The satellites when launched from the earth are not given the orbital velocity initially, a multistage rocket propeller carries the spacecraft up to its orbit and during each stage rocket has been fired to increase the velocity to acquire the desired velocity for a particular orbit. The last stage of the rocket brings the satellite in circular/elliptical (desired) orbit.
Consider a satellite of mass `150 kg` in a low circular orbit. In this orbit, we cannot neglect the effect of air drag. This air opposes the motion of satellite and hence the total mechanical energy of earth-satellite system decreases. That means the total energy becomes more negative and hence the orbital radius decreases which causes the increase in `KE` When the satellite comes in the low enough orbit, excessive thermal energy generation due to air friction may cause the satellite to burn up. Based on the above information, answer the following questions.
It has been mentioned in the passage that as `r` decreases, `E` decreases but `K` increases. The increase in `K` is [`E =` total mechanical energy, `r =` orbital radius, `K=` kinetic energy ]

Options

A. due to increase in gravitational `PE`
B. due to decrease in gravitational `PE`
C. due to work done by air friction force
D. both (b) and (c)

Explanation

To solve the question regarding the relationship between the total mechanical energy, orbital radius, and kinetic energy of a satellite in a low circular orbit, we can follow these steps:

### Step-by-Step Solution

1. **Understanding the System**:
- We have a satellite of mass \( m = 150 \, \text{kg} \) in a low circular orbit around the Earth.
- The effects of air drag are significant, which means that the satellite experiences a force opposing its motion.

2. **Total Mechanical Energy (E)**:
- The total mechanical energy \( E \) of a satellite in orbit is given by the sum of its kinetic energy \( K \) and potential energy \( U \):
\[ E = K + U \]
- In a circular orbit, the potential energy \( U \) can be expressed as:
\[ U = -\frac{GMm}{r} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( r \) is the orbital radius.

3. **Kinetic Energy (K)**:
- The kinetic energy of the satellite is given by:
\[ K = \frac{1}{2}mv^2 \]
- For a satellite in a circular orbit, the orbital velocity \( v \) is related to the gravitational force:
\[ v = \sqrt{\frac{GM}{r}} \]
- Therefore, the kinetic energy can also be expressed in terms of \( r \):
\[ K = \frac{1}{2}m\left(\frac{GM}{r}\right) = \frac{GMm}{2r} \]

4. **Effect of Air Drag**:
- As the satellite moves through the atmosphere, air drag does work on the satellite, which decreases its total mechanical energy \( E \).
- This results in a decrease in the orbital radius \( r \), as the satellite loses energy.

5. **Relationship Between E, r, and K**:
- As \( r \) decreases, the potential energy \( U \) becomes more negative (i.e., \( U \) decreases).
- The decrease in potential energy leads to an increase in kinetic energy \( K \) because the total mechanical energy \( E \) is still conserved in the sense that:
\[ \Delta K + \Delta U = W_{\text{air drag}} \]
- Since the work done by air drag is negative, the increase in kinetic energy \( K \) compensates for the decrease in potential energy \( U \).

6. **Conclusion**:
- Therefore, as the orbital radius \( r \) decreases, the total mechanical energy \( E \) becomes more negative, while the kinetic energy \( K \) increases due to the conversion of potential energy into kinetic energy.

### Final Answer:
The increase in kinetic energy \( K \) occurs because the decrease in potential energy \( U \) (due to the decrease in orbital radius \( r \)) is greater than the work done against air drag, leading to a net increase in \( K \).