For a 98 kg person standing at the equator, what is the magnitude of the angular momentum about Earth's center due to Earth's rotation? Assume the Earth has a radius of 6.4 x 106 m.
Angular Momentum = rmv * sin(theta)
The Attempt at a Solution
r= 6.4*10^6 m
m = 98kg
Vangular (of earth) = 7.27*10^5 rad/s
I multiplied these 3 values together and got 45597.44 kgm^2/s as my answer but it's wrong. Do I have to incorporate anything about the sin of the angle into my equation (I figured it would just be sin(90) = 1 though...)? I don't know what else to do :\