10prob_11

# 10prob_11 - 81.3 r e r M M F 1 F 2 F y F 1 x F 2 x m 81.3 r...

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Unformatted text preview: 81.3 r e r M M F 1 F 2 F y F 1 x F 2 x m 81.3 r e r M M F 1 F 2 F y F 1 x F 2 x m Physics 9A-C Assignment #10 Cole [12.13]UseNewton'ssecondlawandremembertheforcesarevectors.From thesymmetry,the F x willcancel. F y = 2 GMm r 2 cos =ma y where cosoe=6/10 a y = 2 6.673 10 11 N m 2 / kg 2 ( ) 0.26 kg ( ) 0.1 m ( ) 2 6 10 =-2.110-9 m/s 2 . [12.41]a)Dividetheringintosmallpiecesofmass dM . dU = GmdM r Allthesepiecesare thesamedistance r = x 2 + a 2 from m, so G,m and r allareconstantsthatcanbepulled outsidetheintegral . U = Gm x 2 + a 2 dM M = GMm x 2 + a 2 b) im a U = GMm x 2 = GMm x c) F x = dU dx = d dx GMm x 2 + a 2 ( ) 1/ 2 F x = GMm x x 2 + a 2 ( ) 3/2 d) im a F x = GMm x x 2 ( ) 3/ 2 = GMm x 2 e)When x=0,U= U = GMm 2 + a 2 = GMm a Rememberthat U=0 at r= andnotatthe origin.Attheorigin,allofthemassisthesamedistance a awayfrom m, sotheresultmakes sense.Fortheforce,the x inthenumeratormakes F x =0. Bysymmetry,alloftheforcemust canceloutsothisalsomakessense. [12.55]Toremainabovethesamepointontheearthatalltimes,theorbitmustbeacircleinthe equatorialplaneoftheearth(inanellipticalorbitthesatellitewouldspeedupandslowdown, anyorbitotherthantheequatorialorbitwouldhavethesatellitesometimesoverthenorthern hemisphereandsometimesoverthesouthern.a)=2/ T F r = GMm r e + h ( ) 2 = m r e + h ( ) 2 r e + h ( ) 3 = GMT 2 4 2 Theperiodofthe earthis24hr(actually1/365lessthanthis--seepage378,middleofthe secondparagraph). r e + h = 6.673 10 11 N m 2 / kg 2 ( ) 5.97 10 24 kg ( ) 86,164 s ( ) 2 4 2 3 h= 3.5810 7 m=3,580km. Forobserversnorthof cos = r e r oe=81.3,thesatelliteisbelowthehorizon. [12.77]a)Usethesemi-majoraxisoftheorbitinthe F r = GMm a 2 = ma 2 T 2 where a= 1 2 ( 0.410 6 m+410 6 m+2(6.410 6 m))=8.5810 6 m. T = 2 a 3/2 GM =7.910=7....
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## This note was uploaded on 09/28/2011 for the course PHY 9A taught by Professor Svoboda during the Spring '08 term at UC Davis.

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10prob_11 - 81.3 r e r M M F 1 F 2 F y F 1 x F 2 x m 81.3 r...

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