mv2. Equating the two gives 4! 2. The angular velocity is the angle swept per GM (2! )2 4! 2 " 2 = GM . Combining the results we get !

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Download "mv2. Equating the two gives 4! 2. The angular velocity is the angle swept per GM (2! )2 4! 2 " 2 = GM . Combining the results we get !"

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1 Chapte. he net foce on the satellite is F = G Mm and this plays the ole of the centipetal foce on the satellite i.e. mv mv. Equating the two gives = G Mm i.e. v = G M. Fo cicula motion we have that v =! " & and so =! " &. Simplifying gives the esult. Fom poblem we know that = unit time i.e.! = ", i.e. =! ". Substituting, = Fom mv = G Mm we deduce that v = G M. Now = = 6900 km = 6.9! 0 6 m and so v = = 6.67! 0"! 6.0! 0 4 v =! " =! v 4!. he angula velocity is the angle swept pe (! ) 4! " = ". 6.9! 0 6 = 7.6! 0 ms ". he peiod is then found fom = s = 95 min. 4 he peiod has to be one day i.e. 4 hous. hen v =! deduce that v = G M. Combining the esults we get and fom mv = G Mm we = = 6.67 " 0 " 6.0 " 0 4 " (4 " 60 " 60) 4! = 4. " 0 7 m. 4! 5 (a) E P =! eath M moon (b) V =! eath (c) v = eath =! 6.67 " 0! " 5.98 " 0 4 " 7.5 " 0.84 " 0 8 =!7.64 " 0 8 J. =! 6.67 " 0! " 5.98 " " 0 8 =!.04 " 0 6 J kg!. = 6.67! 0"! 5.98! ! 0 8 =.0! 0 m s ". 6 We must plot the function E P =! eathm! moonm giving the gaph in the answes. d! Hee m is the mass of the spacecaft and d the sepaation of the eath and the moon (cente-

2 to-cente). Putting numbes in, E P =! 6.67 " 0! " 5.98 " 0 4 ".0 " 0 4. " 09.5 " 0 7 =!.84 " 0 8!! 6.67 " 0! " 7.5 " 0 ".0 " " 0 8! =. " 09 /.84 " 0 8!.5 " 07 /.84 " 0 8 /.84 " 0 8! /.84 " 0 8. " 00.9 " 08 =! x! x whee x =. In this way the function can be plotted on a calculato to give the 8.84! 0 gaph in the answes in the textbook. 7 (a) V =! 5 e (b) =! 6.67 " 0! " 5.98 " " 6.4 " 0 6 =!. " 0 7 J kg! E P =! m 5 e =! 6.67 " 0! " 5.98 " 0 4 " " 6.4 " 0 6 =!6. " 0 9 J 8 he net foce is the gavitational foce and this must point towads the cente of the eath. his happens only fo obit. 9 s shown in the text the eaction foce fom the spacecaft floo is zeo giving the impession of weightlessness. Moe simply, both spacecaft and astonaut ae in fee fall with the same acceleation. 0 he diffeence is!u = " m + h " " m & ( = m " m + h = m + h " ( + h) " h!u = m ( + h) &. When h is small compaed to this expession is appoximately!u = m h = m h = mgh. he wok done by an extenal agent in moving an object fom = a to = b at a small constant speed. & (.

3 (a) he potential at the suface of the planet is V =! =! 6.67 " 0! " M.0 " 0 5 =!4.9 " 0 J kg!. Hence M =.5! 0 8 kg. (b) he escape speed is obtained fom mv! m = 0, i.e. v =. But V =! and so =!V. Substituting, v =!V. (c) v =!V = " 4.9 " 0 =. " 0 6 m s! (d) he wok equied is W = m!v. his is W = 500! (".0! 0 " ("4.9! 0 )) = 5.8! 0 5 J. (e) We have that mv + mv = mv! v = (V " V ). his gives v = (!. " 0! (!4.9 " 0 )) =. " 0 6 m s!. (a) t = 0.75, g = P (0.75d)! m (0.5d) = 0. Hence M P = (0.75d) = 9. (b) he pobe M m (0.5d) must have enough enegy to get to the maximum of the gaph. Fom then on the moon will pull it in. hen W = mv = m!v " v =!V = (0. 0 (6.4 0 )) = m s. 4 (a) We deduced many times that v = and so = mv! m = m! m =! m. (b) =! 6.67 " 0! ".0 " 0 0 " 6.0 " 0 4 ".5 " 0 =!.7 " 0 J. 5 Using E K = m, E P =! m and =! m we deduce that (a) B has the lage kinetic enegy, (b) has the lage potential enegy and (c) has the lage total enegy. 6 he speed in obit is given fom mv then E K = mv = m is = m! m = G Mm to be v = G M. he potential enegy is E P =! m =! m. Since = 5, =! m 0.. he kinetic enegy is and so the total enegy 7 Fom poblem 6, =! m. Since we ae told that =! m 5 conseved,! m =! m 5 " = 5. and enegy is

4 8 he space station is in an obit with obit adius e and so has speed v = with e espect to the eath. Let the satellite be launched with speed u with espect to the space station. hen the speed with espect to the eath is u + v. Its total enegy is theefoe m(u + v)! m. t the escape speed this enegy must be zeo and so e u = e! v = e! e ". 0 ms!. 9 (a) he engines do positive wok inceasing the total enegy of the satellite. Since =! m it follows that the obit adius will incease. (b) Since the kinetic enegy is given by E K = m and the obit adius has inceased the speed in the new cicula obit will decease. (c) he fiing of the ockets when the satellite is in the lowe obit make the satellite move on an elliptical obit. fte half a evolution the satellite will be at and futhe fom the eath than in the oiginal position at P. s the satellite gets to its kinetic enegy is educed and the potential enegy inceases. t the speed is too low fo the new cicula obit and the engines must again be fied to incease the speed to that appopiate to the new obit. (If the engines ae not fied at then the satellite will emain in the elliptical obit and will etun to P.) 0 he tangential component at is in the diection of velocity and so the planet inceases its speed. t B it is opposite to the velocity and so the speed deceases. he nomal component does zeo wok since the angle between foce and displacement is a ight angle and cos90! = 0. he potential enegy is given by E P =! m. his is least when the distance to the sun is the smallest (emembe E P is negative). heefoe since the total enegy is conseved, the kinetic enegy and hence the speed ae geatest at P. (a) he atio is

5 B F moon F B sun! F moon! F sun this gives B F moon F B sun =! F moon! F sun = moon m (d moon! e )! m moon (d moon + e ) sun m (d sun! e )! = sunm (d sun + e ) M moon (d moon! e )! M moon (d moon + e ) 7.5 " 0 (.84 " 0 8! 6.4 " 0 6 )! 7.5 " 0 (.84 " " 0 6 ).99 " 0 0 (.5 " 0! 6.4 " 0 6 )!.99 " 0 0 (.5 " " 0 6 ) indicates the elative impotance of the moon. M sun (d sun! e )! M sun (d sun + e ). Numeically.. (b) his he escape speed is obtained fom mv! m and so = g. Substituting, v = g. = 0, i.e. v =. But g = 4 (a) he net gavitational field stength at the indicated position has magnitude g = G6m (4d / 5)! Gm (d / 5) = G5m! G5m = 0. (b) d d V =! G6m (4d / 5)! Gm (d / 5) =! G0m! G5m =! 5Gm. d d d 5 (a) See poblem. (b) = 4! M = 4!" 4!. Substituting, = G. Now! and! = M 4" 4!" =! G". 6 (a) We must use the fomula = 4! Now g =! = g. Substituting, = 4! g =!.4 " = M 4". Hence, that we have deived many times aleady. = 4! g. Hence =! g. (b) = 5.5 " 0 s = 9 min. (c) Fom = 4! we deduce that = hence 9 40 = (.4! 06 ) and so = 4.5! 0 6 m. he height is theefoe h = 4.5! 0 6 ".4! 0 6 =.! 0 6 m.

6 7 (a) F = 4 = = "!. (b) 4 4 &. Hence = 6! = Mv and so v = 4. But v =. (c) "! & and so 6! (.0 " 0 9 ) 6.67 " 0 ".5 ".0 " 0 0 =.8 " 04 s = 7.8 h. (d) = Mv + Mv!. Since v = 4 we have that E = M 4! " = 4 " = " 4. (e) Since enegy is being lost the total enegy will decease. his implies that the distance will decease. (Fom the peiod fomula in (b) the peiod will decease as well.) (f) (i) he total enegy is =! 4 and the peiod is = 6!. Combining the two gives =! 4 & o E / =!c!/ whee c is a constant. Woking as we do with 6" ( popagation of uncetainties (o using calculus) we have that! =! o! = is theefoe!. (ii)! =!.9! 0 "9 y " =.6! 08 y. = " 7 " 06 s y.8 " 0 4 s =.9 " 0 9 y (g) he lifetime 8 (a) he patten is not symmetical and so the masses must be diffeent. he spheical equipotential sufaces of the ight mass ae much less distoted and so this is the lage mass.(b) he gavitational field lines ae nomal to the equipotential sufaces. (c) Fom fa away it looks like we have a single mass of magnitude equal to the sum of the two individual masses. he equipotential sufaces of a single point mass ae spheical. 9 (a) he magnitude of the gavitational field stength is the slope of a potential distance gaph. Dawing a tangent to the cuve at d = 0.0 we find a slope of appoximately 4.7 N kg!. (b) t = 0.75 the slope of the gaph is zeo. But the slope of a potential d distance gaph is the magnitude of the gavitational field stength. Hence g is zeo at this point. (c) Since g = 0 =! Gm (d! ) = (0.75d)! Gm it follows that (d! 0.75d) M m = (0.75d) (0.5d) = 9.

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