A satellite of mass 220 kg is launched from a site on Earth's equator into an orbit at 180 km above the surfac

Question

A satellite of mass 220 kg is launched from a site on Earth's equator into an orbit at 180 km above the surface of Earth.
(a) Assuming a circular orbit, what is the orbital period of this satellite?
s
(b) What is the satellite's speed in its orbit?
m/s
(c) What is the minimum energy necessary to place the satellite in orbit, assuming no air friction?
J

Answer 1

(a) 1 hour, 28 minutes, 5 seconds
(b) 7796 m/sec
(c) Since e = 1/2 m v^2, orbital energy is .5* 20*7796*7796 = 6685747583 Joules.
Potential energy needed to lift the satellite to an altitude of 180 km is more complicated, but we will simplify things considerably by neglecting the loss of gravitational potential as the satellite moves farther away from Earth's center, i.e., we will assume no gravitational gradient. In that case, potential energy is given by PE= mgh, so PE = 220*9.8*180,000 = 388080000 Joules.

Total energy needed to orbit the satellite is then
6685747583 + 388080000 = 7073827583 Joules, or
7073.8 MJ.