A satellite revolves in a circular orbit with speed, `V=1/(sqrt(3))V_(e)`. If satellite is suddenly stopped and allowed to fall freely onto the earth, the speed with which it hits the earth's surface is

Options

A. `sqrt(gR)`
B. `sqrt((gR)/(3)`
C. `sqrt(2gR)`
D. `sqrt((2)/(3)gR)`

Explanation

To solve the problem, we need to find the speed with which a satellite, initially revolving in a circular orbit with a speed of \( V = \frac{1}{\sqrt{3}} V_e \) (where \( V_e \) is the escape velocity), will hit the Earth's surface after being suddenly stopped and allowed to fall freely.

### Step-by-Step Solution:

1. **Understanding the Given Velocity**:
The satellite's velocity is given as:
\[ V = \frac{1}{\sqrt{3}} V_e \]
The escape velocity \( V_e \) is given by:
\[ V_e = \sqrt{\frac{2GM}{R}} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

2. **Finding the Orbital Velocity**:
The orbital velocity \( V_o \) for a satellite in a circular orbit is given by:
\[ V_o = \sqrt{\frac{GM}{r}} \]
where \( r \) is the distance from the center of the Earth to the satellite.

Since we know that \( V = \frac{1}{\sqrt{3}} V_e \), we can express it in terms of \( V_o \):
\[ V = \frac{1}{\sqrt{3}} \sqrt{\frac{2GM}{R}} = \sqrt{\frac{2}{3}} \sqrt{\frac{GM}{R}} \]

3. **Relating Orbital Velocity to Escape Velocity**:
We know that the escape velocity is related to the orbital velocity:
\[ V_e = \sqrt{2} V_o \]
Thus, we can express \( V \) in terms of \( V_o \):
\[ V = \frac{1}{\sqrt{3}} \sqrt{2} V_o = \sqrt{\frac{2}{3}} V_o \]

4. **Finding the Distance of the Satellite from the Earth's Center**:
Since \( V_o = \sqrt{\frac{GM}{r}} \), we can find \( r \):
\[ V^2 = \frac{2}{3} V_o^2 \Rightarrow \frac{2GM}{r} = \frac{2}{3} \frac{GM}{R} \]
This implies:
\[ r = \frac{3R}{2} \]
Therefore, the satellite is at a distance of \( \frac{3R}{2} \) from the center of the Earth.

5. **Calculating Potential Energy**:
The potential energy \( U \) at a distance \( r \) is given by:
\[ U = -\frac{GMm}{r} \]
At \( r = \frac{3R}{2} \):
\[ U_i = -\frac{GMm}{\frac{3R}{2}} = -\frac{2GMm}{3R} \]

6. **Final Potential Energy at Earth's Surface**:
When the satellite reaches the Earth's surface (at distance \( R \)):
\[ U_f = -\frac{GMm}{R} \]

7. **Using Conservation of Energy**:
The change in potential energy will equal the kinetic energy gained:
\[ U_f - U_i = \frac{1}{2} mv^2 \]
Substituting the values:
\[ -\frac{GMm}{R} - \left(-\frac{2GMm}{3R}\right) = \frac{1}{2} mv^2 \]
Simplifying:
\[ -\frac{GMm}{R} + \frac{2GMm}{3R} = \frac{1}{2} mv^2 \]
\[ -\frac{3GMm}{3R} + \frac{2GMm}{3R} = \frac{1}{2} mv^2 \]
\[ -\frac{GMm}{3R} = \frac{1}{2} mv^2 \]

8. **Solving for \( v \)**:
Cancel \( m \) from both sides:
\[ -\frac{GM}{3R} = \frac{1}{2} v^2 \]
\[ v^2 = \frac{2GM}{3R} \]
Taking the square root:
\[ v = \sqrt{\frac{2GM}{3R}} \]

9. **Expressing in terms of \( g \)**:
We know \( g = \frac{GM}{R^2} \), thus:
\[ v = \sqrt{\frac{2gR}{3}} \]

### Final Answer:
The speed with which the satellite hits the Earth's surface is:
\[ v = \sqrt{\frac{2gR}{3}} \]