Question

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?

Answer 1

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.

$\frac{mV^{2}}{R} = \frac{GMm}{R^{2}}$

$V = \sqrt{\frac{GM}{R}}$

$V = \sqrt{\frac{(6.67\ast 10^{-11})(3.85\ast 10^{29})}{8.35\ast 10^{10}}}$

V = 17536.787 m/s

Orbital period of the planet = T

VT = 2$\pi$R

(17536.787)T = 2$\pi$(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.

2.991 10 T 86400

T = 346.18 days

Orbital period of the planet = 346.18 days