The time period of a $1500 \mathrm{~kg}$ satellite is equal to the time period of rotation of the earth. The…

The time period of a $1500 \mathrm{~kg}$ satellite is equal to the time period of rotation of the earth. The altitude of the satellite is nearly
  1. $42,211 \mathrm{~km}$
  2. $35,840 \mathrm{~km}$
  3. $6,400 \mathrm{~km}$
  4. $13,800 \mathrm{~km}$

Solution

Mass of satellite, $\mathrm{m}=1500 \mathrm{~kg}$ Time period of satellite, $\mathrm{T}=26 \times 60 \times 60=93,600 \mathrm{sec}$ $\begin{aligned} & \mathrm{T}=2 \pi \sqrt{\frac{\mathrm{R}^3}{\mathrm{GM}}} \\ & \mathrm{R}^3=\frac{\mathrm{T}^2 \mathrm{GM}}{4 \pi^2} \\ & \text { Altitude, } \mathrm{A}=\left(\frac{\mathrm{T}^2 \mathrm{GM}}{4 \pi^2}\right)^{\frac{1}{3}}-\mathrm{R}\end{aligned}$

Asked in: AP EAMCET 2023 (16 May Shift 1)

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