Q134 / 150 · MHT CET 2024 (16 May Shift 1) · Physics

A satellite is orbiting just above the surface of the planet of density ' $\rho$ ' with periodic time ' $T$ '. The quantity $\mathrm{T}^2 \rho$ is equal to ( $\mathrm{G}=$ universal gravitational constant)

  • A
    $\frac{4 \pi^2}{\mathrm{G}}$
  • B
    $\frac{3 \pi^2}{\mathrm{G}}$
  • Correct answer
    C
    $\frac{3 \pi}{\mathrm{G}}$
  • D
    $\frac{\pi}{G}$

Solution

Time period of a nearby satellite is given by, $\begin{aligned} & T=2 \pi \sqrt{\frac{R}{g}} \\ \therefore & T^2=4 \pi^2 \times \frac{R}{g} \\ & \text { But, } g=\frac{4}{3} \pi \rho G R . \\ \therefore & T^2=\frac{4 \pi^2 R}{\frac{4}{3} \pi \rho G R} \\ \therefore & T^2 \rho=\frac{3 \pi}{G} \end{aligned}$

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

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