Near the earth's surface time period of a satellite is 1.4 hrs. Find its time period if it is at the distance '4R' from the centre of earth :-

Options

A. 32 hrs.
B. `(1/(8sqrt(2)))` hrs.
C. `8sqrt(2)` hrs.
D. `16` hrs.

Answer

To solve the problem of finding the time period of a satellite at a distance of '4R' from the center of the Earth, we can follow these steps:

### Step-by-Step Solution

1. **Understand the Given Information:**
- The time period of a satellite near the Earth's surface (at distance R from the center) is given as \( T_1 = 1.4 \) hours.
- We need to find the time period \( T_2 \) of the satellite when it is at a distance of \( 4R \) from the center of the Earth.

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

3. **Relate the Time Periods Using Kepler's Third Law:**
- According to Kepler's Third Law, the square of the time period is proportional to the cube of the semi-major axis (distance from the center of the Earth):
\[ \frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3} \]
- Here, \( r_1 = R \) (the radius of the Earth) and \( r_2 = 4R \).

4. **Substitute the Values:**
- Substitute \( r_1 \) and \( r_2 \) into the equation:
\[ \frac{T_1^2}{T_2^2} = \frac{R^3}{(4R)^3} \]
- Simplifying the right side:
\[ \frac{T_1^2}{T_2^2} = \frac{R^3}{64R^3} = \frac{1}{64} \]

5. **Solve for \( T_2 \):**
- Taking the square root of both sides:
\[ \frac{T_1}{T_2} = \frac{1}{8} \]
- Therefore, we can express \( T_2 \) as:
\[ T_2 = 8T_1 \]
- Now substitute \( T_1 = 1.4 \) hours:
\[ T_2 = 8 \times 1.4 = 11.2 \text{ hours} \]

6. **Final Answer:**
- The time period \( T_2 \) of the satellite at a distance of \( 4R \) from the center of the Earth is \( 11.2 \) hours.