SLUTINS T NEPTS HPTE. Gvittionl fce of ttction, F.7 0 0 0.7 0 7 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 00 0 0 5 G 0 40 5 G (0 89.) G G 40. 4 4. ) 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.
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 4 4 4 5. 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 7.4 0 (740 000) 0 v 4 49.58 0. 740 740 0 hpte 49.58 0 5.8 0 0.5 /s 0.75 0 7. 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.7 0 0 0 Initil P.E..4 0 9 J When seption is 0.5, D
.4 0 9 + 0.4 0 (/ ).4 0 9 -.8 0 9 + 5 v + 0 v.4 0 9 -.8 0 9 + 0 v 9 + (/) 0 v + (/) 0 v () v 9.4 0 4.44 0 0 0 v. 0 5 /s. So, v 4. 0 5 /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.
. 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 ) 5 900 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. 0. 4 0. ( ) 49 P n P / d F / d.4
hpte 0. 0.4 ( ) ( ) ( ) ( + ) 0.8 fo kg ss..44 7. 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, G0 0. 0. 0..7 0 0 9. E (5 N/kg) î + ( N/kg) ĵ ) F E.7 0 0 J kg [(5 N/kg) î + ( N/kg) ĵ ] (0 N) î + ( N) ĵ 0kg 00g 0c F 00 57 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 00 00 0 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
hpte. Let the height be h (/) ( h) ( + h) + h h ( ). Let g be the cceletion due to gvity on ount eveest. h g g 79 9.8 9.8 ( 0.007) 9.77 /s 400000 4. Let g be the cceletion due to gvity in ine. d Then g g 40 9.8 9.8 0.9999 9.799 /s 400 0 5. Let g be the cceletion due to gvity t eqution & tht of pole g g g 9.8 (7. 0 5 ) 400 0 9.8 0.04 9.77 /s g kg 9.77 /s 9.77 N 0.997 kg The body will weigh 0.997 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 5 7. 0 400 0 9.8 7. The ppent g t equt becoes zeo. i.e. g g 0 g T g 9.8 400 0.4.5 0.5 0 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.
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 5.84 0 7..4.7 0 5.4.84 0.7.7.5 0.7 0 (7.) 5 (.4) (.84) 0.0 0 4 kg ss of eth is found to be.0 0 4 kg.. T 9.4 0 0 7540.4.7 0 (7540) (.8) 9.4 0. ) V.7 0.7 0 8 (.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 ).8 0 0 J c) P.E. ( h).7 0 0 4 (400 000) 0 ( h) d) T V 0.9 0.4 8400 0 40 0 8400 4.7 0 0 J 7. 0 sec. hou.7
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) 9 9.8 (400) 0 9 (400 h) 0 (400 h) 0 4 0 4 9 7 0 (400 + h) 7 4 0 4 400 + h (7 4 0 4 ) / h (7 4 0 4 ) / 400 400 c. b) Tie tken fo nth pole to equt (/) t (/).8 (400 400) 0 (400) 0.4 (497) 0 (4) 0 ( 00) (.4) 497 497 497.4 hou. 5 4 4 0 4. F geo sttiony stellite, 4. 0 4 k h. 0 4 k Given g 0 N gh g h 400 0 0 400 0 00 5. T g T 4 g g 4 T 0 409 0. N 7980 cceletion due to gvity of the plnet is. The colttitude is given by. 90 gin 400 8 sin 4000 5 sin 8 5 4 T sin 0.5. oltitude.8
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.7 0 400 0 0 4 40.0 0.4 0. 0 7 0. 0 8 v 8 () () ( h ) 0. 0 0.79 0 4 /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.7 0 0 400 0 v 5 0 40.0 0 8 v 5 0. 0 8 0 8 (.05) v.0 0 4 /s 0 k/s 9. The n of the sphee 0 4 kg. Escpe velocity 0 8 /s V c V c.7 0 8 0 0 4 4 80.0 0 8.89 0 9. 9.9