A planet of mass m = 7.85 × 1024 kg is orbiting in a circular path a star of mass M = 3.85 × 1029 kg. The radius of the orbit is R = 8.35 × 107 km. What is the orbital period (in Earth days) of the planet Tplanet?
Gravitational constant = G = 6.67 x 10-11 N.m2/kg2
Mass of the star = M = 3.85 x 1029 kg
Mass of the planet = m = 7.85 x 1024 kg
Radius of the orbit = R = 8.35 x 107 km = 8.35 x 1010 m
Orbital speed of the planet = V
The gravitational force between the star and the planet provides the planet with the necessary centripetal force for the circular motion.
V = 17536.787 m/s
Orbital period of the planet = T
VT = 2R
(17536.787)T = 2(8.35x1010)
T = 2.991 x 107 sec
1 Day = 24 Hours = 24 x 3600 sec = 86400 sec
Converting the orbital period from seconds to days.
T = 346.18 days
Orbital period of the planet = 346.18 days
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