A geostationary communications satellite orbits the earth directly above the equator at an altitude of 36800 km. Calculate the time it would take a cell phone signal to travel from a point on the equator to the satellite and back.
A geostationary communications satellite orbits the earth directly above the equator at an altitude of 36800 km. Calculate the time it would take a cell phone signal to travel from a point on the equator to the satellite and back.
Answer:
0.24898 seconds
Explanation:
v = Speed of light = [tex]3\times 10^8\ m/s[/tex]
[tex]R_e[/tex] = Radius of Earth = 6371000 m
[tex]R_s[/tex] = Radius of satellite = 36800 km
Distance to the satellite from the surface
[tex]R=\sqrt{R_e^2+R_s^2}\\\Rightarrow R=\sqrt{6371000^2+36800000^2}\\\Rightarrow R=37347418.13\ m[/tex]
Distance the signal will travell is two times
[tex]d=37347418.13+37347418.13=74694836.26\ m[/tex]
Time is given by
[tex]T=\dfrac{d}{v}\\\Rightarrow T=\dfrac{74694836.26}{3\times 10^8}\\\Rightarrow T=0.24898\ s[/tex]
The time taken is 0.24898 seconds