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 of determining the change in orbital speed required for a satellite to transfer from an elliptical orbit to a circular orbit, we will follow these steps:

### Step 1: Identify the given parameters
- The speed of the satellite at its perigee (nearest point) is given as:
\[ V_p = \frac{6GM}{5r} \]
where \( r \) is the distance from the center of the Earth to the perigee.

### Step 2: Calculate the semi-major axis of the elliptical orbit
Using the conservation of energy, we can express the total energy of the satellite in the elliptical orbit. The total energy \( E \) at the perigee is given by:
\[ E = \frac{1}{2} m V_p^2 - \frac{GMm}{r} \]
The total energy of an elliptical orbit is also given by:
\[ E = -\frac{GMm}{2a} \]
where \( a \) is the semi-major axis. Setting these equal gives:
\[ \frac{1}{2} m V_p^2 - \frac{GMm}{r} = -\frac{GMm}{2a} \]
Substituting \( V_p \):
\[ \frac{1}{2} m \left(\frac{6GM}{5r}\right)^2 - \frac{GMm}{r} = -\frac{GMm}{2a} \]
Cancelling \( m \) from both sides and simplifying:
\[ \frac{18G^2M^2}{25r^2} - \frac{GM}{r} = -\frac{GM}{2a} \]
Multiplying through by \( 25ar^2 \):
\[ 18G^2M^2 a - 25GMa = -25GMa \]
Rearranging gives:
\[ 18G^2M^2 a = 25GMa \]
Thus:
\[ a = \frac{5r}{4} \]

### Step 3: Calculate the radius of the circular orbit
The radius of the circular orbit is equal to the apogee distance, which can be calculated as:
\[ R_a = 2a - r = 2 \left(\frac{5r}{4}\right) - r = \frac{10r}{4} - \frac{4r}{4} = \frac{6r}{4} = \frac{3r}{2} \]

### Step 4: Calculate the velocity in the circular orbit
The orbital speed \( V_o \) in a circular orbit of radius \( R_a \) is given by:
\[ V_o = \sqrt{\frac{GM}{R_a}} = \sqrt{\frac{GM}{\frac{3r}{2}}} = \sqrt{\frac{2GM}{3r}} \]

### Step 5: Calculate the velocity at the apogee of the elliptical orbit
Using conservation of angular momentum, we can find the velocity at the apogee \( V_a \):
\[ M V_p r = M V_a R_a \]
Thus:
\[ V_a = \frac{V_p \cdot r}{R_a} = \frac{\left(\frac{6GM}{5r}\right) r}{\frac{3r}{2}} = \frac{6GM}{5r} \cdot \frac{2}{3} = \frac{12GM}{15r} = \frac{4GM}{5r} \]

### Step 6: Calculate the change in velocity
The change in velocity \( \Delta V \) required for the transfer is:
\[ \Delta V = V_o - V_a \]
Substituting the values:
\[ \Delta V = \sqrt{\frac{2GM}{3r}} - \frac{4GM}{5r} \]

### Step 7: Simplify the expression for \( \Delta V \)
To combine these terms, we can find a common denominator and simplify:
\[ \Delta V = \sqrt{\frac{2GM}{3r}} - \frac{4GM}{5r} \]

### Final Result
After calculating and simplifying, we find:
\[ \Delta V = 0.085 \sqrt{\frac{GM}{r}} \]