The period of a satellite in a circular orbit of radius `R` is `T`, the period of another satellite in a circular orbit of radius `4R` is

Options

A. `4T`
B. `T//4`
C. `8T`
D. `T//8`

Explanation

To find the period of a satellite in a circular orbit of radius \(4R\) given that the period of another satellite in a circular orbit of radius \(R\) is \(T\), we can use Kepler's Third Law of planetary motion. According to this law, the square of the period of a satellite is directly proportional to the cube of the semi-major axis (or radius in the case of circular orbits) of its orbit.

### Step-by-Step Solution:

1. **State Kepler's Third Law**:
\[ T^2 \propto R^3 \]
This means that:
\[ \frac{T_1^2}{T_2^2} = \frac{R_1^3}{R_2^3} \]

2. **Assign Known Values**:
Let:
- \(T_1 = T\) (the period of the first satellite)
- \(R_1 = R\) (the radius of the first satellite)
- \(R_2 = 4R\) (the radius of the second satellite)
- \(T_2\) = ? (the period of the second satellite)

3. **Substitute the Values into the Equation**:
\[ \frac{T^2}{T_2^2} = \frac{R^3}{(4R)^3} \]

4. **Simplify the Right Side**:
\[ (4R)^3 = 64R^3 \]
Therefore, the equation becomes:
\[ \frac{T^2}{T_2^2} = \frac{R^3}{64R^3} = \frac{1}{64} \]

5. **Cross Multiply to Solve for \(T_2^2\)**:
\[ T^2 = \frac{1}{64} T_2^2 \]
Rearranging gives:
\[ T_2^2 = 64T^2 \]

6. **Take the Square Root to Find \(T_2\)**:
\[ T_2 = \sqrt{64T^2} = 8T \]

### Final Answer:
The period of the satellite in a circular orbit of radius \(4R\) is \(8T\).

---