An exo-planet orbits its local star, of mass 2.00x10^30kg, in a steady circular orbit of radius 8.00x10^8km. Calculate the orbital period of the star, in years.

Key idea is equating gravitational force between planet and star (right hand side) with the centripetal force necessary to maintain the orbit (left hand side):(mv²)/r = (GMm)/r²Notice how the mass of the planet cancels here; only the mass of the star is needed. Thus this can be rearranged to find the orbital velocity.v² = GM/r.The orbital velocity is related to the angular frequency by v = ωr and the angular frequency, ω, can be related to the period by ω = 2π/T. Hence we get:2π/T = (GM/r³)^(1/2)And hence T = 2π((r³/GM))^(1/2).Substitute in the values for M, G and r to calculate T, the orbital period. Answers should be given to 2sf, given the precision in the data provided. Be sure to convert the answer into years from seconds, and be aware that r is given in kilometres and will need to be converted to metres.T = 2π((8.00x10¹¹m)³/6.67x10^-11Nm²kg^–2 x2.00x10³⁰kg)^(1/2) = 3.89x10⁸s or 12.3 years.