A satellite moves round the earth in a circular orbit of radius R making one revolution per day. A second satellite moving in a circular orbit, moves round the earth one in 8 days. The radius of the orbit of the second satellite is

Options

A. 8 R
B. 4 R
C. 2 R
D. R

Explanation

To solve the problem, we will use Kepler's Third Law of planetary motion, which states that the square of the orbital period (T) of a planet is directly proportional to the cube of the semi-major axis (R) of its orbit. This can be expressed mathematically as:

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

### Step-by-Step Solution:

1. **Identify the Given Data:**
- For the first satellite:
- Orbital period \( T_1 = 1 \) day
- Orbital radius \( R_1 = R \)
- For the second satellite:
- Orbital period \( T_2 = 8 \) days
- Orbital radius \( R_2 = ? \)

2. **Apply Kepler's Third Law:**
According to Kepler's Third Law:
\[ \frac{T_2^2}{T_1^2} = \frac{R_2^3}{R_1^3} \]

3. **Substitute the Known Values:**
Substituting the values of \( T_1 \) and \( T_2 \):
\[ \frac{(8 \text{ days})^2}{(1 \text{ day})^2} = \frac{R_2^3}{R^3} \]
This simplifies to:
\[ \frac{64}{1} = \frac{R_2^3}{R^3} \]

4. **Cross-Multiply to Solve for \( R_2^3 \):**
\[ R_2^3 = 64 R^3 \]

5. **Take the Cube Root:**
To find \( R_2 \), take the cube root of both sides:
\[ R_2 = (64 R^3)^{1/3} = 4R \]

6. **Conclusion:**
The radius of the orbit of the second satellite is:
\[ R_2 = 4R \]

### Final Answer:
The radius of the orbit of the second satellite is \( 4R \).

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