Question

Two planets P1 and P2 orbit around a star S in circular orbits with speeds v1 = 43.8 km/s, and v2 = 58.6 km/s. A) If the period of the first planet

P₁ is 780 years what is the mass, in kg, of the star it orbits around?B) Determine the orbital period, in years, of P₂. Please show each step of the process

Answer 1

For a planet
F = GMm/R^2 = mv^2/R ............... (1)
where G is the universal gravitational constant, M is the mass of the star, m is the mass of the planet, R is the orbital radius and v is orbital speed.
For the slower planet
T = 2.R/v
R = Tv/(2) = 780 x 365x 24 x 60 x 60 x 43.8 x 1000/(2) = 1.714 x 10¹⁴ m
Going back to (1)
GM/R = v^2
M = Rv^2/G = 1.714 x 10¹⁴ x (43.8 x 1000)² / 6.673 x 10^-11 = 4.929 x 10³³ kg

For the faster planet (from (1))
GM/R = v^2
R = GM/v^2 = 6.673 x 10^-11 x 4.929 x 10³³/(58.6 x 1000)^2 = 9.579 x 10¹³ m
T = 2R/v = 2 x 9.579 x 10¹³/(58.6 x 1000) = 1.027 x 10¹⁰ s = 325.68 yr