A geostationary communications satellite orbits the earth directly above the equator at an altitude of 36800 km. Calculate the time it would

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.

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  1. 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

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