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Cambridge Physics for the IB Diploma Answers to Coursebook questions – Chapter 2.11 1 Mm and this plays the role of the centripetal force on R2 mv 2 mv 2 Mm M G 2 , i.e. v 2 G . the satellite, i.e. . Equating the two gives R R R R The net force on the satellite is F G 2 2 GM 2πR 2πR . and so R T T 2 For circular motion we have that v Simplifying gives the result. 2 From Q1 we know that R 3 GMT 2 . The angular velocity is the angle swept per unit time, i.e. 4π 2 2π 2π , i.e. T . T GM (2π)2 GM 2 . Substituting, R3 4π 2 2 3 mv 2 Mm M G 2 we deduce that v 2 G . Now, From R R R R 6400 500 6900 km 6.9 106 m and so 6.67 10 11 6.0 10 24 v 7.6 10 3 m s 1 . 6 6.9 10 2πR 2πR The period is then found from v T 5.7 103 s 95 min . T v GM R 4 The period has to be one day, i.e. 24 hours. Then v deduce that v 2 G R 5 3 mv 2 Mm 2πR G 2 we and from T R R M . Combining the results we get R GMT 2 3 6.67 1011 6.0 1024 (24 60 60)2 4.2 107 m . 2 2 4π 4π a EP b V c v GM earth M moon 6.67 1011 5.98 1024 7.35 1022 7.64 1028 J . 8 r 3.84 10 GM earth 6.67 1011 5.98 1024 1.04 106 J kg 1 . r 3.84 108 GM earth 6.67 1011 5.98 1024 1.02 103 m s 1 . 8 r 3.84 10 Copyright Cambridge University Press 2011. All rights reserved. Page 1 of 8 Cambridge Physics for the IB Diploma 6 We must plot the function EP GM earth m GM moon m giving the graph in the r d r answers. Here m is the mass of the spacecraft and d the separation of the earth and the moon (centre-to-centre). Putting numbers in, 6.67 10 11 5.98 10 24 3.0 10 4 6.67 10 11 7.35 10 22 3.0 10 4 r 3.84 10 8 r 1.2 1019 1.5 1017 r 3.84 10 8 r 1.2 1019 / 3.84 10 8 1.5 1017 / 3.84 10 8 r / 3.84 10 8 1 r / 3.84 10 8 3.1 1010 3.9 10 8 x 1 x r where x . In this way the function can be plotted on a calculator to give the graph 3.84 10 8 EP in the answers (see page 798 in Physics for the IB Diploma). 7 a V GM 5Re 6.67 10 11 5.98 10 24 5 6.4 10 6 1.2 10 7 J kg 1 b EP GMm 5Re 6.67 10 11 5.98 10 24 500 5 6.4 10 6 6.2 10 9 J 8 The net force is the gravitational force and this must point towards the centre of the earth. This happens only for orbit 2. 9 As shown in the text, the reaction force from the spacecraft floor is zero, giving the impression of weightlessness. More simply, both spacecraft and astronaut are in free fall with the same acceleration. 10 The difference is U GMm GMm GMm GMm R h R GMm . R(R h) Rh R R Rh h U GMm . When h is small compared to R this expression is approximately R(R h) U GMm GM h m 2 h mgh . 2 R R Copyright Cambridge University Press 2011. All rights reserved. Page 2 of 8 Cambridge Physics for the IB Diploma 11 The work done by an external agent in moving an object from r = a to r = b at a small constant speed. 12 a The potential at the surface of the planet is GM 6.67 10 11 M V 4.9 1012 J kg 1 . 5 R 2.0 10 Hence M 1.5 10 28 kg . b The escape speed is obtained from But V 13 2GM . R GM and so GM VR . Substituting, v 2V . R c v 2V 2 4.9 1012 3.1 10 6 m s1 d The work required is W mV . This is W 1500 (1.0 1012 (4.9 1012 )) 5.8 1015 J . e We have that 1 2 mv mV1 mV2 v 2(V2 V1 ) . 2 This gives v 2(2.2 1012 (4.9 1012 )) 2.3 10 6 m s1 . a At r = 0.75, g M P (0.75d)2 GM P GM m 9. . Hence 0 M m (0.25d)2 (0.75d)2 (0.25d)2 b The probe must have enough energy to get to the maximum of the graph. From then on the moon will pull it in. Then W 14 1 2 GMm mv 0 , i.e. v 2 R 1 2 mv mV v 2V 2(0.2 1012 (6.42 1012 )) 3.5 106 m s 1. 2 We deduced many times that v2 GM r and so ET 1 2 GMm GMm GMm GMm mv 2 r 2r r 2r . ET 6.67 1011 2.0 1030 6.0 1024 2.7 1033 J. 11 2 1.5 10 Copyright Cambridge University Press 2011. All rights reserved. Page 3 of 8 Cambridge Physics for the IB Diploma 15 Using EK GMm GMm GMm , EP and ET we deduce that 2r r 2r a B has the larger kinetic energy b A has the larger potential energy and c A has the larger total energy. mv 2 Mm M G 2 to be v 2 G . r r r 1 2 1 GMm The kinetic energy is then EK mv . 2 2 r GMm The potential energy is EP r 1 GMm GMm GMm and so the total energy is ET . 2 r r 2r GMm Since r 5R , ET . 10R 16 The speed in orbit is given from 17 From Q16, ET 18 The space station is in an orbit with orbit radius 2Re and so has speed v GMm GMm . Since we are told that ET and energy is conserved, 2r 5R GMm GMm 5R . r 2r 5R 2 GM with respect 2Re to the earth. Let the satellite be launched with speed u with respect to the space station. Then the speed with respect to the earth is u v . Its total energy is therefore 1 GMm . m(u v)2 2 2Re At the escape speed this energy must be zero and so u GM GM GM v 2.3 10 3 ms1 . Re Re 2Re Copyright Cambridge University Press 2011. All rights reserved. Page 4 of 8 Cambridge Physics for the IB Diploma 19 a The engines do positive work increasing the total energy of the satellite. Since ET b GMm it follows that the orbit radius will increase. 2r Since the kinetic energy is given by EK GMm and the orbit radius has increased, the 2r speed in the new circular orbit will decrease. c The firing of the rockets when the satellite is in the lower orbit make the satellite move on an elliptical orbit. After half a revolution the satellite will be at A and further from the earth than in the original position at P. As the satellite gets to A its kinetic energy is reduced and the potential energy increases. At A the speed is too low for the new circular orbit and the engines must again be fired to increase the speed to that appropriate to the new orbit. (If the engines are not fired at A then the satellite will remain in the elliptical orbit and will return to P.) new circular or bit A F old circular orbit P 20 The tangential component at A is in the direction of velocity and so the planet increases its speed. At B it is opposite to the velocity and so the speed decreases. The normal component does zero work since the angle between force and displacement is a right angle and cos 90o 0 . 21 The potential energy is given by EP GMm . r This is least when the distance to the sun r is the smallest (remember EP is negative). Therefore since the total energy is conserved, the kinetic energy and hence the speed are greatest at P. Copyright Cambridge University Press 2011. All rights reserved. Page 5 of 8 Cambridge Physics for the IB Diploma 22 a The ratio is F F F F B moon B sun A moon A sun GM moon m GM moon m M moon M moon 2 2 2 (d Re ) (d moon Re ) (d Re ) (d moon Re ) 2 moon moon GM sun m GM sun m M sun M sun 2 2 2 (dsun Re ) (dsun Re ) (dsun Re ) (dsun Re ) 2 . Numerically this gives B A Fmoon Fmoon B A Fsun Fsun b 23 This indicates the relative importance of the moon. The escape speed is obtained from But g 24 7.35 1022 7.35 1022 (3.84 108 6.4 106 ) 2 (3.84 108 6.4 106 ) 2 2.2 . 1.99 1030 1.99 1030 (1.5 1011 6.4 106 )2 (1.5 1011 6.4 106 ) 2 a 1 2 GMm mv 0 , i.e. v 2 R 2GM . R GM and so GM gR2 . Substituting, v 2gR . R2 The net gravitational field strength at the indicated position has magnitude g G16m Gm G25m G25m 0. 2 2 (4d / 5) (d / 5) d2 d2 b V 25 G16m Gm G20m G5m 25Gm (4d / 5) (d / 5) d d d . a See Q1. b T2 4π 2 r 3 . GM Now r R and M 4 3 3 . Hence, R Substituting, T c Tplanet Tearth M 3M . 3 4πR 4πR3 3 4π 2 3 3π . G 4π G earth 169 earth 3.95 4 . planet planet 85 2 Copyright Cambridge University Press 2011. All rights reserved. Page 6 of 8 Cambridge Physics for the IB Diploma 26 a We must use the formula T 2 4π 2 R3 that we have derived many times already. GM GM GM gR 2 . R2 4π 2 R3 4π 2 R Substituting, T 2 . gR 2 g Now g Hence T 2π R . g b 3.4 106 T 2π 5.5 103 s 91 min . 4.5 c From T 2 T 2 R3 4 2 R 3 we deduce that 12 13 . T2 R2 GM 912 (3.4 10 6 )3 Hence and so R2 4.5 10 6 m . 2 3 140 R2 The height is therefore h 4.5 106 3.4 106 1.1106 m . 27 a F GM 2 4R 2 b GM 2 Mv 2 GM 2 v2 2 4R . 4R R and so 2 2 GM 2πR 2πR But v . and so 4R T T 16π 2 R 3 Hence T 2 . GM 2 c 16π 2 (1.0 109 )3 T 2.8 104 s 7.8 h . 11 30 6.67 10 1.5 2.0 10 d ET 1 1 GM 2 Mv 2 Mv2 2 2 2R . Since v2 GM 4R we have that ET 1 GM GM 2 GM 2 GM 2 GM 2 M 2 2 4R 2R 4R 2R 4R . Copyright Cambridge University Press 2011. All rights reserved. Page 7 of 8 Cambridge Physics for the IB Diploma e f Since energy is being lost, the total energy will decrease. This implies that the distance R will decrease. (From the period formula in b the period will decrease as well.) GM 2 2 3 ET 4R and the period is T 2 16π R . Combining The total energy is GM 2 GM the two gives ET 3/2 GMT 2 4 2 16π i or ET cT 3/2 , where c is a constant. Working as we do with propagation of uncertainties (or using calculus) we have that ET T 3 ET 3 T t . or t ET 2 T ET 2 T ii 28 29 ET T ET 3 T 3 72 106 s y 1 3.9 109 y 1 t 2 t 2 2.8 104 s 1 2.6 108 y. 9 1 3.9 10 y g The lifetime is therefore a The pattern is not symmetrical and so the masses must be different. The spherical equipotential surfaces of the right mass are much less distorted and so this is the larger mass. b The gravitational field lines are normal to the equipotential surfaces. c From far away it looks like we have a single mass of magnitude equal to the sum of the two individual masses. The equipotential surfaces of a single point mass are spherical. a The magnitude of the gravitational field strength is the slope of a potential–distance graph. Drawing a tangent to the curve at r 0.20 we find a slope of approximately d 4.7 N kg1 . b At r 0.75 the slope of the graph is zero. But the slope of a potential–distance graph is d the magnitude of the gravitational field strength. Hence g is zero at this point. c Since g 0 GM Gm GM Gm it follows that 2 2 2 r (d r) (0.75d) (d 0.75d)2 M (0.75d)2 9. m (0.25d)2 Copyright Cambridge University Press 2011. All rights reserved. Page 8 of 8