A satellite is revolving round the earth at a height (R/2) from surface of earth in circular orbit. The change ini speed required for the satellite to excape to infinity for the satellite to escape to infinity is: [Take R as radius of earth and M as mass of earth]

Options

A. `sqrt((2GM)/(3R))`
B. `sqrt((GM)/(3R))(2-sqrt(2))`
C. `sqrt((GM)/(3R))(sqrt(2)-1)`
D. `sqrt((4GM)/(3R))`

Explanation

To solve the problem of determining the change in speed required for a satellite to escape to infinity from its circular orbit at a height of \( \frac{R}{2} \) above the Earth's surface, we can follow these steps:

### Step 1: Determine the distance from the center of the Earth
The radius of the Earth is \( R \). The satellite is at a height of \( \frac{R}{2} \) above the surface, so the distance \( r \) from the center of the Earth to the satellite is:
\[ r = R + \frac{R}{2} = \frac{3R}{2} \]

### Step 2: Calculate the orbital speed of the satellite
The orbital speed \( V_1 \) of the satellite in a circular orbit is given by the formula:
\[ V_1 = \sqrt{\frac{GM}{r}} \]
Substituting \( r = \frac{3R}{2} \):
\[ V_1 = \sqrt{\frac{GM}{\frac{3R}{2}}} = \sqrt{\frac{2GM}{3R}} \]

### Step 3: Determine the escape velocity from the orbit
The escape velocity \( V \) from a distance \( r \) from the center of the Earth is given by:
\[ V = \sqrt{\frac{2GM}{r}} \]
Again substituting \( r = \frac{3R}{2} \):
\[ V = \sqrt{\frac{2GM}{\frac{3R}{2}}} = \sqrt{\frac{4GM}{3R}} \]

### Step 4: Calculate the change in speed required
The change in speed \( \Delta V \) required for the satellite to escape to infinity is the difference between the escape velocity and the orbital speed:
\[ \Delta V = V - V_1 \]
Substituting the expressions for \( V \) and \( V_1 \):
\[ \Delta V = \sqrt{\frac{4GM}{3R}} - \sqrt{\frac{2GM}{3R}} \]
Factoring out \( \sqrt{\frac{GM}{3R}} \):
\[ \Delta V = \sqrt{\frac{GM}{3R}} \left( \sqrt{4} - \sqrt{2} \right) = \sqrt{\frac{GM}{3R}} \left( 2 - \sqrt{2} \right) \]

### Final Result
Thus, the change in speed required for the satellite to escape to infinity is:
\[ \Delta V = \sqrt{\frac{GM}{3R}} (2 - \sqrt{2}) \]