SOLUTIONS TO CONCEPTS CHAPTER 11

Transcription

1 SLUTINS T NEPTS HPTE. Gvittionl fce of ttction, F N (0.). To clculte the gvittionl fce on t unline due to othe ouse. F D G 4 ( / ) 8G E F I F G ( / ) G ( / ) G 4G 4 D F F G ( / ) G esultnt F F G 4 G G 0 esultnt F E G 4 The net esultnt fce will be, G 4 G 5 G G G F G G (0 89.) G G ) if is plced t id point of side then F 4G in diection 4G F in diection Since equl & opposite cncel ech othe F oc G / 4G Net gvittionl fce on b) If plced t (centoid) G G the F ( / ) in diection 4G.

2 F 4. G esultnt F G G G G Since F, equl & opposite to F, cncel Net gvittionl fce 0 G G F cos 0î sin0 ĵ 4 4 G G F cos 0î sin0 ĵ 4 4 F F + F G G G sin0 ĵ Fce on t due to gvittionl ttction.. G F ĵ F D î 4 F cos 45 ĵ sin 45 ĵ 4 4 So, esultnt fce on, F F + F + FD î ĵ 4 4 F 4 F oving long the cicle, F 4 h.7 0 V V ( ) 0 v hpte /s The line oentu of bodies is 0 initilly. Since gvittionl fce is intenl, finl oentu is lso zeo. So (0 kg)v (0 kg) v v v () Since P.E. is conseved Initil P.E J When seption is 0.5, D

3 (/ ) v + 0 v v 9 + (/) 0 v + (/) 0 v () v v. 0 5 /s. So, v /s. 8. In the seicicle, we cn conside, sll eleent of d then d (/L) d d. d F L df df since sin d L d / / F sind cos 0 L L L 0 ( ) L L L L / L 9. sll section of od is consideed t distnce ss of the eleent (/L). d d G(d) de d de esultnt de de sin G(d) d d d Totl gvittionl field E L / 0 L d Gd d / d d d L d Integting the bove eqution it cn be found tht, E d L 4d 0. The gvittionl fce on due to the shell of is 0. is t distnce Then the gvittionl fce due to is given by ( / 4 ( ). n of eth (4/) n of the iginy sphee, hving dius, (4/) Gvittionl fce on F F de d hpte de d d.

4 . Let d be the distnce fo cente of eth to n then D 4 (/) 4 be the ss of the eth, the ss of the sphee of dius d/. Then (4/) (4/)d d Gvittionl fce is, G Gd d F d d So, Nl fce eeted by the wll F cos. d (theefe I think nl fce does not depend on ) d. ) is plced t distnce fo. If <,, Let s conside thin shell of n d (4 / ) Thus d 4 Gd G / Then gvittionl fce F b) < <, then F is due to only the sphee. G F G c) if >, then Gvittionl fce is due to both sphee & shell, then due to shell, F G due to the sphee G So, esultnt fce + 4. t P, Gvittionl field due to sphee t P, Gvittionl field is due to sphee & shell, ( 4 ) + (4 ) hpte 5. We know in the thin spheicl shell of unif density hs gvittionl field t its intenl point is zeo. t nd point, field is equl nd opposite nd cncel ech othe so Net field is zeo. Hence, E E. Let 0. kg n is fo kg ss nd ( ) fo 4 kg ss ( ) 49 P n P / d F / d.4

5 hpte ( ) ( ) ( ) ( + ) 0.8 fo kg ss Initilly, the ide of is To incese it to, G G G wk done 8. Wk done ginst gvittionl fce to tke wy the pticle fo sphee, G E (5 N/kg) î + ( N/kg) ĵ ) F E J kg [(5 N/kg) î + ( N/kg) ĵ ] (0 N) î + ( N) ĵ 0kg 00g 0c F N b) V E t (, 0), V (0 J/kg) î V 0 J t (0, 5 ), V (0 J/kg) ĵ V 0 J c) V (,,5) d (0,0) E (,5) ( 0N)î (4N) ĵ (0 J î + 0 J î ) 40 J 0,5 (0Nî 4Nĵ) d) v,0 0 ĵ + 0 î 0 0. ) V (0 N/kg) ( + y) LT L 0 L T 0 L T L.H.S.H.S b) E (,y) c) F E L T L 0(N/kg) î 0(N/kg) ĵ L T (0,0) 0.5kg [ (0 N/kg) î (0 N/kg) ĵ 0N î - 0 N ĵ F N. E î + ĵ The field is epesented s tn / 5/ j gin the line y + 5 cn be epesented s tn / 5/ j Since, the diection of field nd the displceent e pependicul, is done by the pticle on the line..5

6 hpte. Let the height be h (/) ( h) ( + h) + h h ( ). Let g be the cceletion due to gvity on ount eveest. h g g ( 0.007) 9.77 /s Let g be the cceletion due to gvity in ine. d Then g g /s Let g be the cceletion due to gvity t eqution & tht of pole g g g 9.8 ( ) /s g kg 9.77 /s 9.77 N kg The body will weigh kg t equt.. t equt, g g () Let t h height bove the south pole, the cceletion due to gvity is se. h Then, hee g g () g - g h h g h g The ppent g t equt becoes zeo. i.e. g g 0 g T g sec..4 hou.0 8. ) Speed of the ship due to ottion of eth v b) T 0 g g T 0 g c) If the ship shifts t speed v T g 5 N 0K (ppoitely). 0 d/s. To.

7 T 0 - v T 0 v v T T 0 + v 9. ccding to Keple s lws of plnety otion, T T e T s es s es.88 hpte s (.88) /.5 es 0. T (7.) 5 (.4) (.84) kg ss of eth is found to be kg.. T (7540) (.8) ) V (.8) (9.4) 0 h 9.8 (400 0 (7540) g h 0 (.4 ) ).5 0 kg..9 0 /s.9 k/s b) K.E. (/) v (/) 000 (47. 0 ) J c) P.E. ( h) ( ) 0 ( h) d) T V J 7. 0 sec. hou.7

8 hpte. ngul speed f eth & the stellite will be se T e Ts 4 00 g ( h) I 00.4 g ( h) g ( h) ( 00) (.4) (400) 0 9 (400 h) 0 (400 h) (400 + h) h ( ) / h ( ) / c. b) Tie tken fo nth pole to equt (/) t (/).8 ( ) 0 (400) 0.4 (497) 0 (4) 0 ( 00) (.4) hou F geo sttiony stellite, k h. 0 4 k Given g 0 N gh g h T g T 4 g g 4 T N 7980 cceletion due to gvity of the plnet is. The colttitude is given by. 90 gin sin sin T sin 0.5. oltitude.8

9 hpte 7. The pticle ttin iu height 400 k. n eth s sufce, its P.E. & K.E. E e (/) v + In spce, its P.E. & K.E. E s + 0 h E s Equting () & () v (/) v v v 8 () () ( h ) /s 7.9 k/s. 8. Initil velocity of the pticle 5k/s Let its speed be v t intestell spce. (/) [(5 0 ) v ] d (/) [(5 0 ) v ] (/) [(5 0 ) v ] 5 0 v v v (.05) v /s 0 k/s 9. The n of the sphee 0 4 kg. Escpe velocity 0 8 /s V c V c