Two satellites `S_1 and S_2` revole round a planet in coplanar circular orbits in the same sanse. Their periods of revolution are 1 hour and 8 hour respectively. The radius of the orbit of `S_1` is `10^4 km,` When `S_2` is closest to `S_1` find
(i) the speed of `S_2` relative to `S_1`
(ii) the angular speed of `S_2` as actually observed by an astronaut is `S_1.`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the radius of the orbit of satellite \( S_2 \)

Using Kepler's Third Law, we know that:
\[ \frac{T_1^2}{T_2^2} = \frac{R_1^3}{R_2^3} \]
Where:
- \( T_1 = 1 \) hour = \( 3600 \) seconds
- \( T_2 = 8 \) hours = \( 28800 \) seconds
- \( R_1 = 10^4 \) km

We need to find \( R_2 \).

Rearranging the formula gives:
\[ R_2^3 = R_1^3 \cdot \left(\frac{T_2^2}{T_1^2}\right) \]

Substituting the values:
\[ R_2^3 = (10^4)^3 \cdot \left(\frac{28800^2}{3600^2}\right) \]
Calculating \( \frac{28800^2}{3600^2} = 64 \):
\[ R_2^3 = (10^{12}) \cdot 64 \]
Taking the cube root:
\[ R_2 = 4 \times 10^4 \text{ km} \]

### Step 2: Calculate the speed of satellites \( S_1 \) and \( S_2 \)

The speed \( V \) of a satellite in circular orbit is given by:
\[ V = \frac{2 \pi R}{T} \]

#### For satellite \( S_1 \):
\[ V_1 = \frac{2 \pi R_1}{T_1} = \frac{2 \pi (10^4)}{3600} \]
Calculating \( V_1 \):
\[ V_1 = \frac{2 \pi \times 10^4}{3600} \approx 17.45 \text{ km/s} \]

#### For satellite \( S_2 \):
\[ V_2 = \frac{2 \pi R_2}{T_2} = \frac{2 \pi (4 \times 10^4)}{28800} \]
Calculating \( V_2 \):
\[ V_2 = \frac{2 \pi \times 4 \times 10^4}{28800} \approx 8.73 \text{ km/s} \]

### Step 3: Calculate the relative speed of \( S_2 \) with respect to \( S_1 \)

The relative speed \( V_{rel} \) is given by:
\[ V_{rel} = V_2 - V_1 \]
Substituting the values:
\[ V_{rel} = 8.73 - 17.45 = -8.72 \text{ km/s} \]
This indicates that \( S_2 \) is moving slower than \( S_1 \).

### Step 4: Calculate the angular speed of \( S_2 \) as observed from \( S_1 \)

The angular speed \( \omega \) is given by:
\[ \omega = \frac{V_{rel}}{R_{avg}} \]
Where \( R_{avg} \) is the average distance between the two satellites when they are closest. Since \( S_1 \) and \( S_2 \) are in coplanar orbits, we can use:
\[ R_{avg} = R_2 - R_1 = 4 \times 10^4 - 10^4 = 3 \times 10^4 \text{ km} \]
Substituting the values:
\[ \omega = \frac{-8.72 \times 10^3 \text{ m/s}}{3 \times 10^7 \text{ m}} = -2.91 \times 10^{-4} \text{ rad/s} \]

### Final Answers:
(i) The speed of \( S_2 \) relative to \( S_1 \) is approximately \( -8.72 \text{ km/s} \) (indicating direction).
(ii) The angular speed of \( S_2 \) as observed by an astronaut in \( S_1 \) is approximately \( -2.91 \times 10^{-4} \text{ rad/s} \).

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