A satellite is in circular orbit at an altitude of 1500 km above the surface of a nonrotating planet with an orbital speed of 3.9 km/s. The minimum speed needed to escape from the surface of the planet is 9.6 km/s, and G = 6.67 × 10-11 N · m2/kg2. The orbital period of the satellite is closest to
54 min.37 min.49 min.43 min.60 min.
60 min is the correct answer.
given
h = 1500 km = 1.5*10^6 m
vo = 3.9 km/s = 3900 m/s
ve = 9.6 km/s = 9600 m/s
let M is the mass of planet and R is the radius of
planet
we know, orbital speed, vo = sqrt(G*M/(R+h))
escape speed, ve = sqrt(2*G*M/R)
ve/vo = sqrt(2*(R+h)/R)
(ve/vo)^2 = 2*(R + h)/R
(9600/3900)^2 = 2*(R + 1.5*10^6)/R
on solving the above equation we get
R = 7.39*10^5 m
radius of orbit, r = R + h
= 7.39*10^5 + 1.5*10^6
= 2.239*10^6 m
orbital periode, T = 2*pi*r/vo
= 2*pi*2.239*10^6/3900
= 3607 s
= 3607/(60)
= 60.1 min <<<<<--------------ANswer
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