The earth's radius is 6.37×10^6m; it rotates once every 24 hours. with the angular speed of 7.3 x 10^-5 what is the speed of a point on the earth's surface located at 2/3 of the length of the arc between the equator and the pole, measured from equator? (hint: what is the radius of the circle in which the point moves? )

The speed of the earth's surface located at 2/3 of the length of the arc between the pole which measure from the equator is 232.5 m/s.

Solution:
So the givens are, earth's radius = 6.37X10^6m, and the angular distance from the pole is 90 degrees. So 60 degrees is the 2/3.

r = 6.37x10^6 * cos(60) = 3.185x10^6m
since v = wr
v = 7.3x10^-5 * 3.185x10^6

v - 232.5 m/s

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