Question

A planet of mass m = 6.25 × 1024 kg is orbiting in a circular path a star of mass M = 7.85 × 1029 kg. The radius of the orbit is R = 5.65 × 107 km. What is the orbital period (in Earth days) of the planet Tplanet?

Answer 1

M = 7.85 x 10²⁹ kg
m = 6.25 x 10²⁴ kg
R = 5.65 x 10¹⁰ m
G = 6.67 x 10^-11 m³ kg^-1 s^-2.

Using Kepler's third law,
T = sqrt[4 $\pi$ ²R³/G(M + m)]
= sqrt{[4 $\pi$ ²(5.65 x 10¹⁰)³] / [(6.67 x 10^-11) x (7.85 x 10²⁹ + 6.25 x 10²⁴)]}
= 11.66 x 10⁶ s
= [(11.66 x 10⁶) / (24 x 60 x 60)]
= 134.97 days