Near the earth's surface time period of a satellite is 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, given that the time period near the Earth's surface is 4 hours, 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. **Understanding the relationship**: According to Kepler's Third Law, the square of the time period \(T\) of a satellite is directly proportional to the cube of the semi-major axis \(r\) of its orbit. This can be expressed mathematically as:
\[ T^2 \propto r^3 \]
or
\[ \frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3} \]

2. **Identifying the variables**:
- Let \(T_1\) be the time period near the Earth's surface, which is given as \(4\) hours.
- Let \(T_2\) be the time period at a distance of \(4R\) from the center of the Earth.
- The radius at the Earth's surface \(r_1\) is \(R\), and the radius at the distance \(4R\) is \(r_2 = 4R\).

3. **Setting up the equation**: We can substitute the known values into the equation:
\[ \frac{T_1^2}{T_2^2} = \frac{R^3}{(4R)^3} \]

4. **Calculating the right side**: Simplifying the right side:
\[ (4R)^3 = 64R^3 \]
Thus, we have:
\[ \frac{T_1^2}{T_2^2} = \frac{R^3}{64R^3} = \frac{1}{64} \]

5. **Taking the square root**: Taking the square root of both sides gives:
\[ \frac{T_1}{T_2} = \frac{1}{8} \]
Therefore, we can express \(T_2\) in terms of \(T_1\):
\[ T_2 = 8T_1 \]

6. **Substituting the value of \(T_1\)**: Now substituting \(T_1 = 4\) hours:
\[ T_2 = 8 \times 4 = 32 \text{ hours} \]

### Final Answer:
The time period of the satellite at a distance of \(4R\) from the center of the Earth is **32 hours**.