HW5_prob_1&amp;2_Tunnel

# HW5_prob_1&amp;amp;2_Tunnel - Tunnel Problem The traversal...

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Tunnel Problem The traversal time t 12 canbeexpressedas t 12 = Z 2 1 ds v , (1) where ds is a di f erential displacement along the tunnel and v is the particle velocity. By symmetry the tunnel always lies in a plane that includes the center of the earth. Thus, we can use plane polar coordinates to describe the trajectory, and ds is given by ds = p dr 2 + r 2 d θ 2 = R q 1+ u 2 · θ 2 du , (2) where u = r/R and · θ = d θ /du . The radial coordinate u is measured from the earth’s center ( R is the earth’s radius), and θ is measured along the trajectory from an axis joining the earth’s center and the initial point of the tunnel on the earth’s surface. The potential energy V of a particle of mass m inside the earth is V = mMG 2 R (3 u 2 )= mgR 2 (3 u 2 ) , (3) where g is the acceleration of gravity at the earth’s surface ( = MG/R 2 ). The particle velocity can be found as a function of u by employing the conservation of energy, T + V = mgR , assuming that the particle velocity is zero at the time of its release into the tunnel at the earth’s surface ( u =1 ). Since T = mv 2 / 2 , we f nd v = p gR (1 u 2 ) . (4) After substituting Eqs.(2) and (4) into Eq.(1), we

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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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HW5_prob_1&amp;amp;2_Tunnel - Tunnel Problem The traversal...

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