A satellite is revolving in a circular orbit at a height 'h' from the earth's surface (radius of earth R, h ltltR). The minimum increase in its orbital velocity required, So that the satellite could escape from the erth's gravitational field, is close to :(Neglect the effect of atomsphere.)

Options

A. `sqrt(gR//2)`
B. `sqrt(gR(sqrt(2-1)))`
C. `sqrt(2gR)`
D. `sqrt(gR)`

Explanation

To solve the problem, we need to find the minimum increase in the orbital velocity of a satellite so that it can escape from the Earth's gravitational field. Let's break it down step by step:

### Step 1: Understand the Orbital Velocity
The orbital velocity \( v_o \) of a satellite in a circular orbit at a height \( h \) above the Earth's surface is given by the formula:
\[ v_o = \sqrt{\frac{GM}{R + h}} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

### Step 2: Understand the Escape Velocity
The escape velocity \( v_e \) from the surface of the Earth is given by:
\[ v_e = \sqrt{\frac{2GM}{R}} \]
However, since the satellite is at a height \( h \), the escape velocity from that height is:
\[ v_e = \sqrt{\frac{2GM}{R + h}} \]

### Step 3: Calculate the Minimum Increase in Velocity
To find the minimum increase in the orbital velocity required for the satellite to escape, we need to find the difference between the escape velocity and the orbital velocity:
\[ \Delta v = v_e - v_o \]
Substituting the expressions for \( v_e \) and \( v_o \):
\[ \Delta v = \sqrt{\frac{2GM}{R + h}} - \sqrt{\frac{GM}{R + h}} \]
Factoring out \( \sqrt{\frac{GM}{R + h}} \):
\[ \Delta v = \sqrt{\frac{GM}{R + h}} \left( \sqrt{2} - 1 \right) \]

### Step 4: Simplifying the Expression
Since \( h \) is much smaller than \( R \) (i.e., \( h \ll R \)), we can approximate \( R + h \approx R \):
\[ \Delta v \approx \sqrt{\frac{GM}{R}} \left( \sqrt{2} - 1 \right) \]
Recognizing that \( \sqrt{\frac{GM}{R}} \) is the orbital velocity at the surface of the Earth, we can express it as:
\[ \Delta v \approx v_o \left( \sqrt{2} - 1 \right) \]

### Conclusion
Thus, the minimum increase in the orbital velocity required for the satellite to escape from the Earth's gravitational field is approximately:
\[ \Delta v \approx v_o \left( \sqrt{2} - 1 \right) \]