Question

Newton’s version of Kepler’s third law is P^2 = 4pi^2/(G(M_1+M_2)) a^3. Since the square of the period P varies inversely with the sum of the masses (M_1 + M_2), the period itself depends on the inverse square root of the object masses. If a solar system has a star that is 4.0 times as massive as our Sun, and if that solar system has Earth's exact twin, what is the orbital period of that planet?

(a) 6 months
(b) 10 months
(c) 9 months
(d) 12 months

Answer 1

P^2=k1/4

period=k/2 six months
period itself depends on the inverse square root of the object masses.

Answer 2

Thanks a lot!