Question

The satellite s travels around the earth in a circular path with a constant speed of 20 mm>h. If the acceleration is 2.5 m>s2, determine the altitude h. Assume the earth's diameter to be 12 713 km.

Answer 1

Answer:

The altitude of planet = 5989.198 km

Explanation:

    We know that centripetal acceleration, $a = v^2/r$, where v is the velocity and r is the radius. In this case both acceleration and velocity is given, so we can find radius of satellite.

    v = $20 Mm/hr = 20*10^6/3600 = 5555.56 m/s$

    a = $2.5 m/s^2$

 So,  $2.5 = 5555.56^2/r\\ \\ r = 12345698.77 m = 12345.698km$

 We have diameter of earth = 12713 km

        Radius of earth = 12713/2 = 6356.5 km

   The altitude of planet = Radius of planet - radius of earth

                                         = 12345.698 - 6356.5

                                         = 5989.198 km