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 of determining the height above the Earth's surface at which the gravitational force on a satellite is reduced to half its value at the Earth's surface, we can follow these steps:

### Step 1: Understand the gravitational force formula
The gravitational force \( F \) acting on a satellite of mass \( m \) 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,
- \( r \) is the distance from the center of the Earth to the satellite.

### Step 2: Determine the force at the Earth's surface
At the Earth's surface, the distance \( r \) is equal to the radius of the Earth \( R \):
\[ F_{\text{surface}} = \frac{G \cdot M \cdot m}{R^2} \]

### Step 3: Set up the equation for half the force
We want to find the height \( h \) above the Earth's surface where the gravitational force is half of that at the surface:
\[ F = \frac{1}{2} F_{\text{surface}} = \frac{1}{2} \left( \frac{G \cdot M \cdot m}{R^2} \right) \]

### Step 4: Express the distance from the center of the Earth
At height \( h \) above the Earth's surface, the distance \( r \) from the center of the Earth is:
\[ r = R + h \]
Substituting this into the gravitational force equation gives:
\[ F = \frac{G \cdot M \cdot m}{(R + h)^2} \]

### Step 5: Set the two expressions for force equal
Now we can set the two expressions for force equal:
\[ \frac{G \cdot M \cdot m}{(R + h)^2} = \frac{1}{2} \left( \frac{G \cdot M \cdot m}{R^2} \right) \]

### Step 6: Simplify the equation
Cancel out \( G \), \( M \), and \( m \) from both sides:
\[ \frac{1}{(R + h)^2} = \frac{1}{2R^2} \]
Cross-multiplying gives:
\[ 2R^2 = (R + h)^2 \]

### Step 7: Expand and rearrange the equation
Expanding the right side:
\[ 2R^2 = R^2 + 2Rh + h^2 \]
Rearranging gives:
\[ R^2 = 2Rh + h^2 \]

### Step 8: Rearranging to form a quadratic equation
Rearranging further gives:
\[ h^2 + 2Rh - R^2 = 0 \]

### Step 9: Solve the quadratic equation
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 10: Substitute the radius of the Earth
Given \( R = 6400 \) km:
\[ h = 6400(\sqrt{2} - 1) \approx 6400(1.414 - 1) \approx 6400(0.414) \approx 2649.6 \text{ km} \]

### Final Answer
The height above the Earth's surface at which the gravitational force on the satellite is reduced to half its value at the Earth's surface is approximately **2649.6 km**.

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