Chapter 4. Answers to examination-style questions. Answers Marks Examiner s tips. 1 (a) (i) r = 2GM

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(a) (i) r = 2GM = 2 6.7 0 6.0 0 24 c 2 (3.0 0 8 ) 2 = 8.93 0 3 m = 8.9 mm (ii) Volume of sphere of radius 9 mm = 4πr3 3 = 4π(9.0 0 3 ) 3 3 = 3.05 0 6 m 3 Density ρ = _ mass 6.0 024 = volume 3.

1 (a) (i) r = 2GM = c 2 ( ) 2 = m = 8.9 mm (ii) Volume of sphere of radius 9 mm = 4πr3 3 = 4π( ) 3 3 = m 3 Density ρ = _ mass = volume = kg m 3 (b) (i) Michell s idea was put fward as a hypothesis because it was an untested idea, a scientific hypothesis is a suggestion, prediction untested idea. Einstein used mathematics to predict his General They of elativity, Schwarzschild predicted (using General elativity) that light could not escape from a sufficiently massive object. (ii) Stars photographed near the Sun during a total solar eclipse were displaced relative to each other. The light from each star was bent by the Sun s gravitational field as it skimmed the Sun. This observation confirmed the prediction by Einstein. AQA Physics A A2 Level Nelson Thnes Ltd 2009

2 (c) FO Either Scientific knowledge is expanded in scientific projects, f example space explation has vastly increased our knowledge of the Solar System New scientific projects can lead to imptant new discoveries, f example HST led to the discovery of many stars and galaxies New scientific projects can lead to new technologies, f example me powerful space rockets used to reach the Moon made geostationary satellites possible AGAINST Either Improved living conditions in poer countries would benefit many me people than a space project would, f example provision of clean water Cost of the project would be met by taxpayers and there would be no direct immediate benefits Scientific expertise could be better used, f example on developing renewable energy sources to combat global warming 5 There are 5 marks available so 5 points to be made. There are two marks f each of the two views, one f identifying a relevant argument and the other f outlining it adequately. egardless of which of the two views you suppt, the fifth mark is f providing a valid argument against the view you do not suppt. AQA Physics A A2 Level Nelson Thnes Ltd

3 If FO Either Improved living conditions in poer countries would benefit many me people than a space project would satellite communication has brought benefits to everyone, f example immediate disaster relief Cost of the project would be met by taxpayers me people would be employed as scientific firms would be contracted to supply parts Scientific expertise could be better used the demand f scientific expertise could be met by training me scientists If AGAINST: Either Scientific knowledge is expanded in scientific projects there may be cheaper and less risky ways of pursuing knowledge of deep space (f example me advanced and bigger terrestrial telescopes linked to me powerful computers) New scientific projects can lead to imptant new discoveries as above, New scientific projects can lead to new technologies as above Explanation of why you would/would not welcome such a project. 2 (a) (i) The gravitational field strength at a point is the fce per unit mass acting on a small test mass placed at that point in the field. The mark would be awarded f g = m F provided F and m were defined. Fce on a kg mass would also be acceptable. (ii) N kg (not m s 2 ) g has two distinct meanings, and has the same numerical value in both meanings. Treated as an acceleration, the unit is m s 2. Treated as a field strength, the unit is N kg. (b) (i) Since g P = g Q, it follows that This solution relies on the relationship GM = GM Q between g and the other quantities 2 2 Q involved in a radial field: g = GM. You 2 M Q = M Q 2 _ = M(3)2 since are told in the question that, at the surface 2 2 Q = 3 of the planets, g P = g Q. giving mass of Q = 9 M AQA Physics A A2 Level Nelson Thnes Ltd

4 (ii) Line drawn on graph: Starts at 3, with same initial value of g as existing curve; is a curve of decreasing negative gradient; shows the crect r 2 relationship, checked from points (f example at 6, value should be exactly.0 vertical square; at 2 it should be 0.25 of a vertical square). 3 Planet Q has 9 times the mass of planet P. At any given radius, g Q will be 9 times greater than g P, since g M. Your curve must start at 3, since this is the surface radius of Q, and show a r 2 relationship crectly f full marks. 3 (a) elevant points about the geosynchronous satellite include: It bits over the Equat. It maintains a fixed position in relation to the surface of the Earth. It has a period of 24 hours (the same as the Earth s period of rotation on its axis). elevant comparisons with the satellite in a low polar bit include: The geosynchronous satellite enables uninterrupted communication between a transmitter and a receiver whereas the other satellite does not. Unlike the other satellite, the geosynchronous satellite does not require the use of a steerable dish. any 3 Full marks ought to be obtained easily in part (a) if a few facts have been committed to memy. In your answer, take care over the use of language: geosynchronous satellites are not stationary, even though they appear to be at rest when viewed from the rotating Earth! They are imptant as a maj means of communicating data. Satellite television relies on their use, so TV aerial dishes can be in fixed positions. A satellite in low polar bit has a fairly sht time period, scanning the Earth several times during the day. (b) (i) Using the given symbols, the centripetal fce equation is GMm = m ω ( + h) 2 2 ( + h) (ii) Substitution of ω = 2π T leads to GM = 4π2 ( + h) 3 T 2 and rearrangement of this gives T 2 = 4π2 ( + h) 3 GM (iii) The limiting case is when the satellite bits the Earth at zero height; h = 0 T 2 = 4π2 ( + h) 2 GM _ 4π 2 ( ) 3 = gives T = 5060 s ( = 84.3 min) Your answer must be in terms of ω to satisfy the question. The bit radius is ( + h), because the satellite is at height h above the surface of a planet radius. The physical basis of this answer is the same as that in Question 2(b), but here the they is applied to a satellite rather than a planet. The final result is again along the lines of Kepler s 3rd law, T 2 3, which applies to satellites biting planets as well as to planets biting the Sun. A lower bit produces a shter period. The lowest conceivable bit is a satellite grazing the surface of the Earth. The period of this would be around 85 min, so any real satellite must have a longer period. Values f and M are taken from the Data Booklet. AQA Physics A A2 Level Nelson Thnes Ltd

5 (c) The speed of the satellite increases. A lower satellite travels at a faster speed It loses potential energy but gains kinetic and has a shter period. Alternatively energy. v 2 GMm, from = m v2 r r 2 r 4 (a) (i) Gravitational field strength at the The equation relating the field strength g surface of the Earth g S = GM to follows from considering the 2 gravitational fce acting on a mass m. Gravitational field strength at height h This is given by either mg by GMm. GM g = 2 ( + h) 2 Equating the two values produces the result. The Earth may be considered as a g gs = 2, giving g = g ( + h) 2 s ( + h ) 2 point mass M placed at its centre, which is a distance ( + h) from the point at height h above the surface (ii) g = 9.8 ( 6 ( ) + ( )) 2 Direct substitution into the equation GM = 9.22 N kg g = from (a)(i) produces a swift ( + h) 2 solution. (b) elevant points include: the fce of gravity on the astronaut is still mg, where g is the local value of the field strength within the spacecraft; this fce provides the centripetal fce to keep the astronaut in bit; the astronaut is in free fall, as is the spacecraft; the astronaut appears weightless because he she is not suppted. any 3 If weight means the pull of gravity acting on an object, then the astronaut is certainly not weightless. The situation is comparable to a lift where all the suppting cables have broken. Both the lift and its contents would be in free fall. Because the contents are unsuppted they appear weightless, but they are actually falling to Earth at the acceleration of free fall! 5 (a) (i) Area indicated between the curve and the distance axis, which starts at distance = 700 km ends at distance = 5700 km 2 Just as the area under a fce distance graph represents wk done, the area under a graph of g against distance represents the wk done on a kg mass. AQA Physics A A2 Level Nelson Thnes Ltd

6 (ii) By counting squares, the area in (a)(i) is 90 ± 0 squares Area of one square = 200 km 0.N kg = J kg Wk done on kg = = (.8 ± 0.2) 0 6 J change of gravitational potential energy of satellite = = (8. ±.0) 0 8 J (b) elevant points include: The satellite must be raised to a point beyond that at which the resultant gravitational field strength of the Earth/ Moon is zero This is much closer to the Moon than to the Earth The Moon s mass is much less than the Earth s mass At their surfaces, g Moon is only _ 6 of g Earth (.7 N kg «9.8 N kg ) The escape speed from the Moon is much less than that from Earth On returning, the gravitational potential energy needed is much less so the kinetic energy at launch can be much less The total rocket weight at launch is much greater on Earth The wk done on a kg mass is the change in gravitational potential, V. This is represented by the area you were expected to mark in part (a)(i). The equation W = m V indicates that the wk done on a 450 kg mass will be 450 times greater. Alternatively: From the graph, when r = 700 km, g =.77 N kg. Substitution in g = GM gives r 2 GM = , allowing you to find V by evaluating V = GM r f r = 700 km and 4000 km. Application of W = m V then leads to the same result. 5 Once the returning Moon probe passes the neutral point, it will fall to Earth under the influence of the Earth s gravitational pull. The fuel carried must be sufficient to get it beyond this point. The gravitational pull of the Moon is much less than that of the Earth. Both the fce and the distance involved are much smaller, so much less wk has to be done to escape from the Moon than from Earth. This means that much less fuel will be required f the returning journey. Multi stage rockets are used to escape from the Earth, so that some weight can be jettisoned as the fuel is used up. 6 (a) Gravitational potential at a point is the wk done per unit mass (by an external agent)... to move a small test mass from infinity to the point. If a test mass is released from a point an infinite distance from another body, wk is done on the test mass by the gravitational field of the body as the mass comes closer. Therefe the external agent does not have to provide positive wk, indeed the gravitational potential is always negative. (Its value is taken to be zero at infinity.) AQA Physics A A2 Level Nelson Thnes Ltd

7 (b) At surface of Earth V E = GM E E At surface of Moon V M = GM M M V M = G M E = 3.7 E 8 GM E E = V E gives gravitational potential at surface of Moon V M = ( 63) = 2.88 MJ kg 2.9 MJ kg (c) Line drawn on graph: Starts at (surface of Earth, V = 63) and ends at V = 2.9, and begins as a curve of decreasing positive gradient... rises to a value close to (but below) zero... then falls to surface of Moon... from a point which is much closer to the Moon than to Earth. In der to satisfy the requirements of the question, you are only allowed to use the data given in the question itself. The first mark is f recognising how to apply the general equation f gravitational potential, V = GM, to the Earth and Moon. The second mark is f appreciating how to rearrange the algebra in the two equations, and the third is f evaluating the gravitational potential. Note that M M = M E 8 and that M = E 3.7. any 3 Gravitational potential is a scalar quantity. The total potential at any point along a line joining the Earth and Moon is the sum of the potentials produced by the Earth and Moon separately. The turning point (maximum) of the graph is the neutral point, where the resultant gravitational field strength is zero. See Question 5(b), which is about what happens when a Moon probe is brought back to Earth over this potential hill. 7 (a) The fce between two point masses is proptional to the product of the masses and inversely proptional to the square of their separation. (b) (i) Since V = GM, it follows that V, V should be constant Calculation of V f the three lines in the table, e.g ; (all three values should be ) V (ii) It is clear from the fmulae g = GM 2 and V = ( ) GM, both of which apply in a radial field, that the value of g is given by g = V in this case. Alternatively since V = GM S, S GM S = = This is Newton s law of gravitation. The question asks you to show inverse proption. The final part of your answer should end with the conclusion that, since V has been shown to be constant, the potential is inversely proptional to the distance from the centre of the Sun. It is necessary to show how the equation g = V is arrived at from the equations f g and V, because it is not a general fmula that can be applied to all fields. (The general fmula is g = ΔV Δr.) AQA Physics A A2 Level Nelson Thnes Ltd

8 From the first line of the table, g = V 9 00 = Alternatively G = GM S 2 = S (7 0 8 ) 2 = 27 N kg 270 N kg (iii) At the Earth s position, V S = GM S S = = J kg Potential energy of Earth = M E V S = ( ) = J Potential energy needed to escape = 0 ( ) = J J (iv) Orbital speed v of Earth around Sun = 2π T = 2π = m s E K = _ 2 mv2 = _ ( ) 2 = J Energy needed = ( ) 0 33 = J J The value is to be determined f g near the surface of the Sun. Only the first line in the table is relevant; the other two lines apply at larger distances from the Sun. Start by finding the gravitational potential due to the Sun in the position of the Earth s bit. The Earth s absolute potential energy is M E V S. If it is removed to infinity (where the gravitational attraction of the Sun would be zero), the Earth s absolute potential energy would be 0. The Earth takes one year to bit the Sun. The biting Earth already has a large amount of kinetic energy. Suppose the Earth stops when it arrives at infinity. Its E K has then all been converted to E P as a contribution to the minimum total energy needed to escape. 8 (a) (i) Gravitational potential energy of rocket = GMm (ii) In escaping from the Earth, potential energy gained = kinetic energy lost GMm = _ m 2 v2 Mass m of rocket cancels, giving escape speed v = 2GM (b) From the above equation v 2 M ( v P ve ) 2 = M P E P M = 4M E E 2 E E M = 2 E escape speed from planet v P = 2 ve = 2.2 = 5.8 km s This is (gravitational potential at the Earth s surface) (mass of rocket). When the rocket escapes, it is moved an infinite distance from Earth so that the there is no longer any gravitational attraction. At infinity the gravitational potential energy of the rocket is zero (V = 0 by definition at infinity) and we presume that the rocket stops moving when it arrives there. Alternatively: escape speed from planet v = 2GM v = v = m s (5.6 km s ). (But this method does not make use of the value of.2 km s given in the question.) AQA Physics A A2 Level Nelson Thnes Ltd

9 (c) elevant points: Wk is done against air resistance whilst in the atmosphere Some of the rocket s kinetic energy is converted into thermal energy in heating the rocket and the atmosphere 2 Since some of the rocket s kinetic energy is not effective in raising its gravitational potential energy, it will need me kinetic energy at the start. This means that the escape speed would need to be greater than that calculated from the equation in (a)(ii). Note, however, that it is somewhat misleading to apply this they to a space rocket. ockets are not thrown into space as a stone might be; they are accelerated as they rise. Nelson Thnes is responsible f the solution(s) given and they may not constitute the only possible solution(s). AQA Physics A A2 Level Nelson Thnes Ltd

Chapter 12 Gravity. Copyright 2010 Pearson Education, Inc.

Chapter 12 Gravity. Copyright 2010 Pearson Education, Inc. Chapter 12 Gravity Units of Chapter 12 Newton s Law of Universal Gravitation Gravitational Attraction of Spherical Bodies Kepler s Laws of Orbital Motion Gravitational Potential Energy Energy Conservation