Answers to test yourself questions

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1 Answes to test youself questions opic Descibing fields Gm Gm Gm Gm he net field t is: g ( d / ) ( 4d / ) d d Gm Gm Gm Gm Gm Gm b he net potentil t is: V d / 4d / d 4d d d V e J kg 7 7 Gm d b E E m e J 4 ethm moon J b V eth eth c v J kg m s ethm moonm 4 We must plot the function E giving the gph in the nswes Hee m is the mss d of the spcecft nd d the seption of the eth nd the moon (cente-to-cente) utting numbes in, E / 4 / 4 / 4 / 4 9 x x HYSICS FO HE IB DILOMA CAMBIDGE UNIVESIY ESS ANSWES O ES YOUSELF QUESIONS

2 whee x In this wy the function cn be plotted on clculto to give the gph shown hee 4 /d 4 E/ J m At 7, g Hence M ( 7d ) 9 ( 7d ) ( d ) M m ( d ) b he pobe must hve enough enegy to get to the mximum of the gph Fom then on the moon will pull it in hen W mv m V v V ( ( 4 )) m s he tngentil component t A is in the diection of velocity nd so the plnet inceses its speed At B it is opposite to the velocity nd so the speed deceses he noml component does zeo wok since the ngle between foce nd displcement is ight ngle nd cos 9 7 he wok done by n extenl gent in moving n object fom to b t smll constnt speed he ptten is not symmeticl nd so the msses must be diffeent he spheicl equipotentil sufces of the ight mss e much less distoted nd so this is the lge mss b he gvittionl field lines e noml to the equipotentil sufces c Fom f wy it looks like we hve single mss of mgnitude equl to the sum of the two individul msses he equipotentil sufces of single point mss e spheicl kq kq 4kq 9 V + d / d / d kq b V kq d / d / A digm is: Q Q he potentil t is V 4 V 4 9 kq 99 V 4 whee m Hence V 4 V b E c he potentil t the cente hs mximum vlue At mximum vlue the deivtive is zeo he wok done is W q V q kq 9 9 kq J b No ANSWES O ES YOUSELF QUESIONS HYSICS FO HE IB DILOMA CAMBIDGE UNIVESIY ESS 4

3 he wok done on the electon is W q V q kq ( ) ( ) J 4 he wok done (W q V ) is equl to the chnge in kinetic enegy mv Hence 9 9 v ( ) v 9 9 m s 9 he foces e oughly s follows F F F hey hve mgnitudes: F 9 F 9 F 7 9 N 4 4 N 99 N We must find the components of F : F x F cos 4 N nd F y F sin 4 N So the net foce hs components: F x 7 N nd F y 4 N he net foce is then F ( 4) + ( ) he diection of the net foce is ctn 4 7 b he distnce of the cente of the sque fom ech of the vetices is + 4 cm So the potentil t the cente is V kq 9 kq kq kq V V ( c he wok done is W q V q( V ) 9 4 J Chge will move until both sphees e t the sme potentil hen kq ) kq By consevtion of chge, q + q Q whee Q is the chge on the one sphee oiginlly hus q q q q nd q + q Hence q µc nd q µc b σ 4 C m nd σ 4 C m 4π 4π kq 9 c E 4πkσ 4π N C nd E 9 4πkσ 4π N C d he electic field is lgest fo the sphee with the lge chge density he wie hs to be long so tht the chge of one sphee will not ffect the chge distibution on the othe so tht both e unifomly chged N HYSICS FO HE IB DILOMA CAMBIDGE UNIVESIY ESS 4 ANSWES O ES YOUSELF QUESIONS

4 7 You must dw lines tht e noml to the equipotentils he potentil distnce x fom the bottom plte is given by ( ) V + x ( + x) V nd so t x cm, V ( + ) V heefoe the electic potentil enegy of the chge is E qv ( ) ( ) mj b he potentil t x cm is V ( + ) V nd hence E qv ( ) mj c he wok done must be W q V E J 9 he kinetic enegy of the electon E K mv 9 ( 9 ) J gets conveted to electic potentil enegy ev t the point whee the electon stops Hence the potentil t is V 7 9 V 9 b V kq Q V ( 7 9) 9 C 9 k 9 he field due to ech of the chges hs the diection shown It is cle tht the net field will point in the negtive y diection θ he mgnitude of the field due to one of the chges is E kq kq he y component is + d kq E kq kq kq y sinθ nd so the net field is E + d + d + d ( + d ) / net ( + d ) / b Fo two negtive chges: θ he net field is clely diected to the left It hs mgnitude kq kq d kqd Enet Ex cosθ + d + d + d ( + d ) / kq c We hve E kq ( + d ) / d / + kqd nd E kq d kq d / b ( + d ) / d / / d + + he plots e (the veticl xis is in units of kq ): 4 ANSWES O ES YOUSELF QUESIONS HYSICS FO HE IB DILOMA CAMBIDGE UNIVESIY ESS 4

5 E 4 d / E b d / he initil potentil enegy of the thee potons is zeo When t the vetices of the tingle of side the potentil k e e enegy is E ( )( ) ke since thee e thee pis of chges distnce pt his evlutes to E 99 ( ) J MeV his is the enegy tht must be supplied Fields t wok m mv v Substituting vlues: v 7 m s + π π π ( ) b Fom v s 94 9 min v 7 We know tht m m v v But v π 4 nd so we deduce tht π 4 Fom the pevious poblem, heefoe 4 7 ( 4 ) 7 4 m he distnce fom the sufce is 4π 4π 7 4 theefoe 4 km b No, it hs to be bove the equto he net foce is the gvittionl foce nd this must point towds the cente of the eth his hppens only fo obit HYSICS FO HE IB DILOMA CAMBIDGE UNIVESIY ESS 4 ANSWES O ES YOUSELF QUESIONS

6 As shown in the text the ection foce fom the spcecft floo is zeo giving the impession of weightlessness Moe simply, both spcecft nd stonut e in fee fll with the sme cceletion 7 Apply enegy consevtion to get: totl enegy t the point the fuel uns out is E m mv m m m At the highest point the kinetic enegy is zeo nd so 4 m m leding to 4 4 b he totl enegy of the ocket t the point whee the fuel uns out is negtive so the ocket cnnot escpe, it will fll bck down c Apply enegy consevtion gin between the points whee the fuel uns out nd the csh point to get: m m mv leding to 4 v 4 4 v d Fom enegy consevtion, when the ocket is distnce fom the cente of the plnet: m m mv his simplifies to v (whee 4 ) We need to plot this 4 function It is best to wite the equivlent fom: v 4 he gph is then: v m We deduced mny times tht v nd so E mv m m m 4 7 b E 7 J 4 m 9 Using E m K, E nd E m we deduce tht B hs the lge kinetic enegy b A hs the lge potentil enegy c A hs the lge totl enegy he totl enegy is negtive so the stellite cnnot escpe m b Fom poblem, E m Since we e told tht E nd enegy is conseved, m m m he engines do positive wok incesing the totl enegy of the stellite Since E it follows tht the obit dius will incese ANSWES O ES YOUSELF QUESIONS HYSICS FO HE IB DILOMA CAMBIDGE UNIVESIY ESS 4

7 m A bit moe: Since the kinetic enegy is given by E K nd the obit dius hs incesed the speed in the new cicul obit will decese he fiing of the ockets when the stellite is in the lowe obit mkes the stellite move on n ellipticl obit Afte hlf evolution the stellite will be t A nd futhe fom the eth thn in the oiginl position t As the stellite gets to A its kinetic enegy is educed nd the potentil enegy inceses At A the speed is too low fo the new cicul obit nd the engines must gin be fied to incese the speed to tht ppopite to the new obit (If the engines e not fied t A then the stellite will emin in the ellipticl obit nd will etun to ) new cicul obit F A old cicul obit m he potentil enegy is given by E his is lest when the distnce to the sun,, is smllest (emembe, E is negtive) heefoe since the totl enegy is conseved, the kinetic enegy nd hence the speed e getest t he escpe speed is vesc At the sufce of the plnet, g g Substituting: g v esc g 4 We hve done this befoe 4 b π M M Now nd ρ 4π 4π Hence, M 4 πρ 4π π Substituting, G 4πρ Gρ c plnet eth ρ ρ eth eth ρplnet ρplnet We must use the fomul π tht we hve deived mny times ledy Now g g 4π 4π Substituting, Hence π g g g 4 b π s 9 min 4 4 c Fom π we deduce tht hence 9 ( 4 ) nd so 4 m he height 4 is theefoe h 4 4 m F 4 b Mv nd so v 4 4 But v π nd so 4 π π Hence c 9 π ( ) 7 4 s 7 h HYSICS FO HE IB DILOMA CAMBIDGE UNIVESIY ESS 4 ANSWES O ES YOUSELF QUESIONS 7

8 d E Mv + Mv Since v 4 we hve tht E M e Since enegy is being lost the totl enegy will decese his implies tht the distnce will decese (Fom the peiod fomul in (b) the peiod will decese s well) f i he totl enegy is E π nd the peiod is Combining the two gives 4 E / o E / c whee c is constnt Woking s we do with popgtion of 4 π E uncetinties (o using clculus) we hve tht E o t t E E ii E E t 7 s y 4 t s g he lifetime is theefoe 9 y y 7 Foce towds the cente of the cicle q b We equte the electic foce to the centipetl foce to get: m v Solving fo the speed gives the nswe 4πε q c he totl enegy is kinetic plus electic potentil enegy: E mv Using the pevious esult fo 4πε speed: v q q q q q gives E m 4 πε m m 4 4 πε πε πε 4πε q q d he chnge in enegy is n incese of E + πε πε πε y q πε As in the pevious poblem v k e Using lso v π we get 4 4 π π k e m m m ke b 4π 9 99 ( ) 9 9 ( 97 ) 4 s ke c he chnge in enegy is E In the fist obit this evlutes to ( ) E J In the othe obit this becomes ( ) E 7 9 J he chnge is 7 J q ANSWES O ES YOUSELF QUESIONS HYSICS FO HE IB DILOMA CAMBIDGE UNIVESIY ESS 4