Question

The four satellites are lying close to the earth at distances $h_1, h_2, h_3$, and $h_4$ meters respectively, away from the center of the earth. If the values of $h_i$'s are given in terms of $R$ (radius of the earth), write the time period of the satellite in increasing order:
$h_1 = R/3$
$h_2 = R/4$
$h_3 = R/5$
$h_4 = R/2$
Choose the correct answer from the options given below:

• A. (C), (B), (A), (D)
• B. (B), (A), (C), (D)
• C. (B), (A), (D), (C)
• D. (C), (B), (D), (A)

Hint:

According to Kepler’s third law, the orbital period increases as the distance from the Earth increases. This is important when considering satellite motions and orbital mechanics.

Answer 1

The Correct Option is A

The orbital period \(T\) of a satellite is inversely proportional to the square root of the mean radius \(r\) (which includes Earth's radius \(R\) plus altitude \(h\)). The relationship is given by Kepler's third law:
\[T \propto \frac{1}{\sqrt{r}}, \quad \text{where} \quad r = R + h\]
The larger the altitude \(h\), the larger the distance \(r\) from the center of Earth, and the longer the orbital period \(T\) will be. Thus, we arrange the satellites in increasing order of \(r\) (and increasing orbital period \(T\)) based on their given values of \(h\):
- \(h_3 = R/5\) corresponds to the smallest altitude, hence the shortest period.
- \(h_2 = R/4\) gives the second smallest altitude.
- \(h_1 = R/3\) corresponds to a larger altitude and hence a larger period.
- \(h_4 = R/2\) gives the largest altitude and the longest period.
Thus, the correct order is: \((\text{C}), (\text{B}), (\text{A}), (\text{D})\).