A spaceship is returning to Earth with its engine turned off. Consider only the gravitational field of Earth and let M be the mass of Earth, m be the mass of the spaceship, and R be the radius of Earth. In moving from position 1 to position 2 the kinetic energy of the spaceship increases by:

Options

A. `GMm//R_2`
B. `GMm//R_2^2`
C. `GMm(R_1-R_2)//R_1^2`
D. `GMm(R_1-R_2)//R_1R_2`

Explanation

To solve the problem of how the kinetic energy of a spaceship increases as it moves from position 1 to position 2 while returning to Earth, we will apply the principle of conservation of energy. Here are the steps to derive the solution:

### Step 1: Understand the Conservation of Energy
The total mechanical energy (kinetic energy + potential energy) of the spaceship remains constant if only gravitational forces are acting on it. Thus, any change in potential energy will result in an equal and opposite change in kinetic energy.

### Step 2: Define Potential Energy
The gravitational potential energy (U) of the spaceship at a distance \( r \) from the center of the Earth is given by the formula:
\[ U = -\frac{GMm}{r} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( m \) is the mass of the spaceship.

### Step 3: Calculate Potential Energy at Positions 1 and 2
Let \( R \) be the radius of the Earth, \( R_1 \) be the distance from the center of the Earth at position 1, and \( R_2 \) be the distance at position 2. The potential energies at these positions are:
- At position 1:
\[ U_1 = -\frac{GMm}{R_1} \]
- At position 2:
\[ U_2 = -\frac{GMm}{R_2} \]

### Step 4: Find the Change in Potential Energy
The change in potential energy (\( \Delta U \)) as the spaceship moves from position 1 to position 2 is:
\[ \Delta U = U_2 - U_1 = -\frac{GMm}{R_2} + \frac{GMm}{R_1} \]
This simplifies to:
\[ \Delta U = GMm \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

### Step 5: Relate Change in Potential Energy to Change in Kinetic Energy
According to the conservation of energy, the change in kinetic energy (\( \Delta K \)) is equal to the negative change in potential energy:
\[ \Delta K = -\Delta U = -GMm \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Thus, we can express the increase in kinetic energy as:
\[ \Delta K = GMm \left( \frac{1}{R_2} - \frac{1}{R_1} \right) \]

### Step 6: Final Expression for Kinetic Energy Increase
The increase in kinetic energy as the spaceship moves from position 1 to position 2 is:
\[ \Delta K = GMm \left( \frac{1}{R_2} - \frac{1}{R_1} \right) \]

### Conclusion
The kinetic energy of the spaceship increases by the amount:
\[ \Delta K = GMm \left( \frac{1}{R_2} - \frac{1}{R_1} \right) \]