A satellite is placed in a circular orbit around the earth at an altitude of $1000 \mathrm{~km}$. The time…
A satellite is placed in a circular orbit around the earth at an altitude of $1000 \mathrm{~km}$. The time period of the satellite in minutes is approximately
(mass of the earth $=6 \times 10^{24} \mathrm{~kg}$, radius of the earth $=6.4$ $\times 10^6 \mathrm{~m}, \mathrm{G}=6.67 \times 10^{-11} \mathrm{Nm}^2 \mathrm{~kg}^{-2}$ )
$105$
$200$
$120$
$62$
Solution
height, $\mathrm{h}=1000 \mathrm{~km}=10^6 \mathrm{~m}$
Radius of Earth, $\mathrm{R}=6.4 \times 10^6 \mathrm{~m}$
Mass of Earth, $M=6 \times 10^{24} \mathrm{~kg}$
Time period of satelite is given as
$\begin{aligned} & \mathrm{T}=2 \pi \sqrt{\frac{\mathrm{r}^3}{\mathrm{GM}}}=2 \pi \sqrt{\frac{(\mathrm{R}+\mathrm{h})^3}{\mathrm{GM}}} \\ & \mathrm{T}=2 \times 3.14 \times \sqrt{\frac{\left(6.4 \times 10^6+10^6\right)^3}{6.67 \times 10^{-11} \times 6 \times 10^{24}}} \\ & =2 \times 3.14 \times \sqrt{\frac{7.4 \times 7.4 \times 7.4 \times 10^{18}}{6.67 \times 6 \times 10^{13}}} \\ & \mathrm{~T}=6.28 \times \sqrt{\frac{405.224 \times 10^5}{40.02}} \\ & \mathrm{~T}=6280 \mathrm{sec}\end{aligned}$
$\begin{aligned} & =\left(\frac{6280}{60}\right) \text { minutes } \\ & T=104.67 \\ & T \simeq 105 \text { minutes }\end{aligned}$