^6 \times 0.414 \]
\[ h \approx 2.6 \times 10^6 \, \text{m} \]

### Final Answer:
The height above the surface of the Earth at which the acceleration due to gravity is half the acceleration due to gravity at the surface is approximately:
\[ \boxed{2.6 \times 10^6 \, \text{m}} \]