Question

A geostationary communications satellite orbits the earth directly above the equator at an altitude of 34800 km .Calculate the time it would take a cell phone signal to travel from a point on the equator to the satellite and back.Would this delay be noticeable in a conversation?

Answer 1

Radius of earth is, \(6371 \mathrm{~km}\) so the distance to satellite is,

$$ \begin{aligned} \sqrt{R_{s}^{2}+R_{s}^{2}} &=\sqrt{(6371)^{2}+(34800)^{2}} \\ &=35378.38 \mathrm{~km} \end{aligned} $$

The total distance of the signal travels is,

$$ \begin{aligned} D &=35378.38+35378.38 \\ &=70756.76 \mathrm{~km} \end{aligned} $$

$$ \begin{aligned} \text { time } &=\frac{\text { distance }}{\text { speed }} \\ &=\frac{70756.76 \mathrm{~km}}{300000 \mathrm{~km} / \mathrm{s}} \\ &=0.24 \mathrm{~s} \end{aligned} $$