Fowles06

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 mV r ρ ρπ == 1 3 3 4 s m r πρ  =   () 22 4 2 33 3 2 44 43 2 s Gmm Gm G Fm m r   =     2 1 2 3 3 4 FFG m m Wm g g 2 11 2 2 3 6 3 1 3 3 23 6.672 10 4 11.35 1 10 1 49 . 8 3 1 0 1 F N m kg g cm kg cm kg 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. 1 4 Rr GmM r R Fd Rr s + =+ s 22 22 2 4 GmM R r R r Rr Rr R rR r R r + +− 2 20 4 GmM Fr 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 →+ . ψ θ P Q R r s 2 2 R Gd rR φ + =− s 2 2 R GR r R rR + − r 2 4 R M GG R R For , < is independent of r. It is constant inside the spherical shell. 6.3 2 ˆ r GMm Fe 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 ˆˆ 3 rr FG m r e k r e πρ =− G The force is a linear restoring force and induces simple harmonic motion. r F G 23 22 4 m T kG ππ ω == = The period depends on the earth’s density but is independent of its size. At the surface of the earth, 3 4 3 e ee GMm Gm mg R RR 4 3 e Gg R = 6 2 6.38 10 1 1 . 9.8 3600 e R mh r Th gm s s × ×≈ 4 r 6.4 2 ˆ g r GMm Fe r G , where 3 4 3 M r = 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 xg FF θ = The net force on the particle is … 4 ˆ cos 3 FiGm r ρθ G cos rx = 4 3 xi k 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 rr = so 2 GM v r = for a circular orbit r , v is constant. 2 r v T π = 22 2 23 2 44 r Tr vG M ππ == ∝ 3 r 6.6 (a) 2 r v T = 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 gR = 1 3 2 4 e Tg R r = 1 1 2 2 2 2 3 3 6 24 3600 9.8 4 4 6.38 10 ee rT g h r s h r m s RR m −− ×⋅ × == × 2 6.62 7 e r R =≈ (b) () 3 3 3 2 2 11 26 0 60 2 2 e e R R r T g = 1 36 2 2 2 2 2 2 60 6.38 10 2 9.8 3600 24 m m s s hr hr day ×× = ⋅× ⋅ × 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 earth’s surface is … e R = vg 2 2 R R T 3

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This is the same expression as derived in Problem 6.3 for a particle dropped into a hole drilled through the earth. T 1.4 hours.
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This note was uploaded on 03/09/2008 for the course PHYS 301 taught by Professor Mokhtari during the Fall '04 term at UCLA.

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

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