The time period of a satellite of earth is 24 hours. If the separation between the earth and the satellite is decreased to one fourth of the previous value, then its new time period will become

Options

A. `12 hours`
B. 6 hours
C. `4 hours`
D. `3 hours`

Answer

To solve the problem, we need to use Kepler's third law of planetary motion, which states that 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. Mathematically, this can be expressed as:

\[ T^2 \propto r^3 \]

This means:

\[ \frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3} \]

Where:
- \( T_1 \) is the initial time period (24 hours),
- \( T_2 \) is the new time period,
- \( r_1 \) is the initial radius (distance from the center of the Earth to the satellite),
- \( r_2 \) is the new radius.

Given that the separation between the Earth and the satellite is decreased to one-fourth of the previous value, we can express this as:

\[ r_2 = \frac{r_1}{4} \]

Now, substituting this into the equation:

1. Start with the relationship from Kepler's law:
\[ \frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3} \]

2. Substitute \( r_2 \):
\[ \frac{T_1^2}{T_2^2} = \frac{r_1^3}{\left(\frac{r_1}{4}\right)^3} \]

3. Simplify the right side:
\[ \frac{T_1^2}{T_2^2} = \frac{r_1^3}{\frac{r_1^3}{64}} = 64 \]

4. Thus, we have:
\[ \frac{T_1^2}{T_2^2} = 64 \implies T_2^2 = \frac{T_1^2}{64} \]

5. Taking the square root of both sides:
\[ T_2 = \frac{T_1}{8} \]

6. Substitute \( T_1 = 24 \) hours:
\[ T_2 = \frac{24}{8} = 3 \text{ hours} \]

Thus, the new time period of the satellite is **3 hours**.