Question

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 44.2 km/s and 57.5 km/s. The slower planet's orbital period is 7.50 years.

(a) What is the mass of the star?
kg

(b) What is the orbital period of the faster planet, in years?

Answer 1

a)

Force of Gravity:
Fg = (Gm1m2)/r^2

G is universal gravitational constant: 6.67 x 10^(-11)
m1, m2 are the two masses that attract each other (star and selected planet)
Fg is the attractive force between the two masses
r is the distance between the planet and the star

F also equals [M(v^2)]/r, where M is the mass of the planet of interest

Lets make m1 = mass of star and m2 = M, Equate the forces:

(Gm1m2)/r^2 = [m2(v^2)]/r

Reduces to (Gm1)/r = (v^2)

Gm1/r = v^2, for the slower planet

6.67 x 10^(-11)m1/r = 42.7^2 = 1823

We can solve for r, knowing that 2 ? r is the circumference of the orbit

8.14 years x (31.5 x 10^6) sec/year x 42.7 km/sec = 1.1 x 10^10 km
Divide by 2 ? and you get:

r = 1.75 x 10^9 km

We established that 6.67 x 10^(-11)m1/r = 42.7^2 = 1823

Solving for m1 gives us [1824/6.67 x 10^(-11)]r = [1824/6.67 x 10^(-11)] 1.75 x 10^9

m1 = 4.79 x (10^22) kg

b)
We need to know the radius of orbit for the faster planet

We have established that (Gm1)/r = (v^2)

So r = (Gm1)/(v^2)

[(6.67 x 10^(-11))(4.79 x (10^22))]/(63.1^2)

=8 x 10^8 km, so the orbit circumference is 2 ? (8 x (10^8)) = 5.03 x 10^9 km

If you divide by the orbital speed of 63.1 km/s we get a period of 7.966 x (10^7) seconds per orbit

This equals 2.53 years.