HW9_prob1_GPS21_G22

# HW9_prob1_GPS21_G22 - Problem GPS 4.21(2nd ed 4.22 If we...

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Problem GPS 4.21 (2 nd ed. 4.22) If we neglect the “centrifugal force” term in GPS 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 aremeasuredinabody 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 0 + v 0 t + 1 2 g t 2 +( v 0 × ω ) t 2 + 1 3 ( g × ω ) t 3 , (2) where the subscript 0 denotes an initial value. Since the initial velocity is upward (vertical), we can write v 0 = v 0 e z , (3) and gravity acts downward, so g = g e z . (4) With r 0 =0 ,the f rst three terms of Eq.(2) are seen to determine the height h and total time t tot of the particle’s trajectory as t tot =2 t h =2 v 0 /g , (5) h = 1 2 gt 2 h , (6) where t h is the time required for the particle to fall to the earth when it is released at the height h above the surface. The fourth and f fth terms in Eq.(2) are responsible for the de F ection of the particle from the vertical. Because ω lies, generally, in the body y - z plane, the cross products generate vectors along the body x axis, i.e., e z × ω = ω y e x = ω sin θ e x , (7) where θ is the polar angle, or colatitude, measured from the earth’s rotation axis. (The equator is at θ = π/ 2 .) Thus

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## This note was uploaded on 12/19/2010 for the course PHYSICS ph 409 taught by Professor Wilemski during the Fall '10 term at Missouri S&T.

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HW9_prob1_GPS21_G22 - Problem GPS 4.21(2nd ed 4.22 If we...

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