Question

The orbital period of a satellite is 1 hr 34 min. It orbits at an altitude of 480 km above the Earth's surface.

a) Determine the initial velocity in order to launch the satellite in that circular orbit.

b) Find the speed of the satellite once its circular orbit has been achieved. Assume a uniform circular motion.

Answer 1

a)

r = orbital radius = radius of earth + altitute = 6.4 x 10⁶ + 0.48 x 10⁶ = 6.88 x 10⁶ m

T = 1 hr 34 min = 60 + 34 min = 94 min = 94 x 60 sec = 5640 sec

orbital speed is given as

v = 2r/T

v = 2 (3.14) (6.88 x 10⁶) /(5640)

v = 7.6 x 10³ m/s

using conservation of energy

(0.5) m v_i² - GMm/R = (0.5) m v_f² - GMm/r

(0.5) v_i² - GM/R = (0.5) v_f² - GM/r

(0.5) v_i²- ((6.67 x 10^-11) (5.98 x 10²⁴))/(6.4 x 10⁶) = (0.5) (7.6 x 10³)² - ((6.67 x 10^-11) (5.98 x 10²⁴))/(6.88 x 10⁶)

v_i = 8.2 x 10³ m/s

b)

v = 7.6 x 10³ m/s