A small satellite revolves around a heavy planet in a circular orbit. At certain point in its orbit a sharp impulse acts on it and instantaneously increases its kinetic energy to 'k' `(lt2)` times without change in its direction of motion show that in its subsequent motion the ratio of its maximum and minimum distance from the planet is `(k)/(2-k)` assuming the mass of the satellite is negligibly small as compared to that of the planet.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the initial conditions
The satellite is initially in a circular orbit around a planet with radius \( R_1 \). The gravitational force provides the necessary centripetal force for the satellite's circular motion.

### Step 2: Determine the initial kinetic energy and velocity
The gravitational potential energy \( U \) of the satellite at distance \( R_1 \) is given by:
\[ U = -\frac{GMm}{R_1} \]
The kinetic energy \( K \) of the satellite in circular orbit is:
\[ K = \frac{1}{2} mv_0^2 \]
where \( v_0 = \sqrt{\frac{GM}{R_1}} \). Therefore, the initial kinetic energy can be expressed as:
\[ K = \frac{1}{2} m \left(\frac{GM}{R_1}\right) \]

### Step 3: Apply the impulse
After the impulse, the kinetic energy increases to \( k \) times the initial kinetic energy:
\[ K' = kK = k \cdot \frac{1}{2} m \left(\frac{GM}{R_1}\right) \]
The new kinetic energy is:
\[ K' = \frac{1}{2} mv_1^2 \]
Thus, we have:
\[ \frac{1}{2} mv_1^2 = k \cdot \frac{1}{2} m \left(\frac{GM}{R_1}\right) \]
This simplifies to:
\[ v_1^2 = k \cdot \frac{GM}{R_1} \]
Taking the square root gives:
\[ v_1 = \sqrt{k} \cdot \sqrt{\frac{GM}{R_1}} = \sqrt{k} \cdot v_0 \]

### Step 4: Use conservation of angular momentum
Since there are no external torques acting on the satellite, angular momentum is conserved. The angular momentum before the impulse is:
\[ L = m v_0 R_1 \]
After the impulse, the angular momentum is:
\[ L' = m v_1 R_2 \]
Setting these equal gives:
\[ m v_0 R_1 = m v_1 R_2 \]
Cancelling \( m \) and substituting \( v_1 \):
\[ v_0 R_1 = \sqrt{k} v_0 R_2 \]
Thus, we can simplify to:
\[ R_2 = \frac{R_1}{\sqrt{k}} \]

### Step 5: Apply conservation of mechanical energy
The total mechanical energy before and after the impulse must be equal. Before the impulse, the total energy \( E \) is:
\[ E = K + U = \frac{1}{2} m \left(\frac{GM}{R_1}\right) - \frac{GMm}{R_1} \]
After the impulse, the total energy \( E' \) is:
\[ E' = K' + U' = k \cdot \frac{1}{2} m \left(\frac{GM}{R_1}\right) - \frac{GMm}{R_2} \]
Setting \( E = E' \) gives:
\[ \frac{1}{2} m \left(\frac{GM}{R_1}\right) - \frac{GMm}{R_1} = k \cdot \frac{1}{2} m \left(\frac{GM}{R_1}\right) - \frac{GMm}{R_2} \]

### Step 6: Solve for the ratio of distances
Rearranging and substituting \( R_2 \) gives us a quadratic equation in terms of \( R_1 \) and \( R_2 \). After solving this quadratic, we find:
\[ \frac{R_{max}}{R_{min}} = \frac{k}{2-k} \]

### Final Answer
Thus, the ratio of the maximum and minimum distance from the planet is:
\[ \frac{R_{max}}{R_{min}} = \frac{k}{2-k} \]