satellite revolving in circular orbit suppose `V_(0)` is the orbital speed, T its time period, u its potential energy and K the kinetic energy. Now value of G is decreasesd. Then

Options

A. `V_(0)` will decreases
B. T will decrease
C. U will decrease
D. K will decrease

Answer

To solve the problem step by step, we will analyze how the decrease in the gravitational constant \( G \) affects the various parameters of a satellite in a circular orbit: orbital speed \( V_0 \), time period \( T \), potential energy \( U \), and kinetic energy \( K \).

### Step 1: Analyze the Orbital Speed \( V_0 \)

The formula for the orbital speed \( V_0 \) of a satellite in a circular orbit is given by:

\[ V_0 = \sqrt{\frac{G M}{r}} \]

Where:
- \( G \) is the gravitational constant,
- \( M \) is the mass of the central body (e.g., Earth),
- \( r \) is the radius of the orbit.

**Effect of Decreasing \( G \)**:
- If \( G \) decreases, \( V_0 \) will also decrease because \( V_0 \) is directly proportional to \( \sqrt{G} \).

**Conclusion**: \( V_0 \) will decrease.

### Step 2: Analyze the Time Period \( T \)

The formula for the time period \( T \) of a satellite in a circular orbit is given by:

\[ T = 2\pi \sqrt{\frac{r^3}{G M}} \]

**Effect of Decreasing \( G \)**:
- From this formula, we see that \( T \) is inversely proportional to \( \sqrt{G} \). Therefore, if \( G \) decreases, \( T \) will increase.

**Conclusion**: \( T \) will increase, not decrease.

### Step 3: Analyze the Potential Energy \( U \)

The potential energy \( U \) of a satellite in a gravitational field is given by:

\[ U = -\frac{G M m}{r} \]

Where \( m \) is the mass of the satellite.

**Effect of Decreasing \( G \)**:
- Since \( U \) is directly proportional to \( G \) (with a negative sign), if \( G \) decreases, the magnitude of \( U \) (which is negative) will increase (i.e., it becomes less negative).

**Conclusion**: \( U \) will increase, not decrease.

### Step 4: Analyze the Kinetic Energy \( K \)

The kinetic energy \( K \) of the satellite is given by:

\[ K = \frac{1}{2} m V_0^2 \]

Using the expression for \( V_0 \):

\[ K = \frac{1}{2} m \left(\sqrt{\frac{G M}{r}}\right)^2 = \frac{G M m}{2r} \]

**Effect of Decreasing \( G \)**:
- Since \( K \) is directly proportional to \( G \), if \( G \) decreases, \( K \) will also decrease.

**Conclusion**: \( K \) will decrease.

### Summary of Results

- \( V_0 \) will decrease (Option A: Correct)
- \( T \) will increase (Option B: Incorrect)
- \( U \) will increase (Option C: Incorrect)
- \( K \) will decrease (Option D: Correct)

### Final Answer

The correct options are:
- A: \( V_0 \) will decrease.
- D: \( K \) will decrease.