A satellite of mass m is in an elliptical orbit around the earth of mass `M(M gt gt m)` the speed of the satellite at its nearest point ot the earth (perigee) is `sqrt((6GM)/(5R))` where R= its closest distance to the earth it is desired to transfer this satellite into a circular orbit around the earth of radius equal its largest distance from the earth. Find the increase in its speed to be imparted at the apogee (farthest point on the elliptical orbit).

Explanation

To solve the problem of finding the increase in speed to be imparted at the apogee of a satellite in an elliptical orbit around the Earth, we will follow these steps:

### Step 1: Identify the given parameters
- Mass of the satellite: \( m \)
- Mass of the Earth: \( M \)
- Closest distance to the Earth (perigee): \( R \)
- Speed at perigee: \( v_1 = \sqrt{\frac{6GM}{5R}} \)

### Step 2: Determine the apogee distance
In an elliptical orbit, the largest distance from the Earth (apogee) can be denoted as \( r_a \). For the given elliptical orbit, we need to find this distance. The semi-major axis \( a \) and semi-minor axis \( b \) can be related to the perigee and apogee distances.

Using the relationship:
\[ r_a = 2a - R \]
However, we need to find \( r_a \) directly from the given parameters.

### Step 3: Use conservation of angular momentum
The angular momentum \( L \) at perigee can be expressed as:
\[ L = m \cdot v_1 \cdot R \]
At apogee, the angular momentum can be expressed as:
\[ L = m \cdot v_2 \cdot r_a \]
Setting these equal due to conservation of angular momentum:
\[ m \cdot v_1 \cdot R = m \cdot v_2 \cdot r_a \]
This simplifies to:
\[ v_1 \cdot R = v_2 \cdot r_a \]
From this, we can express \( v_2 \):
\[ v_2 = \frac{v_1 \cdot R}{r_a} \]

### Step 4: Calculate the total mechanical energy
The total mechanical energy \( E \) in an elliptical orbit is given by:
\[ E = K + U = \frac{1}{2} m v_1^2 - \frac{GMm}{R} \]
At apogee, the total mechanical energy will be:
\[ E = \frac{1}{2} m v_2^2 - \frac{GMm}{r_a} \]

### Step 5: Set up the energy conservation equation
Setting the energies equal gives:
\[ \frac{1}{2} m v_1^2 - \frac{GMm}{R} = \frac{1}{2} m v_2^2 - \frac{GMm}{r_a} \]
Cancelling \( m \) and rearranging:
\[ \frac{1}{2} v_1^2 + \frac{GM}{R} = \frac{1}{2} v_2^2 + \frac{GM}{r_a} \]

### Step 6: Solve for \( v_2 \)
Substituting \( v_1 = \sqrt{\frac{6GM}{5R}} \) into the equation and solving for \( v_2 \):
1. Calculate \( v_1^2 \):
\[ v_1^2 = \frac{6GM}{5R} \]
2. Substitute into the energy equation:
\[ \frac{1}{2} \cdot \frac{6GM}{5R} + \frac{GM}{R} = \frac{1}{2} v_2^2 + \frac{GM}{r_a} \]
3. Solve for \( v_2 \).

### Step 7: Calculate the increase in speed
The increase in speed \( \Delta v \) required at apogee is given by:
\[ \Delta v = v_{\text{final}} - v_2 \]
Where \( v_{\text{final}} \) is the speed required for a circular orbit at radius \( r_a \).

### Step 8: Final expression for increase in speed
By substituting the values and simplifying, we can find the exact increase in speed required.