Q62 / 90 · JEE Main 2024 (05 Apr Shift 2) · Physics

A satellite revolving around a planet in stationary orbit has time period 6 hours. The mass of planet is one-fourth the mass of earth. The radius orbit of planet is : $\left(\right.$ Given $=$ Radius of geo-stationary orbit for earth is $4.2 \times 10^4 \mathrm{~km}$ )

  • A
    $1.4 \times 10^4 \mathrm{~km}$
  • Correct answer
    B
    $1.05 \times 10^4 \mathrm{~km}$
  • C
    $8.4 \times 10^4 \mathrm{~km}$
  • D
    $1.68 \times 10^5 \mathrm{~km}$

Solution

$\begin{aligned} & \mathrm{T}=\frac{2 \pi \mathrm{r}^{3 / 2}}{\sqrt{\mathrm{GM}}} \\ & \frac{\mathrm{T}_1}{\mathrm{~T}_2}=\left(\frac{\mathrm{r}_1}{\mathrm{r}_2}\right)^{3 / 2}\left(\frac{\mathrm{M}_2}{\mathrm{M}_1}\right)^{1 / 2} \\ & \frac{6}{24}=\frac{\left(\mathrm{r}_1\right)^{3 / 2}}{\left(4.2 \times 10^4\right)^{3 / 2}}\left(\frac{\mathrm{M}}{\mathrm{M} / 4}\right)^{1 / 2} \\ & \mathrm{r}_1=1.05 \times 10^4 \mathrm{~km}\end{aligned}$

Hence, the correct answer is option: 'B'.

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