omg physics question!! help me!?

Question

What is the Average radius of the earths orbit with the sun.
Given Mass of sun - 2.00x10^30kg
Equations.. Fg = G x (m1xm2) / r^2
Fg = Fc
Fc = M x Centripetal Acceleration (which is noted as Ac)
Ac - v^2 / r

Answer 1

P = 365.259 days
G = 6.6742*10^-11 m^3s^-2kg^-1
M = 2*10^30 kg

v^2/r = GM/r^2
(rω)^2/r = GM/r^2
r^3 = GM/ω^2
r^3 = GM/(2π/P)^2
r^3 = [(365.259 days)(24 hr/day)(3600 s/h)]^2(6.6742*10^-11 m^3s^-2kg^-1)(1 km/1,000 m)^3(2*10^30 kg)/(4π^2)
r^3 = (9.95931 19674 41817 6*10^14 s^2)(6.6742*10^-20 km^3s^-2kg^-1)(5*10^29 kg)/π^2
r^3 = (6.64704 39933*10^-5 km^3kg^-1)(5.06605 91821*10^28 kg)
r^3 = (33.674 318 262*10^21 km^3
r ≈ 149,887,794 km ≈ 93,621,358 mi

Rounded to 3 significant digits, you have

r ≈ 150,000,000 km ≈ 93,600,000 mi

Answer 2

Gm1m2/r^2=m2v^2/r

simplified after algebraic canceling of m2 and one r becomes

Gm1/r=v^2.......
we will come back to this

the earth takes 365 days to orbit, or 365 days to travel one circumference. Since v=d/t ......
v=2(pi)(r)/t.....t must be in seconds...so convert years to seconds.

combining equations and squaring the velocity we get

GM1/r=2^2(pi)^2(r)^2/t^2....after simplifying
(GM1t^2)/(4pi^2)=r....plug in your numbers to solve

Answer 3

You don't need gravitational forces to determine that.

You need:
Diameter of sun
Distance to sun from earth
Diameter of earth