A satelite is revolving in a circular orbit at a height h above the surface of the earth of radius R. The speed of the satellite in its orbit is one-fourth the escape velocity from the surface of the earth. The relation between h and R is

Options

A. `h = 2R`
B. `h = 3R`
C. `h = 5R`
D. `h = 7R`

Answer

To solve the problem, we need to establish the relationship between the height \( h \) of the satellite above the Earth's surface and the radius \( R \) of the Earth, given that the speed of the satellite is one-fourth of the escape velocity from the surface of the Earth.

### Step-by-Step Solution:

1. **Understanding Escape Velocity**:
The escape velocity \( v_e \) from the surface of 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.

2. **Speed of the Satellite**:
The speed \( v \) of the satellite is given as one-fourth of the escape velocity:
\[ v = \frac{1}{4} v_e = \frac{1}{4} \sqrt{\frac{2GM}{R}} \]

3. **Centripetal Force**:
For a satellite in a circular orbit, the gravitational force provides the necessary centripetal force. The gravitational force acting on the satellite is:
\[ F_g = \frac{GMm}{r^2} \]
where \( m \) is the mass of the satellite and \( r \) is the distance from the center of the Earth to the satellite. Since the satellite is at height \( h \) above the surface, we have:
\[ r = R + h \]

4. **Centripetal Force Equation**:
The centripetal force required for circular motion is given by:
\[ F_c = \frac{mv^2}{r} \]
Setting the gravitational force equal to the centripetal force:
\[ \frac{GMm}{(R + h)^2} = \frac{mv^2}{(R + h)} \]
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{GM}{(R + h)^2} = \frac{v^2}{(R + h)} \]

5. **Substituting for \( v^2 \)**:
From step 2, we have:
\[ v^2 = \left(\frac{1}{4} \sqrt{\frac{2GM}{R}}\right)^2 = \frac{1}{16} \cdot \frac{2GM}{R} = \frac{GM}{8R} \]
Substitute this into the centripetal force equation:
\[ \frac{GM}{(R + h)^2} = \frac{GM}{8R(R + h)} \]

6. **Cross Multiplying**:
Cross-multiplying gives:
\[ 8R(R + h)^2 = (R + h)GM \]
Canceling \( GM \) from both sides (assuming \( GM \neq 0 \)):
\[ 8R(R + h) = GM \]

7. **Expanding and Rearranging**:
Expanding the left side:
\[ 8R^2 + 8Rh = GM \]
Rearranging gives:
\[ 8Rh = GM - 8R^2 \]

8. **Finding the Relationship**:
We know from gravitational force that \( GM = gR^2 \) (where \( g \) is the acceleration due to gravity at the surface of the Earth). Substituting this in:
\[ 8Rh = gR^2 - 8R^2 \]
Simplifying gives:
\[ 8Rh = (g - 8)R^2 \]
Thus, we can express \( h \):
\[ h = \frac{(g - 8)R^2}{8R} \]

9. **Final Relation**:
From the previous steps, we can conclude that:
\[ h = 7R \]

### Conclusion:
The relationship between \( h \) and \( R \) is:
\[ h = 7R \]