A satellite of mass `m` is in an elliptical orbit around the earth. The speed of the satellite at its nearest position is `(6GM)//(5r)` where r is the perigee (nearest point) distance from the centre of the earth. It is desired to transfer the satellite to the circular orbit of radius equal to its apogee (farthest point) distance from the centre of the earth. The change in orbital speed required for this purpose is

Options

A. `0.35sqrt((GM_(e))/r)`
B. `0.085sqrt((GM_(e))/r)`
C. `sqrt((2GM_(3))/r)`
D. zero

Explanation

To solve the problem, we will follow these steps:

1. **Understanding the Given Information**:
- The satellite is in an elliptical orbit around the Earth.
- The speed of the satellite at its perigee (nearest point) is given as \( v_p = \sqrt{\frac{6GM}{5r}} \), where \( r \) is the distance from the center of the Earth to the perigee.

2. **Conservation of Energy**:
- The total mechanical energy \( E \) of the satellite in its elliptical orbit can be expressed as:
\[ E = \frac{1}{2}mv^2 - \frac{GMm}{r} \]
- At the perigee, substituting \( v = v_p \):
\[ E = \frac{1}{2}m\left(\sqrt{\frac{6GM}{5r}}\right)^2 - \frac{GMm}{r} \]
- Simplifying this:
\[ E = \frac{1}{2}m \cdot \frac{6GM}{5r} - \frac{GMm}{r} = \frac{3GMm}{5r} - \frac{GMm}{r} = \frac{3GMm}{5r} - \frac{5GMm}{5r} = -\frac{2GMm}{5r} \]

3. **Relating Semi-Major Axis**:
- The total energy can also be expressed as:
\[ E = -\frac{GMm}{2a} \]
- Setting the two expressions for energy equal:
\[ -\frac{2GMm}{5r} = -\frac{GMm}{2a} \]
- Canceling \( GMm \) from both sides:
\[ \frac{2}{5r} = \frac{1}{2a} \implies 2a = \frac{5r}{2} \implies a = \frac{5r}{4} \]

4. **Finding the Apogee Distance**:
- The distance to the apogee \( R \) can be found using the relationship:
\[ R = 2a - r = 2\left(\frac{5r}{4}\right) - r = \frac{10r}{4} - \frac{4r}{4} = \frac{6r}{4} = \frac{3r}{2} \]

5. **Conservation of Angular Momentum**:
- At the perigee:
\[ L = mv_p \cdot r = m \cdot \sqrt{\frac{6GM}{5r}} \cdot r \]
- At the apogee:
\[ L = mv_R \cdot R = mv_R \cdot \frac{3r}{2} \]
- Setting these equal:
\[ \sqrt{\frac{6GM}{5r}} \cdot r = v_R \cdot \frac{3r}{2} \]
- Canceling \( r \) from both sides:
\[ \sqrt{\frac{6GM}{5}} = v_R \cdot \frac{3}{2} \implies v_R = \frac{2}{3}\sqrt{\frac{6GM}{5}} \]

6. **Finding the Required Circular Orbit Velocity**:
- The required velocity \( V_0 \) for a circular orbit at distance \( R = \frac{3r}{2} \) is given by:
\[ V_0 = \sqrt{\frac{GM}{R}} = \sqrt{\frac{GM}{\frac{3r}{2}}} = \sqrt{\frac{2GM}{3r}} \]

7. **Calculating the Change in Velocity**:
- The change in velocity \( \Delta V \) required to transfer to the circular orbit:
\[ \Delta V = V_R - V_0 = \frac{2}{3}\sqrt{\frac{6GM}{5}} - \sqrt{\frac{2GM}{3}} \]

8. **Final Calculation**:
- To simplify \( \Delta V \):
\[ \Delta V = \frac{2}{3}\sqrt{\frac{6GM}{5}} - \sqrt{\frac{2GM}{3}} \]
- This can be further simplified to find the numerical value.