Two satellites `S_(1)` and `S_(2)` are revolving round a planet in coplanar and concentric circular orbit of radii `R_(1)` and `R_(2)` in te same direction respectively. Their respective periods of revolution are 1 hr and 8 hr. the radius of the orbit of satellite `S_(1)` is equal to `10^(4)`km. Find the relative speed in kmph when they are closest.

Options

A. `pi//2xx10^(4)`
B. `pixx10^(4)`
C. `2pixx10^(4)`
D. `4pixx10^(4)`

Answer

To solve the problem step by step, we will follow these guidelines:

### Step 1: Identify Given Values
We have the following information:
- Period of satellite \( S_1 \) (T1) = 1 hour
- Period of satellite \( S_2 \) (T2) = 8 hours
- Radius of orbit of satellite \( S_1 \) (R1) = \( 10^4 \) km

### Step 2: Use Kepler's Third Law to Find R2
According to Kepler's Third Law, the square of the period of revolution of a satellite is proportional to the cube of the radius of its orbit. This can be expressed as:
\[ \frac{T_1^2}{T_2^2} = \frac{R_1^3}{R_2^3} \]

Substituting the known values:
\[ \frac{1^2}{8^2} = \frac{(10^4)^3}{R_2^3} \]
\[ \frac{1}{64} = \frac{10^{12}}{R_2^3} \]

Cross-multiplying gives:
\[ R_2^3 = 64 \times 10^{12} \]
Taking the cube root:
\[ R_2 = \sqrt[3]{64} \times 10^{4} = 4 \times 10^{4} \text{ km} \]

### Step 3: Calculate the Speeds of the Satellites
The speed of a satellite in circular orbit is given by the formula:
\[ V = R \omega \]
where \( \omega = \frac{2\pi}{T} \).

#### For Satellite \( S_1 \):
\[ V_1 = R_1 \cdot \omega_1 = R_1 \cdot \frac{2\pi}{T_1} \]
Substituting the known values:
\[ V_1 = 10^4 \cdot \frac{2\pi}{1} = 2\pi \times 10^4 \text{ km/hr} \]

#### For Satellite \( S_2 \):
\[ V_2 = R_2 \cdot \omega_2 = R_2 \cdot \frac{2\pi}{T_2} \]
Substituting the known values:
\[ V_2 = 4 \times 10^4 \cdot \frac{2\pi}{8} = \frac{2\pi \times 4 \times 10^4}{8} = \pi \times 10^4 \text{ km/hr} \]

### Step 4: Calculate the Relative Speed
The relative speed when the satellites are closest is given by:
\[ V_{\text{relative}} = V_1 - V_2 \]
Substituting the values we calculated:
\[ V_{\text{relative}} = (2\pi \times 10^4) - (\pi \times 10^4) = \pi \times 10^4 \text{ km/hr} \]

### Final Answer
The relative speed when the satellites are closest is:
\[ \pi \times 10^4 \text{ km/hr} \]

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