A satellite having time period same as that of the earth's rotation about its own axis is orbiting the earth at a height 8R above the surface of earth. Where R is radius of earth. What will be the time period of another satellite at a height 3.5 R from the surface of earth ?

Options

A. `2 sqrt(2)` h
B. `4 sqrt(2)` h
C. `6 sqrt(2)` h
D. `3 sqrt(2)` h

Explanation

To solve the problem, we will use Kepler's third law of planetary motion, which relates the period of orbiting satellites to their distance from the center of the Earth.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- The time period of the first satellite (T1) is the same as the Earth's rotation period, which is 24 hours.
- The height of the first satellite above the Earth's surface is 8R, where R is the radius of the Earth.

2. **Calculating the Distance from the Center of the Earth**:
- The distance from the center of the Earth for the first satellite (R1) is:
\[ R1 = R + 8R = 9R \]

3. **Finding the Time Period of the Second Satellite**:
- The height of the second satellite above the Earth's surface is 3.5R.
- The distance from the center of the Earth for the second satellite (R2) is:
\[ R2 = R + 3.5R = 4.5R \]

4. **Applying Kepler's Third Law**:
- According to Kepler's third law:
\[ \left( \frac{T1}{T2} \right)^2 = \frac{R1^3}{R2^3} \]
- Substituting the values we have:
\[ \left( \frac{T1}{T2} \right)^2 = \frac{(9R)^3}{(4.5R)^3} \]

5. **Simplifying the Equation**:
- This simplifies to:
\[ \left( \frac{T1}{T2} \right)^2 = \frac{729R^3}{91.125R^3} = \frac{729}{91.125} \]
- Further simplifying gives:
\[ \left( \frac{T1}{T2} \right)^2 = 8 \]
- Taking the square root:
\[ \frac{T1}{T2} = 2\sqrt{2} \]
- Thus,
\[ T2 = \frac{T1}{2\sqrt{2}} \]

6. **Substituting the Value of T1**:
- Since T1 = 24 hours:
\[ T2 = \frac{24}{2\sqrt{2}} = \frac{12}{\sqrt{2}} = 6\sqrt{2} \text{ hours} \]

### Final Answer:
The time period of the second satellite at a height of 3.5R from the surface of the Earth is \( 6\sqrt{2} \) hours.

---