Answers to Coursebook questions Chapter 2.11
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1 Answes to Couseook questions Chapte 11 1 he net foe on the satellite is F = G Mm and this plays the ole of the entipetal foe on the satellite, ie mv mv Equating the two gives π Fo iula motion we have that v = Simplifying gives the esult Fom Q1 we know that 3 = 4π π π ω =, ie = ω 3 (π) Sustituting, = 4π ω = ω and so 3 Fom mv = G Mm we dedue that v = G M Now, = = 6900 km = 69 6 m and so = G Mm, ie v = G M π = he angula veloity is the angle swept pe unit time, ie v = = = 76 3 ms π π 3 he peiod is then found fom v = 57 s 95 min = v = = π 4 he peiod has to e one day, ie 4 hous hen v = and fom mv dedue that v = G M Comining the esults we get = G Mm we 4π ( ) 4π = = = 4 m 5 a E P M eath = = = 764 J V eath = = = 4 J kg 6 1 v eath = = = m s 3 1 Copyight Camidge Univesity Pess 011 All ights eseved Page 1 of
2 eathm m 6 We must plot the funtion EP = giving the gaph in the d answes Hee m is the mass of the spaeaft and d the sepaation of the eath and the (ente-to-ente) Putting numes in, E P = = 34 = 1 19 / /34 /34 1 / = x 1 x whee x = In this way the funtion an e plotted on a alulato to give the gaph 34 in the answes (see page 79 in Physis fo the IB Diploma) 7 a V = 5 e = =1 7 J kg 1 E P = m 5 e = =6 9 J he net foe is the gavitational foe and this must point towads the ente of the eath his happens only fo oit 9 As shown in the text, the eation foe fom the spaeaft floo is zeo, giving the impession of weightlessness Moe simply, oth spaeaft and astonaut ae in fee fall with the same aeleation he diffeene is ΔU = m + h m = m m + h = m + h ( + h) h ΔU = m ( + h) When h is small ompaed to this expession is appoximately ΔU = m h = m h = mgh Copyight Camidge Univesity Pess 011 All ights eseved Page of
3 11 he wok done y an extenal agent in moving an ojet fom = a to = at a small onstant speed 1 a he potential at the sufae of the planet is V = = M =49 1 J kg Hene M = 15 kg he esape speed is otained fom 1 mv m = 0, ie v = But V = and so = V Sustituting, v = V v = V = 49 1 = 31 6 m s 1 d he wok equied is W = mδv his is W = 1500 ( ( 49 )) = 5 J e We have that 1 mv + mv 1 = mv v = (V V 1 ) his gives v = ( 1 (49 1 )) = 3 6 m s 1 13 a At = 075, g = P (075d) m (05d) = 0 Hene M P = (075d) M m (05d) = 9 he poe must have enough enegy to get to the maximum of the gaph Fom then on the will pull it in hen W = mv = mδv v = Δ V = ( 0 ( 64 )) = 35 m s 14 We dedued many times that v = and so E = 1 mv m = m m = m E = = J Copyight Camidge Univesity Pess 011 All ights eseved Page 3 of
4 15 Using E K = m, E P = m and E = m we dedue that a B has the lage kineti enegy A has the lage potential enegy and A has the lage total enegy 16 he speed in oit is given fom mv = G Mm to e v = G M he kineti enegy is then E K = 1 mv = 1 m he potential enegy is E P = m and so the total enegy is E = 1 Sine = 5, E = m m m = m 17 Fom Q16, E = m Sine we ae told that E = m 5 m = m 5 = 5 and enegy is onseved, 1 he spae station is in an oit with oit adius e and so has speed v = e to the eath Let the satellite e launhed with speed u with espet to the spae station hen the speed with espet to the eath is u + v Its total enegy is theefoe 1 m(u + v) m e At the esape speed this enegy must e zeo and so u = v = 3 3 ms 1 e e e with espet Copyight Camidge Univesity Pess 011 All ights eseved Page 4 of
5 19 a he engines do positive wok ineasing the total enegy of the satellite Sine E = m it follows that the oit adius will inease Sine the kineti enegy is given y E K = m speed in the new iula oit will deease and the oit adius has ineased, the he fiing of the okets when the satellite is in the lowe oit make the satellite move on an elliptial oit Afte half a evolution the satellite will e at A and futhe fom the eath than in the oiginal position at P As the satellite gets to A its kineti enegy is edued and the potential enegy ineases At A the speed is too low fo the new iula oit and the engines must again e fied to inease the speed to that appopiate to the new oit (If the engines ae not fied at A then the satellite will emain in the elliptial oit and will etun to P) new iula oit F A old iula oit P 0 he tangential omponent at A is in the dietion of veloity and so the planet ineases its speed At B it is opposite to the veloity and so the speed deeases he nomal omponent does zeo wok sine the angle etween foe and displaement is a ight angle and os90 o = 0 1 he potential enegy is given y E P = m his is least when the distane to the sun is the smallest (ememe E P is negative) heefoe sine the total enegy is onseved, the kineti enegy and hene the speed ae geatest at P Copyight Camidge Univesity Pess 011 All ights eseved Page 5 of
6 a he atio is m m M M F F d d d d B A ( e) ( + e) ( e) ( + e) = = B A Fsun F sun sunm sunm M sun M sun ( dsun e) ( dsun + e) ( dsun e) ( dsun + e) Numeially this gives F B B Fsun A 6 6 F (34 64 ) ( ) = A Fsun (15 64 ) ( ) his indiates the elative impotane of the 3 he esape speed is otained fom 1 mv m But g = and so = g Sustituting, v = g = 0, ie v = 4 a he net gavitational field stength at the indiated position has magnitude g = G16m (4d /5) Gm (d /5) = G5m G5m = 0 d d V = G16m (4d /5) Gm (d /5) =G0m G5m = 5Gm d d d 5 a See Q1 3 4π = Now and ρ = M 3M 3 3 4π = 4π 3 M Hene, = 4πρ 3 3 4π 3 3π Sustituting, = G 4πρ = Gρ ρ ρ 169 = = = ρ ρ 5 planet eath eath eath planet planet Copyight Camidge Univesity Pess 011 All ights eseved Page 6 of
7 6 a 3 4π We must use the fomula = that we have deived many times aleady Now g = = g 3 4π 4π Sustituting, g g Hene = π g = π = 55 s = 91 min 45 Fom = 4π 3 we dedue that 1 = Hene 140 = (34 6 ) 3 and so 3 = 45 6 m he height is theefoe h = = 11 m 7 a F = 4 = Mv v = 4 and so 4 π π But v = and so = π Hene = π ( ) = = = s 7 h d E = 1 Mv + 1 Mv Sine v = 4 we have that E = 1 M 4 = 4 = 4 Copyight Camidge Univesity Pess 011 All ights eseved Page 7 of
8 e Sine enegy is eing lost, the total enegy will deease his implies that the distane will deease (Fom the peiod fomula in the peiod will deease as well) E = 3 f i he total enegy is 4 16π and the peiod is = Comining the two gives E = 3/ 4 16π o E = 3/, whee is a onstant Woking as we do with popagation of unetainties (o using alulus) we have that ΔE ΔE = 3 Δ Δ E o Δt = 3 Δt E ii ΔE Δ 6 1 E s y = = = 39 y 4 Δt Δ t s 9 1 g he lifetime is theefoe 1 = y 6 y a he patten is not symmetial and so the masses must e diffeent he spheial equipotential sufaes of the ight mass ae muh less distoted and so this is the lage mass he gavitational field lines ae nomal to the equipotential sufaes Fom fa away it looks like we have a single mass of magnitude equal to the sum of the two individual masses he equipotential sufaes of a single point mass ae spheial 9 a he magnitude of the gavitational field stength is the slope of a potential distane gaph Dawing a tangent to the uve at d = 00 we find a slope of appoximately 47 N kg 1 At d = 075 the slope of the gaph is zeo But the slope of a potential distane gaph is the magnitude of the gavitational field stength Hene g is zeo at this point Sine g = 0 = Gm (d ) = (075d) Gm it follows that (d 075d) M m = (075d) (05d) = 9 Copyight Camidge Univesity Pess 011 All ights eseved Page of
F g. = G mm. m 1. = 7.0 kg m 2. = 5.5 kg r = 0.60 m G = N m 2 kg 2 = = N
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