An artificial satellite is moving in a circular orbit around the earth with a speed equal to half the magnitude of escape velocity from the surface of earth. R is the radius of earth and g is acceleration due to gravity at the surface of earth. (R=6400 km).
The time period of revolution of satellite in the given orbit is

Options

A. `2pi sqrt((2R)/(g))`
B. `2pi sqrt((4R)/(g))`
C. `2pi sqrt((8R)/(g))`
D. `2pi sqrt((6R)/(g))`

Answer

To solve the problem, we need to find the time period of revolution of a satellite moving in a circular orbit around the Earth, given that its speed is half the escape velocity from the surface of the Earth.

### Step-by-Step Solution:

1. **Understanding Escape Velocity**:
The escape velocity (\(v_{escape}\)) from the surface of the Earth is given by the formula:
\[ v_{escape} = \sqrt{2gR} \]
where \(g\) is the acceleration due to gravity at the surface of the Earth and \(R\) is the radius of the Earth.

2. **Given Speed of Satellite**:
The speed of the satellite (\(v\)) is half of the escape velocity:
\[ v = \frac{1}{2} v_{escape} = \frac{1}{2} \sqrt{2gR} = \frac{\sqrt{2gR}}{2} \]

3. **Centripetal Force and Gravitational Force**:
For a satellite in circular motion, the gravitational force provides the necessary centripetal force. Thus, we have:
\[ F_g = F_c \]
where:
- Gravitational force, \(F_g = \frac{GMm}{(R+h)^2}\)
- Centripetal force, \(F_c = \frac{mv^2}{R+h}\)

Here, \(M\) is the mass of the Earth, \(m\) is the mass of the satellite, and \(h\) is the height above the Earth's surface.

4. **Setting Up the Equation**:
Since \(F_g = F_c\), we can write:
\[ \frac{GMm}{(R+h)^2} = \frac{mv^2}{R+h} \]
Canceling \(m\) from both sides and multiplying through by \((R+h)\):
\[ \frac{GM}{(R+h)} = v^2 \]

5. **Substituting for \(v\)**:
Now substitute \(v = \frac{\sqrt{2gR}}{2}\):
\[ \frac{GM}{(R+h)} = \left(\frac{\sqrt{2gR}}{2}\right)^2 \]
Simplifying the right side:
\[ \frac{GM}{(R+h)} = \frac{2gR}{4} = \frac{gR}{2} \]

6. **Rearranging to Find \(h\)**:
Rearranging gives:
\[ GM = \frac{gR}{2}(R+h) \]
Substituting \(GM = gR^2\) (from the gravitational force formula):
\[ gR^2 = \frac{gR}{2}(R+h) \]
Canceling \(gR\) (assuming \(g \neq 0\)):
\[ R = \frac{1}{2}(R+h) \]
Solving for \(h\):
\[ 2R = R + h \implies h = R \]

7. **Finding the Radius of Orbit**:
The radius of the orbit \(r\) is:
\[ r = R + h = R + R = 2R \]

8. **Calculating the Time Period**:
The time period \(T\) of revolution is given by:
\[ T = \frac{2\pi r}{v} \]
Substituting \(r = 2R\) and \(v = \frac{\sqrt{2gR}}{2}\):
\[ T = \frac{2\pi (2R)}{\frac{\sqrt{2gR}}{2}} = \frac{4\pi R}{\frac{\sqrt{2gR}}{2}} = \frac{8\pi R}{\sqrt{2gR}} = \frac{8\pi \sqrt{R}}{\sqrt{2g}} \]

9. **Final Simplification**:
Simplifying further:
\[ T = 2\pi \sqrt{\frac{2R}{g}} \]

### Final Answer:
The time period of revolution of the satellite in the given orbit is:
\[ T = 2\pi \sqrt{\frac{2R}{g}} \]