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. 2R
C. 3 R
D. 4 R

Answer

To solve the problem of finding the height of an artificial satellite above the surface of the Earth, we can follow these steps:

### Step 1: Understand the relationship between escape velocity and orbital velocity
The escape velocity \( V_e \) from the Earth is given by the formula:
\[ V_e = \sqrt{\frac{2GM}{R}} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

The orbital velocity \( V \) of a satellite in a circular orbit at a distance \( r \) from the center of the Earth is given by:
\[ V = \sqrt{\frac{GM}{r}} \]

### Step 2: Relate the orbital velocity to the escape velocity
According to the problem, the speed of the satellite is half the escape velocity:
\[ V = \frac{1}{2} V_e \]
Substituting the expression for escape velocity:
\[ V = \frac{1}{2} \sqrt{\frac{2GM}{R}} = \frac{1}{2} \sqrt{2} \sqrt{\frac{GM}{R}} = \frac{1}{\sqrt{2}} \sqrt{\frac{GM}{R}} \]

### Step 3: Set up the equation for orbital velocity
Since the orbital velocity is also given by:
\[ V = \sqrt{\frac{GM}{r}} \]
we can equate the two expressions for \( V \):
\[ \frac{1}{\sqrt{2}} \sqrt{\frac{GM}{R}} = \sqrt{\frac{GM}{r}} \]

### Step 4: Square both sides to eliminate the square roots
Squaring both sides gives:
\[ \frac{1}{2} \frac{GM}{R} = \frac{GM}{r} \]

### Step 5: Cancel out common terms
We can cancel \( GM \) from both sides (assuming \( GM \neq 0 \)):
\[ \frac{1}{2R} = \frac{1}{r} \]

### Step 6: Solve for \( r \)
Cross-multiplying gives:
\[ r = 2R \]

### Step 7: Find the height \( h \) above the surface of the Earth
The orbital radius \( r \) is related to the height \( h \) above the surface of the Earth by:
\[ r = R + h \]
Substituting \( r = 2R \) into this equation:
\[ 2R = R + h \]

### Step 8: Solve for \( h \)
Rearranging gives:
\[ h = 2R - R = R \]

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

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