Question

The nefarious Dr. X lives on a dark and lonely planet. He attempts to launch a 2000 k control satellite in order to take over his world. The mean radius of the dark and planet is 3 3 times 10^6 m and its mass is M = 7.5 times 10^23 kg. If Dr. X wishes to launch the satellite into a circular orbit 1500 km above the surface of the planet, What would the period of the satellite be once it is in orbit 1500 km the surface? How much energy will it take to launch the satellite from the ground in orbit? Neglect any non-gravitational effects.

Answer 1

m = 2000 kg, M = 7.5*10^23 kg,

R = 3.3*10^6 m , h = 1500 km

(a) From Keplers third law

T^2 = 4pi^2(R+h)^3/(GM)

T^2 = 4*3.14^2*(3300000+1500000)^3/(6.67*10^-11*7.5*10^23)

T = 9337.44 s

T = 9337.44/(60*60)

T = 2.6 hours

(b) potential energy of satellite PE = -GMm/2r

Required energy =W = change in potential energy

W = -GMm/2(R+h) + GMm/2R

W = (6.67*10^-11*7.5*10^23*2000/2)(-1/(3300000+1500000) +1/(3300000))

W = 4.74*10^9 J