If an artificial satellite is moving in a circular orbit around earth with speed equal to one fourth of `V_(e)` from earth, then height of the satellite above the surface of the earth is

Options

A. `7R`
B. `4R`
C. `3R`
D. `R`

Explanation

To solve the problem, we need to find the height of an artificial satellite above the surface of the Earth when it is moving in a circular orbit with a speed equal to one fourth of the escape velocity from the Earth.

### Step-by-Step Solution:

1. **Understand the given information:**
- The speed of the satellite, \( V_0 = \frac{1}{4} V_e \)
- The escape velocity from the Earth, \( V_e = \sqrt{\frac{2GM}{R}} \)
- \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

2. **Relate the orbital speed to the gravitational force:**
The orbital speed \( V \) of a satellite in a circular orbit is given by:
\[ V = \sqrt{\frac{GM}{r + h}} \]
where \( r \) is the radius of the Earth and \( h \) is the height of the satellite above the Earth's surface.

3. **Substitute the expression for escape velocity:**
Since \( V_0 = \frac{1}{4} V_e \), we can substitute for \( V_e \):
\[ V_0 = \frac{1}{4} \sqrt{\frac{2GM}{R}} \]

4. **Set the two expressions for speed equal:**
Equating the two expressions for speed:
\[ \frac{1}{4} \sqrt{\frac{2GM}{R}} = \sqrt{\frac{GM}{r + h}} \]

5. **Square both sides to eliminate the square roots:**
\[ \left(\frac{1}{4}\right)^2 \cdot \frac{2GM}{R} = \frac{GM}{r + h} \]
Simplifying gives:
\[ \frac{1}{16} \cdot \frac{2GM}{R} = \frac{GM}{r + h} \]

6. **Cancel \( GM \) from both sides:**
\[ \frac{2}{16R} = \frac{1}{r + h} \]
Simplifying further:
\[ \frac{1}{8R} = \frac{1}{r + h} \]

7. **Cross-multiply to solve for \( r + h \):**
\[ r + h = 8R \]

8. **Isolate \( h \):**
\[ h = 8R - r \]
Since \( r = R \) (the radius of the Earth), we have:
\[ h = 8R - R = 7R \]

9. **Convert height to the surface of the Earth:**
The height of the satellite above the surface of the Earth is:
\[ h = 7R \]

### Final Answer:
The height of the satellite above the surface of the Earth is \( 7R \).