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

Question

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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Diễm Thu 4 years 2021-08-22T04:31:35+00:00 1 Answers 49 views 0

Answers ( )

  1. Helga
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    2021-08-22T04:33:19+00:00
    Reply

    Answer:

    0.24898 seconds

    Explanation:

    v = Speed of light = 3\times 10^8\ m/s

    R_e = Radius of Earth = 6371000 m

    R_s = Radius of satellite = 36800 km

    Distance to the satellite from the surface

    R=\sqrt{R_e^2+R_s^2}\\\Rightarrow R=\sqrt{6371000^2+36800000^2}\\\Rightarrow R=37347418.13\ m

    Distance the signal will travell is two times

    d=37347418.13+37347418.13=74694836.26\ m

    Time is given by

    T=\dfrac{d}{v}\\\Rightarrow T=\dfrac{74694836.26}{3\times 10^8}\\\Rightarrow T=0.24898\ s

    The time taken is 0.24898 seconds

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