An artificial satellite is moving in a circular orbit around the earth with a speed equal to half the escape velocity from the earth of radius R. The height of the satellite above the surface of the earth is

Options

A. R
B. R/2
C. 3 R
D. 6 R

Explanation

To solve the problem step by step, we will analyze the motion of the artificial satellite and apply the relevant formulas.

### Step 1: Understand the given information
- The speed of the satellite (V) is half of the escape velocity (V_escape) from the Earth.
- The radius of the Earth (R) is given.
- We need to find the height (h) of the satellite above the Earth's surface.

### Step 2: Write the formula for escape velocity
The escape velocity (V_escape) from the surface of the Earth is given by:
\[ V_{\text{escape}} = \sqrt{2gR} \]
where \( g \) is the acceleration due to gravity at the Earth's surface.

### Step 3: Calculate the speed of the satellite
Since the speed of the satellite is half the escape velocity:
\[ V = \frac{1}{2} V_{\text{escape}} = \frac{1}{2} \sqrt{2gR} = \frac{\sqrt{gR}}{\sqrt{2}} \]

### Step 4: Write the formula for orbital speed
The orbital speed (V) of a satellite in a circular orbit at a distance \( r \) from the center of the Earth is given by:
\[ V = \sqrt{\frac{GM}{r}} \]
where \( G \) is the gravitational constant and \( M \) is the mass of the Earth.

### Step 5: Relate the two speeds
Since both expressions represent the speed of the satellite, we can set them equal to each other:
\[ \frac{\sqrt{gR}}{\sqrt{2}} = \sqrt{\frac{GM}{r}} \]

### Step 6: Substitute \( g \) in terms of \( G \) and \( M \)
We know that:
\[ g = \frac{GM}{R^2} \]
Substituting this into the equation gives:
\[ \frac{\sqrt{\frac{GM}{R^2} R}}{\sqrt{2}} = \sqrt{\frac{GM}{r}} \]
This simplifies to:
\[ \frac{\sqrt{GM}}{\sqrt{2R}} = \sqrt{\frac{GM}{r}} \]

### Step 7: Square both sides
Squaring both sides results in:
\[ \frac{GM}{2R} = \frac{GM}{r} \]

### Step 8: Cancel \( GM \) and rearrange
Cancelling \( GM \) from both sides (assuming \( GM \neq 0 \)):
\[ \frac{1}{2R} = \frac{1}{r} \]
Cross-multiplying gives:
\[ r = 2R \]

### Step 9: Find the height of the satellite
The height \( h \) of the satellite above the Earth's surface is given by:
\[ h = r - R \]
Substituting \( r = 2R \):
\[ h = 2R - R = R \]

### Final Answer
The height of the satellite above the surface of the Earth is:
\[ \boxed{R} \]