A satellite with a time period of 24 hours is orbiting around the earth at a height of 4R above the earth's surface. Calculate the time period of another satellite at a height of 2R from the surface of earth. Here R is radius of earth.

Answer

To solve the problem, we need to find the time period of a satellite (let's call it Satellite 2) that is orbiting the Earth at a height of 2R above the Earth's surface, given that another satellite (Satellite 1) with a time period of 24 hours is orbiting at a height of 4R above the Earth's surface.

### Step-by-Step Solution:

1. **Identify the Radius of Orbit for Both Satellites:**
- The radius of the Earth (R) is given.
- For Satellite 1 (S1), which is at a height of 4R above the Earth's surface:
\[ \text{Radius of S1} = R + 4R = 5R \]
- For Satellite 2 (S2), which is at a height of 2R above the Earth's surface:
\[ \text{Radius of S2} = R + 2R = 3R \]

2. **Use the Formula for the Time Period of a Satellite:**
The time period (T) of a satellite is given by the formula:
\[ T = 2\pi \sqrt{\frac{R^3}{GM}} \]
where \( R \) is the radius of the orbit, \( G \) is the gravitational constant, and \( M \) is the mass of the Earth.

3. **Write the Time Period for Both Satellites:**
- For Satellite 1 (T1):
\[ T_1 = 2\pi \sqrt{\frac{(5R)^3}{GM}} = 2\pi \sqrt{\frac{125R^3}{GM}} \]
- For Satellite 2 (T2):
\[ T_2 = 2\pi \sqrt{\frac{(3R)^3}{GM}} = 2\pi \sqrt{\frac{27R^3}{GM}} \]

4. **Set Up the Ratio of Time Periods:**
Since we know \( T_1 = 24 \) hours, we can set up the ratio:
\[ \frac{T_2}{T_1} = \frac{\sqrt{27R^3}}{\sqrt{125R^3}} = \frac{\sqrt{27}}{\sqrt{125}} = \frac{3\sqrt{3}}{5\sqrt{5}} \]

5. **Calculate T2:**
Rearranging gives:
\[ T_2 = T_1 \cdot \frac{3\sqrt{3}}{5\sqrt{5}} = 24 \cdot \frac{3\sqrt{3}}{5\sqrt{5}} \]
Now, we simplify this:
\[ T_2 = \frac{72\sqrt{3}}{5\sqrt{5}} \text{ hours} \]
To find a numerical approximation:
\[ \sqrt{3} \approx 1.732, \quad \sqrt{5} \approx 2.236 \]
\[ T_2 \approx \frac{72 \cdot 1.732}{5 \cdot 2.236} \approx \frac{125.664}{11.18} \approx 11.25 \text{ hours} \]

### Final Answer:
The time period of Satellite 2 at a height of 2R from the surface of the Earth is approximately **11.25 hours**.