# Fowles06 - CHAPTER 6 GRAVITATIONAL AND CENTRAL FORCES 6.1 4...

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CHAPTER 6 GRAVITATIONAL AND CENTRAL FORCES 6.1 3 4 3 s m V r ρ ρ π = = 1 3 3 4 s m r πρ = ( ) 2 2 4 2 3 3 3 2 4 4 4 3 4 3 2 s Gmm Gm G F m m r πρ πρ = = = 2 1 2 3 3 4 4 3 F F Gm m W mg g πρ = = ( ) 2 11 2 2 3 6 3 1 3 3 2 3 6.672 10 4 11.35 1 10 1 4 9.8 3 10 1 F N m kg g cm kg cm kg W m s g m π × × = × × 3 × × 9 2.23 10 F W = × 6.2 (a) The derivation of the force is identical to that in Section 6.2 except here r < R. This means that in the last integral equation, (6.2.7), the limits on u are R – r to R + r. 2 2 2 2 1 4 R r R r GmM r R F d Rr s + = + s ( ) 2 2 2 2 2 4 GmM R r R r R r R r Rr R r R r = + + + ( ) 2 2 0 4 GmM F r R r R r Rr = + + = (b) Again the derivation of the gravitational potential energy is identical to that in Example 6.7.1, except that the limits of integration on s are ( ) ( ) R r R r + . ψ θ P Q R r s 2 2 R r R r R G d rR πρ φ + = − s ( ) 2 2 R G R r R rR πρ = − + r 2 4 R M G G R R π ρ φ = − = − For , r R < φ is independent of r. It is constant inside the spherical shell. 6.3 2 ˆ r GMm F e r = − G 1

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The gravitational force on the particle is due only to the mass of the earth that is inside the particle’s instantaneous displacement from the center of the earth, r. The net effect of the mass of the earth outside r is zero (See Problem 6.2). 3 r π ρ = 4 3 M 4 ˆ ˆ 3 r r F G mre kre πρ = − = − G The force is a linear restoring force and induces simple harmonic motion. r F G 2 3 2 2 4 m T k G π π π ω πρ = = = The period depends on the earth’s density but is independent of its size. At the surface of the earth, 3 2 2 4 3 e e e GMm Gm mg R R R π ρ = = 4 3 e G g R πρ = 6 2 6.38 10 1 2 2 1. 9.8 3600 e R m hr T h g m s s π π × = = × 4 r 6.4 2 ˆ g r GMm F e r = − G , where 3 4 3 M r π ρ = The component of the gravitational force perpendicular to the tube is balanced by the normal force arising from the side of the tube. The component of force along the tube is cos x g F F θ = The net force on the particle is … 4 ˆ cos 3 F i G mr πρ θ = − G cos r x θ = 4 ˆ ˆ 3 F i G mx ik πρ = − = − G x As in problem 6.3, the motion is simple harmonic with a period of 1.4 hours. 2
6.5 2 2 GMm mv r r = so 2 GM v r = for a circular orbit r , v is constant. 2 r v T π = 2 2 2 2 3 2 4 4 r T r v GM π π = = 3 r 6.6 (a) 2 r v T π = From Example 6.5.3, the speed of a satellite in circular orbit is … 1 2 2 e gR v r = 3 2 1 2 2 e r T g R π = 1 2 2 3 2 4 e T gR r π = 1 1 2 2 2 2 2 2 3 3 2 2 6 24 3600 9.8 4 4 6.38 10 e e r T g hr s hr m s R R m π π × × = = × 2 6.62 7 e r R = (b) ( ) 3 3 3 2 2 1 1 2 2 2 60 60 2 2 e e e e R R r T g g R g R π π π = = = 1 3 6 2 2 2 2 2 2 2 2 60 6.38 10 2 9.8 3600 24 m m s s hr hr day π × × = × × 27.27 27 T day day = 6.7 From Example 6.5.3, the speed of a satellite in a circular orbit just above the

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• Fall '04
• mokhtari

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