Question

What is the speed of a point on the earth's surface located at 2/5?

Answer 1

Angular distance from equator to pole: 90 degrees  

2/3 that distance is 60 degrees  

radius is 6.37x10^6  

so we draw a right triangle with hypotenuse 6.37x10^6 and an one angle at 60 degrees. The shortest line of the triangle is the radius of your new point.  

r = 6.37x10^6*cos(60) = 3.185x10^6 m  

since angular speed is constant.  

v = wr = 7.3x10^-5*3.185x10^6  

v = 232.5 m/s