Q95 / 180 · NEET 2019 (Odisha) · Physics
The time period of a geo-stationary satellite is $24 \mathrm{~h}$, at a height $6 R_E-R_E$ is the radius of earth) from surface of earth. The time period of another satellite whose height is $2.5 \mathrm{R}_{\mathrm{E}}$ from surface will be
- Correct answerA$6 \sqrt{2} \mathrm{~h}$
- B$12 \sqrt{2} \mathrm{~h}$
- C$\frac{24}{2.5} \mathrm{~h}$
- D$\frac{12}{2.5} \mathrm{~h}$
Solution
From Kepler's third law, the time period of revolution of satellite around earth is
$$
\mathrm{T}^2 \propto \mathrm{r}^3 \text { or } \mathrm{T} \propto \mathrm{r}^{3 / 2}
$$
where, $\mathrm{r}$ is the radius of satellite's orbit.
Here, $\mathrm{r}_1=6 \mathrm{R}_{\mathrm{E}}+\mathrm{R}_{\mathrm{E}}, \mathrm{T}_1=24 \mathrm{~h}$
$\mathrm{r}_2=2.5 \mathrm{R}_{\mathrm{E}}+\mathrm{R}_{\mathrm{E}}, \mathrm{T}_2=$ ?
where $\mathrm{R}_{\mathrm{E}}=$ radius of earth
So, from Eq. (i), we get
$$
\begin{aligned}
\frac{\mathrm{T}_1}{\mathrm{~T}_2} & =\left(\frac{\mathrm{r}_1}{\mathrm{r}_2}\right)^{3 / 2} \\
\frac{24}{\mathrm{~T}_2} & =\left(\frac{6 \mathrm{R}_{\mathrm{E}}+\mathrm{R}_{\mathrm{E}}}{2.5 \mathrm{R}_{\mathrm{E}}+\mathrm{R}_{\mathrm{E}}}\right)^{3 / 2}=\left(\frac{7}{3.5}\right)^{3 / 2} \\
\Rightarrow \quad \mathrm{T}_2 & =\frac{24}{(2)^{3 / 2}}=\frac{24}{2 \sqrt{2}}=\frac{12}{\sqrt{2}}=6 \sqrt{2} \mathrm{~h}
\end{aligned}
$$
Hence, the correct answer is option: 'A'.