Physics question, seems easy but cant solve it

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 44.8 km/s and 50.9 km/s. The slower planet's orbital period is 8.15 years. (a) What is the mass of the star? (b) What is the orbital period of the faster planet, in years?

Well its not a tough problem indeed.... Here is the solution:

a)
The centripetal force of the slower planet = The gravitational force between them
m*v^2/R = GMm/R^2 Here, m= slower planet's mass
or, M = R*v^2/G v= slower planet's velocity
M= Star's Mass
R= Radius of the orbit of the planet
We know V and G(Gravitational Constant), We need R.
v = wR Here, w(omega)= angular velocity of the slower planet
or, v = (2pi/T)*R as w(omega)= 2pi/t where T = Period or, Orbital Period
or, R = v*T/2pi 2pi = 2 * 3.14159

After calculating, R = 1.8325 * 10^12 m

So, M = R*v^2/G

After calculating, M = 5.5118 * 10^31 Kg

B)
Lets guess,
R1=Slower planet's orbit's radius
R2=Faster planet's orbit's radius
V1=Slower planet's velocity
V2=Faster planet's velocity
T1=Slower planet's orbit period
T2=Faster planet's orbit period

We can write, M1*V1^2/R1 = GM(M1)/R1^2
or, V1^2 = G*M/R1........(1)
Similarly, V2^2 = G*M/R2........(2)

So, (1)/(2), (V1/V2)^2 = R2/R1
or, R2 = R1*(V1/V2)^2 = 1.4195 * 10^12 m

From, V2 = (W2)*(R2)
or, V2 = (2pi/T2)*(R2)
or, T2 = 2pi*R2/V2 = 2*3.1416*1.4195 * 10^12/(50.9*1000) = 175237701.5 secs
= 175237701.5/(60*60*24*365) = 5.5567 yrs


I hope u'l undrstnd.........

for the slower planet, you know the velocity and the orbital period, so by finding how far it goes in 8.15 years you find the circumference of the orbit. then divide by pi to get the radius. then you know the acceleration is v^2/r, and is also G*(star mass)/r^2, so set them equal and calculate the mass of the star. then do this backwards to get the orbital period of the faster planet