Question

An Earth satellite moves in a circular orbit 924 km above Earth's surface with a period of 103.3 min. What are (a) the speed and (b) the magnitude of the centripetal acceleration of the satellite?

Answer 1

Given that :

height above the earth surface, h = 924 km

radius of earth, r_E = 6371 km

time-period, T = 103.3 min = 6198 sec

orbital radius which is given by, r = r_E + h

r = (6371 km) + (924 km)

r = 7295 km

r = 7.29 x 10⁶ m

(a) The speed of satellite will be given as ::

using an equation, v = 2r / T                                                     { eq.1 }

inserting the values in eq.1,

v = 2 (3.14) (7.29 x 10⁶ m) / (6198 sec)

v = (45.78 x 10⁶ m) / (6198 sec)

v = 7386.2 m/s

(b) Magnitude of the centripetal acceleration of the satellite which is given as ::

a_r = v² / r                                                                                  { eq.2 }

inserting the values in eq.2,

a_r = (7386.2 m/s)² / (7.29 x 10⁶ m)

a_r = (54555950.4 m²/s²) / (7.29 x 10⁶ m)

a_r = 7.48 m/s²