If the angular velocity of a planet about its own axis is halved, the distance of geostationary satellite of this planet from the centre of the centre of the planet will become

Options

A. `(2)^(1//2)` times
B. `(2)^(3//2)`times
C. `(2)^(2//3)` times
D. 4 times

Answer

To solve the problem, we need to determine how the distance of a geostationary satellite from the center of a planet changes when the planet's angular velocity is halved.

### Step-by-Step Solution:

1. **Understanding Geostationary Satellites**: A geostationary satellite orbits the planet in such a way that it remains fixed above a specific point on the planet's surface. This requires the satellite's orbital period to match the planet's rotation period.

2. **Relating Angular Velocity and Orbital Period**: The angular velocity (\(\omega\)) of the planet is related to its rotational period (\(T\)) by the formula:
\[ \omega = \frac{2\pi}{T} \]
If the angular velocity is halved, then:
\[ \omega_2 = \frac{\omega_1}{2} \]
This implies that the new period \(T_2\) will be:
\[ T_2 = \frac{2\pi}{\omega_2} = \frac{2\pi}{\frac{\omega_1}{2}} = \frac{4\pi}{\omega_1} = 2T_1 \]
Thus, the new period is double the original period.

3. **Using Kepler's Third Law**: According to Kepler's Third Law, the square of the orbital period (\(T\)) of a satellite is proportional to the cube of the semi-major axis (\(r\)) of its orbit:
\[ T^2 \propto r^3 \]
For the initial state, we can write:
\[ T_1^2 \propto r_1^3 \]
For the new state:
\[ T_2^2 \propto r_2^3 \]

4. **Setting Up the Proportionality**: Since \(T_2 = 2T_1\):
\[ (2T_1)^2 \propto r_2^3 \]
This simplifies to:
\[ 4T_1^2 \propto r_2^3 \]

5. **Relating \(r_1\) and \(r_2\)**: From the original relationship:
\[ T_1^2 \propto r_1^3 \]
We can express the proportionality constants:
\[ T_1^2 = k \cdot r_1^3 \quad \text{and} \quad 4T_1^2 = k \cdot r_2^3 \]
Substituting \(T_1^2\) into the second equation gives:
\[ 4k \cdot r_1^3 \propto r_2^3 \]

6. **Finding the Ratio**: Dividing both sides by \(k\):
\[ 4r_1^3 \propto r_2^3 \]
Taking the cube root of both sides, we find:
\[ r_2 \propto r_1 \cdot 2^{2/3} \]

7. **Final Expression for \(r_2\)**: Therefore, the new distance \(r_2\) of the geostationary satellite from the center of the planet is:
\[ r_2 = r_1 \cdot 2^{2/3} \]

### Conclusion:
The distance of the geostationary satellite from the center of the planet becomes \(2^{2/3}\) times the original distance.