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.

Answer 1

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

Answer 2

The time taken to travel will be "0.24898 seconds".

Given:

• Speed of light, $v = 3\times 10^8 \ m/s$
• Radius of earth, $R_e = 6371000 \ m$
• Radius of satellite, $R_s = 36800 \ km$

Now,

From the surface, distance to satellite will be:

→ $R = \sqrt{R_e^2+R_s^2}$

      $= \sqrt{(6371000)^2+(36800000)^2}$

      $= 37347418.13 \ m$

The signal travel in two time will be:

→ $d = 2\times 37347418.13$

     $= 74694836.26 \ m$

hence,

The time will be:

→ $Time = \frac{distance}{speed}$

By substituting the values, we get

            $= \frac{74694836.26}{3\times 10^8}$

            $= 0.24898 \ s$

Thus the above response is correct.      

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