It takes earth 365 days to complete one orbit around the sun. Knowing that the magnitude of the average acceleration of the earth experiences is over half an orbit is 3.78 x 10^-3 m/s^2, what is the average speed with which it travels?

Answer 1

Ac = v^2/R = omega^2 R

omega = 2 pi radians/ 365 days
but 365 days *24 hr/day * 3600 s/hr = 365*24 *3600 seconds
so
omega = 2 pi /(365*24*3600) = 2*10^-7 radians/second
omega^2 = 4 * 10^-14
so
3.78*10^-3 = 4*10^-14 * R
R = .945 * 10^11 = 9.45 *10^10 meters
v = omega * R = 2*10^-7 * 9.45 * 10^10 = 18.9 *10*3 = 18,900 m/s