西南交大 大学物理 英文 试题 答案No.A1-4.pdf

University Physics AI No 4 The Gravitational Force and the Gravitational Field Class Number Name I Choose the Correct Answer 1 The magnitude of the force of gravity between two identical objects is given by F0 If the mass of each object is doubled but the distance between them is halved then the new force of gravity between the objects will be A A 16 F0 B 4 F0 C F0 D F0 2 Solution The magnitude of the gravitational force is 2 r GMm Fgrav according to the problem we get 0 2 2 2 2 1616 2 4 F r Gm r Gm Fgrav 2 A spherical symmetric nonrotating body has a density that varies appreciably with the radial distance from the center At the center of the body the acceleration of free fall is C A Definitely larger than zero B Possibly larger than zero C Definitely equal to zero Solution According to the shell theorem 1 at the center of the body the force acted on the body is zero so the acceleration of the body is zero 3 The acceleration due to gravity in a hole dug into a nonuniform spherically symmetric body C A will increase as you go deeper reaching a maximum at the center B will increase as you go deeper but eventually reach a maximum and then decrease until you reach the center C can increase or decrease as you go deeper D must decrease as you go deeper Solution For a nonuniform spherically symmetric body the force of gravity depends on the distance from the center which is related to how the density of the body changed with respect to the distance from the center for instance according to the Gauss s law of gravity VGMGsg i s d44d rr when 2 rA the acceleration g r will increase as you go deeper when Ar the acceleration g r will decrease as you go deeper II Filling the Blanks 1 Two masses m1 and m2 exert gravitational forces of equal magnitude on each other Show that if the total mass M m1 m2 is fixed the magnitude of the mutual gravitational force on each of the two masses is a maximum when m1 m2 Fill Solution The magnitude of the gravitational force is 2 11 2 21 r mMGm r mGm Fgrav We get 2 0 2 0 d d 1 2 1 1 M m r mMG m F Then 21 mm 2 A mass m is inside a uniform spherical shell of mass M and a mass M is outside the shell as shown in Figure 1 The magnitude of the total gravitational force on m is 22 dRs Mm GF Solution 222 on on onon total 0 dRs GMm r GMm F FF F FFF total mMtotal mM mMmM rr r Q rrr 3 A certain neutron star has a radius of 10 0 km and a mass of 4 00 1030 kg about twice the mass of the Sun The magnitude of the acceleration of an 80 0 kg student foolish enough to be 100 km from the center of the neutron star is 2 67 1010 m s2 The ratio of the magnitude of this acceleration and g is 2 72 109 If the student is in a circular orbit of radius 100 km about the neutron star the orbital period is 3 84 10 5 s Solution a The magnitude of the force is ma r GMm F 2 So 210 25 3011 2 m s1067 2 101 1041067 6 r GM a M s m M R Fig 1 d b The ratio of the magnitude of this acceleration and g is 9 10 1072 2 81 9 1067 2 g a c Since the acceleration is 2 2 2 2 42 T r r T ra The orbital period is s 1012 1 1041067 6 101 14 3 444 2 3011 352322 GM r a r T 4 The magnitude of the total gravitational field at the point P in Figure 2 is 2 37 10 3 m s2 the magnitude of the acceleration experienced by a 4 00 kg salt lick at point P is 2 37 10 3 m s2 the magnitude of the total gravitational force on the salt lick if it is placed at P is 9 48 N Solution a The total gravitational field at the point P is sin cos 22 ji r GM i r GM ggg PE E PM M EMtotal rrr ji jii jii 1012 2 1007 1 92 0 38 0 103 2 1092 1 1016 4 1084 3 1016 4 106 1 1016 4 1098 5 1067 6 106 1 1036 7 1067 6 33 34 8 8 8 8 28 2411 28 2211 The magnitude of the total gravitational field at the point P is 23 m s1037 2 total g b The acceleration has no relation with anything at some point So the magnitude of the acceleration experienced by a 4 00 kg salt lick at point P is also equal to a 23 m s1037 2 total ga c The magnitude of the total gravitational force on the salt lick if it is placed at P is N48 937 24 maF III Give the Solutions of the Following Problems Moon Earth 7 36 1022kg 5 98 1024kg 1 60 108m 4 16 108m 90 P Fig 2 i j 1 Several planets the gas giants Jupiter Saturn Uranus and Neptune possess nearly circular surrounding rings perhaps composed of material that failed to form a satellite In addition many galaxies contain ring like structures Consider a homogeneous ring of mass M and radius R a Find an expression for the gravitational force exerted by the ring on a particle of mass m located a distance x from the center of the ring along its axis See Fig 3 b Suppose that the particle falls from rest as a result of the attraction of the ring of matter Find an expression for the speed with which it passes through the center of the ring Solution a Set up the coordinate system shown in figure 3 Choose a pointlike differential masses dM shown as in figure 3 the field produced by dM is 22 d d xR MG g sin d sindd cos d cosdd 22 22 xR MG gg xR MG gg y x According to the symmetry of the ring like structures 0d y g So the gravitational force is igg x d v According to the graph we get 22 cos xR x so 2 322 d d xR MGx gx Thus the gravitational force is i xR GxM iM xR Gx igg x d d 2 3222 322 v b Apply Newton s second law of motion we have i xR GMmx gmF 2 322 v v Due to x v v t x x v t v xR GMx g d d d d d d d d 2 322 We can get the speed 11 2d d 22 0 2 322 0 xR R GMvx xR GMx vv x v m R M Fig 3 x x dM y 2 A mass M is in the shape of a thin uniform disk of radius R Let the z axis represent the symmetry axis of the disk as indicated in Figure 4 The Milky Way Galaxy is modeled as such a mass disk to a first approximation a Find the gravitational field of the disk at a coordinate z along the symmetry axis of the disk b What is the expression for g r for z R in part a c Let be the surface mass density of the disk the number of kilograms per square meter of the disk so that 2 R M What is the gravitational field in part a as 0 z along the positive z axis Solution a The mass M is a uniform disk so the surface mass density of the disk is 2 R M Choose a circular differential mass dM shown in figure rrMd2d The gravitational field established by the ring at a coordinate z is k rz rrGz k rz MGz g d2 d d 2 3222 322 v Thus the gravitational field of the disk at a coordinate z along the symmetry axis of the disk is 0 1 2 0 1 2 d2 dg 2 1222 2 1222 0 2 322 zk Rz z R M G zk Rz z R M G k rz rrGz g R vv b For Rz the disk can be treated as a point mass so k z GM g 2 r c In part a k zR z R GM g 1 2 2 1 22 2 r When 0 z the gravitational field is k R GM g 2 2 r 0 z R M k Fig 4 d dr r The surface mass density of the disk 2 R M so the gravitational field is constant 2 kGg r 3 The Earth is not in fact a sphere of uniform density A high density core is surrounded by a shell or mantle of lower density material Suppose we model a planet of radius R as indicated in Figure 5 A core of density 2 and radius 3R 4 is surrounded by a mantle of density and thickness R 4 Let M be the mass of the core and m be the mass of the mantle This is not an accurate model of the interior structure of the Earth but makes for an interesting and tractable problem a Find the mass of the core b Find the mass of the mantle c What is the total mass of the planet core mantle d Find the magnitude of the gravitational field at the surface of the planet e What is the magnitude of the gravitational field at the interface between the core and the mantle Solution a The mass of the core is 33 core 8 9 4 3 3 4 2RRVM b The mass of the mantle is 333 mantle 48 37 4 3 3 4 3 4 RRRVm c The total mass of the planet is 333 48 91 48 37 8 9 RRRmM d The magnitude of the gravitational field at the surface of the planet