`S_(1)` and `S_(2)` are two satellites of radii R and 9R revolving around a planet P in circular orbits. If speed of `S_(1)` is 6V, then how many times the speed of `S_(2)` is greater than speed of `S_(2)` ?

Explanation

To solve the problem, we need to find the relationship between the speeds of the two satellites \( S_1 \) and \( S_2 \).

### Step-by-Step Solution:

1. **Understanding Orbital Velocity**:
The orbital velocity \( v \) of a satellite in a circular orbit around a planet is given by the formula:
\[ v = \sqrt{\frac{GM}{r}} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( r \) is the radius of the orbit.

2. **Calculate Velocity of Satellite \( S_1 \)**:
For satellite \( S_1 \) with radius \( R \):
\[ v_1 = \sqrt{\frac{GM}{R}} \]
According to the problem, the speed of \( S_1 \) is given as \( 6V \). Thus, we can write:
\[ v_1 = 6V \]

3. **Calculate Velocity of Satellite \( S_2 \)**:
For satellite \( S_2 \) with radius \( 9R \):
\[ v_2 = \sqrt{\frac{GM}{9R}} = \frac{1}{3} \sqrt{\frac{GM}{R}} = \frac{1}{3} v_1 \]

4. **Relate \( v_1 \) and \( v_2 \)**:
From the previous step, we have:
\[ v_2 = \frac{1}{3} v_1 \]
Substituting \( v_1 = 6V \):
\[ v_2 = \frac{1}{3} \times 6V = 2V \]

5. **Finding the Ratio of Speeds**:
Now, we need to find how many times the speed of \( S_1 \) is greater than the speed of \( S_2 \):
\[ \frac{v_1}{v_2} = \frac{6V}{2V} = 3 \]

### Final Answer:
The speed of satellite \( S_1 \) is 3 times greater than the speed of satellite \( S_2 \).

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