Question

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

• A. 3 hours
• B. 6 hours
• C. 24 hours
• D. 12 hours

Hint:

Kepler's third law says the square of the period is proportional to the cube of the orbital radius.

Answer 1

The Correct Option is A

Step 1: Ratio form
$\dfrac{T'}{T}=\left(\dfrac{r'}{r}\right)^{3/2}$.

Step 2: Insert values
$\dfrac{r'}{r}=\dfrac{1}{4}$, and $\left(\dfrac14\right)^{3/2}=\dfrac18$.

Step 3: Result
$T'=\dfrac{24}{8}=3\text{ h}$.

Final Answer:
The period becomes 3 hours. \[ \boxed{\text{(A) }\text{3 hours}} \]