Two satellites `S_(1)` and `S_(2)` revolve around a planet in coplanar circular orbits in the same sense their periods of revolution are 1 hour and 8hours respectively the radius of the orbit of `S_(1)` is `10^(4)` km when `S_(1)` is closest to `S_(2)` the angular speed of `S_(2)` as observed by an astronaut in `S_(1)` is :

Options

A. `pixx10^(4)`
B. `-pixx10^(4)`
C. `pixx10^(5)`
D. `pixx10^(5)`

Explanation

To solve the problem, we need to find the angular speed of satellite \( S_2 \) as observed from satellite \( S_1 \). Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the given data
- Period of satellite \( S_1 \) (\( T_1 \)) = 1 hour = 3600 seconds
- Period of satellite \( S_2 \) (\( T_2 \)) = 8 hours = 28800 seconds
- Radius of the orbit of \( S_1 \) (\( R_1 \)) = \( 10^4 \) km = \( 10^7 \) m

### Step 2: Use Kepler's Third Law to find the radius of \( S_2 \)
According to Kepler's Third Law:
\[ \frac{T_1^2}{T_2^2} = \frac{R_1^3}{R_2^3} \]
Substituting the known values:
\[ \frac{(3600)^2}{(28800)^2} = \frac{(10^7)^3}{R_2^3} \]
Calculating the left side:
\[ \frac{12960000}{829440000} = \frac{1}{64} \]
Thus, we have:
\[ \frac{(10^7)^3}{R_2^3} = \frac{1}{64} \]
This implies:
\[ R_2^3 = 64 \times (10^7)^3 \]
Taking the cube root:
\[ R_2 = 4 \times 10^7 \text{ m} \]

### Step 3: Calculate the angular velocities of \( S_1 \) and \( S_2 \)
The angular velocity (\( \omega \)) is given by:
\[ \omega = \frac{2\pi}{T} \]
For satellite \( S_1 \):
\[ \omega_1 = \frac{2\pi}{T_1} = \frac{2\pi}{3600} \text{ rad/s} \]
For satellite \( S_2 \):
\[ \omega_2 = \frac{2\pi}{T_2} = \frac{2\pi}{28800} \text{ rad/s} \]

### Step 4: Calculate the angular speed of \( S_2 \) as observed from \( S_1 \)
The angular speed of \( S_2 \) as observed from \( S_1 \) is given by:
\[ \omega_{S_2}^{obs} = \omega_2 - \omega_1 \]
Substituting the values:
\[ \omega_{S_2}^{obs} = \frac{2\pi}{28800} - \frac{2\pi}{3600} \]
Finding a common denominator (28800):
\[ \omega_{S_2}^{obs} = \frac{2\pi}{28800} - \frac{16\pi}{28800} = \frac{-14\pi}{28800} \]
Simplifying:
\[ \omega_{S_2}^{obs} = -\frac{7\pi}{14400} \text{ rad/s} \]

### Conclusion
The angular speed of satellite \( S_2 \) as observed from satellite \( S_1 \) is:
\[ \omega_{S_2}^{obs} = -\frac{7\pi}{14400} \text{ rad/s} \]