A satellite is projected vertically upwards from an earth station. At what height above the earth's surface will the force on the satellite due to the earth be reduced to half its value at the earth station? (Radius of the earth is 6400 km.)

Explanation

To solve the problem, we need to find the height \( h \) above the Earth's surface where the gravitational force on a satellite is reduced to half of its value at the Earth's surface.

### Step-by-Step Solution:

1. **Understand the Gravitational Force**:
The gravitational force \( F \) acting on the satellite at a distance \( r \) from the center of the Earth is given by the formula:
\[ F = \frac{G \cdot M \cdot m}{r^2} \]
where:
- \( G \) is the gravitational constant,
- \( M \) is the mass of the Earth,
- \( m \) is the mass of the satellite,
- \( r \) is the distance from the center of the Earth to the satellite.

2. **Force at the Earth's Surface**:
At the Earth's surface, the distance \( r \) is equal to the radius of the Earth \( R_e \):
\[ F_e = \frac{G \cdot M \cdot m}{R_e^2} \]

3. **Force at Height \( h \)**:
At a height \( h \) above the Earth's surface, the distance from the center of the Earth becomes \( R_e + h \):
\[ F_s = \frac{G \cdot M \cdot m}{(R_e + h)^2} \]

4. **Setting Up the Equation**:
We want the force at height \( h \) to be half of the force at the Earth's surface:
\[ F_s = \frac{1}{2} F_e \]
Substituting the expressions for \( F_s \) and \( F_e \):
\[ \frac{G \cdot M \cdot m}{(R_e + h)^2} = \frac{1}{2} \cdot \frac{G \cdot M \cdot m}{R_e^2} \]

5. **Canceling Common Terms**:
Since \( G \), \( M \), and \( m \) are common on both sides, we can cancel them out:
\[ \frac{1}{(R_e + h)^2} = \frac{1}{2 R_e^2} \]

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

7. **Expanding the Right Side**:
Expanding the right side:
\[ 2 R_e^2 = R_e^2 + 2 R_e h + h^2 \]

8. **Rearranging the Equation**:
Rearranging the equation gives:
\[ R_e^2 = 2 R_e h + h^2 \]

9. **Rearranging into Standard Form**:
This can be rearranged into a standard quadratic form:
\[ h^2 + 2 R_e h - R_e^2 = 0 \]

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

11. **Substituting the Radius of the Earth**:
Given \( R_e = 6400 \) km:
\[ h = 6400(\sqrt{2} - 1) \]
Calculating \( \sqrt{2} \approx 1.414 \):
\[ h \approx 6400(1.414 - 1) \approx 6400 \cdot 0.414 \approx 2649.6 \text{ km} \]

### Final Answer:
The height \( h \) above the Earth's surface where the gravitational force on the satellite is reduced to half its value at the Earth's surface is approximately **2649.6 km**.