The period of revolution of a satellite in an orbit of radius 2R is T. What will be its period of revolution in an orbit of radius 8R ?

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 (T) of a satellite is directly proportional to the cube of the semi-major axis (R) of its orbit. Mathematically, this can be expressed as:

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

### Step-by-step Solution:

1. **Identify the Given Information**:
- The period of revolution of a satellite in an orbit of radius \(2R\) is \(T\).
- We need to find the period of revolution of a satellite in an orbit of radius \(8R\).

2. **Apply Kepler's Third Law**:
- For the first orbit (radius \(2R\)):
\[ T^2 = k \cdot (2R)^3 \]
where \(k\) is a constant.

3. **Calculate for the First Orbit**:
- Expanding the equation:
\[ T^2 = k \cdot 8R^3 \]

4. **For the Second Orbit (radius \(8R\))**:
- We can write:
\[ T'^2 = k \cdot (8R)^3 \]
- Expanding this equation:
\[ T'^2 = k \cdot 512R^3 \]

5. **Set Up the Ratio**:
- Now, we can set up the ratio of the two periods:
\[ \frac{T'^2}{T^2} = \frac{k \cdot 512R^3}{k \cdot 8R^3} \]
- The \(k\) and \(R^3\) cancel out:
\[ \frac{T'^2}{T^2} = \frac{512}{8} = 64 \]

6. **Solve for \(T'\)**:
- Taking the square root of both sides:
\[ \frac{T'}{T} = \sqrt{64} = 8 \]
- Therefore:
\[ T' = 8T \]

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