(a).Find the radius of the circular orbit of a satellite moving with an angular speed equal to the angular speed of earth's rotation. (b). If the satellite is directly above the north pole at some instant, find the time it takes to come over the equatorial plane. Mass of the earth `=6xx10^24kg`

Explanation

To solve the given problem, we will break it down into two parts as specified in the question.

### Part (a): Finding the radius of the circular orbit of a satellite

1. **Understanding Angular Speed**:
The angular speed of the satellite is equal to the angular speed of the Earth. The angular speed of the Earth can be calculated using the formula:
\[ \omega = \frac{2\pi}{T} \]
where \( T \) is the time period of Earth's rotation (24 hours).

2. **Convert Time Period to Seconds**:
Since \( T = 24 \) hours, we convert this into seconds:
\[ T = 24 \times 3600 = 86400 \text{ seconds} \]

3. **Using the Formula for Orbital Radius**:
The time period \( T \) of a satellite in a circular orbit is given by:
\[ T^2 = \frac{4\pi^2 r^3}{GM} \]
where \( G \) is the gravitational constant (\( 6.67 \times 10^{-11} \, \text{m}^3/\text{kg s}^2 \)) and \( M \) is the mass of the Earth (\( 6 \times 10^{24} \, \text{kg} \)).

4. **Rearranging for Radius \( r \)**:
Rearranging the formula gives:
\[ r^3 = \frac{T^2 GM}{4\pi^2} \]

5. **Substituting Values**:
Substitute \( T = 86400 \, \text{s} \), \( G = 6.67 \times 10^{-11} \, \text{m}^3/\text{kg s}^2 \), and \( M = 6 \times 10^{24} \, \text{kg} \):
\[ r^3 = \frac{(86400)^2 \times (6.67 \times 10^{-11}) \times (6 \times 10^{24})}{4\pi^2} \]

6. **Calculating \( r^3 \)**:
Perform the calculations:
\[ r^3 \approx \frac{(7.46496 \times 10^9) \times (6.67 \times 10^{-11}) \times (6 \times 10^{24})}{39.478} \]

7. **Finding \( r \)**:
After calculating \( r^3 \), take the cube root to find \( r \):
\[ r \approx 4.23 \times 10^7 \text{ meters} \quad \text{or} \quad 42300 \text{ kilometers} \]

### Part (b): Finding the time taken to come over the equatorial plane

1. **Understanding the Movement**:
The satellite moves from the North Pole to the equatorial plane. This is a quarter of a full revolution.

2. **Calculating Time for Quarter Revolution**:
Since the total time period for one complete revolution is \( T = 24 \) hours, the time taken to move from the North Pole to the equatorial plane is:
\[ \text{Time} = \frac{1}{4} T = \frac{1}{4} \times 24 \text{ hours} = 6 \text{ hours} \]

### Final Answers:
- (a) The radius of the circular orbit of the satellite is approximately **42300 kilometers**.
- (b) The time taken for the satellite to come over the equatorial plane is **6 hours**.