Asked by Luke
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?
Answers
Answered by
Damon
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
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
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