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Unformatted text preview: Problem GPS 4.22 (2 nd ed. 4.23) If we neglect the very small “centrifugal force” term in Eq.(4-129) [3 rd ed. Eq.(4.91)], the equation of motion for a particle of mass m in the rotating frame of the earth is m d v dt = m g + 2 m v × ω , (1) where v is the particle velocity, ω is the earth’s angular velocity vector, and g is the acceleration of gravity at the earth’s surface. All the vectors and derivatives are measured in a body f xed frame in which the z axis is normal to the earth’s surface and the x- y plane is tangent to the earth’s surface and is located at z = 0 . Assume that we are in the Northern hemisphere, and let the y axis point north and the x axis point east. As we saw in class, the solution to Eq.(1) to f rst order in ω is r = r + v t + 1 2 g t 2 + ( v × ω ) t 2 + 1 3 ( g × ω ) t 3 , (2) where the subscript denotes an initial value. Since the projectile is f red horizontally, its initial velocity has no z compo- nent, and we can write v = v x e x + v...
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