A and B are two satellite revolving around the earth in circular orbits with radii `R_(A) and R_(B)`. Their periods `T_(A) and T_(B)` are 8 h and 1 h respectively. Then the ratio of `(R_(A)//R_(B))` is equal to

Options

A. 4
B. 8
C. `(8)^(1//3)`
D. `sqrt8`

Explanation

To solve the problem, we will use Kepler's Third Law of planetary motion, which states that the square of the period of revolution of a satellite is directly proportional to the cube of the semi-major axis (or radius for circular orbits) of its orbit. Mathematically, this can be expressed as:

\[ T^2 \propto R^3 \]

For two satellites A and B, we can write:

\[ \frac{T_A^2}{T_B^2} = \frac{R_A^3}{R_B^3} \]

Where:
- \( T_A \) and \( T_B \) are the periods of satellites A and B respectively.
- \( R_A \) and \( R_B \) are the radii of the orbits of satellites A and B respectively.

Given:
- \( T_A = 8 \) hours
- \( T_B = 1 \) hour

Now, we can substitute the values into the equation:

1. Calculate \( T_A^2 \) and \( T_B^2 \):
\[ T_A^2 = (8 \text{ hours})^2 = 64 \text{ hours}^2 \]
\[ T_B^2 = (1 \text{ hour})^2 = 1 \text{ hour}^2 \]

2. Substitute these values into the ratio:
\[ \frac{T_A^2}{T_B^2} = \frac{64}{1} = 64 \]

3. Now, using the relationship from Kepler's Third Law:
\[ \frac{R_A^3}{R_B^3} = 64 \]

4. Taking the cube root of both sides to find the ratio of the radii:
\[ \frac{R_A}{R_B} = \sqrt[3]{64} = 4 \]

Thus, the ratio of the radii \( \frac{R_A}{R_B} \) is equal to 4.

### Final Answer:
\[ \frac{R_A}{R_B} = 4 \]