Two newly discovered planets follow circular orbits around a star in a distant part of the galaxy. The orbital speeds of the planets are determined to be 43.5 km/s and 54.0 km/s. The slower planet's orbital period is 6.53 years. (a) What is the mass of the star? (b) What is the orbital period of the faster planet, in years?
To solve this problem, we can use Kepler's Third Law, which states that the ratio of the squares of the orbital periods of two planets is equal to the ratio of the cubes of their average distances from the star. We can then use this information to find the mass of the star.
(a) To find the mass of the star, we need to first find the average distance of the slower planet from the star, which we'll call r1. We can use the formula for centripetal force to relate the orbital speed of the slower planet (v1) to the gravitational force between the planet and the star:
F = (mv^2) / r
where F is the gravitational force, m is the mass of the slower planet, v1 is its orbital speed, and r1 is its distance from the star.
We can rearrange this equation to solve for r1:
r1 = (mv1^2) / F
Now, we want to find the mass of the star (M). The gravitational force between the planet and the star can be expressed as:
F = G * (M * m) / r1^2
where G is the gravitational constant.
Rearranging this equation to solve for M, we have:
M = (F * r1^2) / (G * m)
We can substitute the given values into this equation to find the mass of the star.
(b) Once we have the mass of the star, we can use Kepler's Third Law to find the orbital period of the faster planet. We'll call the average distance of the faster planet from the star r2. Using the same equation as above, but substituting r2 for r1, we can solve for the orbital period of the faster planet (T2):
T2 = sqrt((r2^3 * T1^2) / r1^3)
where T1 is the orbital period of the slower planet, and we know its value.
We can substitute the known values into this equation to find the orbital period of the faster planet.