The time period of a satellite of earth is 5 hours. If the separation between the centre of earth and the satellite is increased to 4 times the previous value, the new time period will become-

Options

A. 10h
B. 80h
C. 40h
D. 20h

Explanation

To solve the problem, we need to find the new time period of a satellite when its separation from the center of the Earth is increased to four times its original value. We can use Kepler's Third Law of planetary motion which relates the time period of a satellite to its orbital radius.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Original time period (T) = 5 hours
- Original separation (r) = r (let's denote it as r)
- New separation (R') = 4r (since it is increased to four times the original value)

2. **Apply Kepler's Third Law:**
According to Kepler's Third Law, the square of the time period (T) is directly proportional to the cube of the semi-major axis (r) of its orbit:
\[ T^2 \propto r^3 \]
This can be expressed as:
\[ \frac{T'^2}{T^2} = \frac{R'^3}{r^3} \]

3. **Substitute the Known Values:**
We know that:
- \( T = 5 \) hours
- \( R' = 4r \)
Therefore:
\[ \frac{T'^2}{(5 \text{ hours})^2} = \frac{(4r)^3}{r^3} \]

4. **Simplify the Right Side:**
\[ \frac{T'^2}{25} = \frac{64r^3}{r^3} = 64 \]

5. **Solve for T':**
Multiply both sides by 25:
\[ T'^2 = 64 \times 25 \]
\[ T'^2 = 1600 \]
Taking the square root of both sides:
\[ T' = \sqrt{1600} = 40 \text{ hours} \]

6. **Conclusion:**
The new time period of the satellite when the separation is increased to four times the original value is **40 hours**.