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 F 0 . 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 F 0 (B) 4 F 0 (C) F 0 (D) F 0 /2 Solution: The magnitude of the gravitational force is 2 r GMm F grav = , according to the problem, we get 02 2 2 2 1616 )2/( 4 F r Gm r Gm F grav ===′ 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: At the center, 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 ρππ rr when 2 rA=ρ , the acceleration g r will increase as you go deeper, when Ar=ρ the acceleration g r will decrease as you go deeper. II. Filling the Blanks 1. Two masses m 1 and m 2 exert gravitational forces of equal magnitude on each other. Show that if the total mass M = m 1 +m 2 is fixed, the magnitude of the mutual gravitational force on each of the two masses is a maximum when m 1 = m 2 ( Fill < or = or > ). Solution: The magnitude of the gravitational force is 2 11 2 21 )( r mMGm r mGm F grav ? == , We get 2 0 )2( 0 d d 12 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 × 10 30 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 2.67×10 10 m/s 2 . The ratio of the magnitude of this acceleration and g is 2. 72×10 9 . 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 ππ ω = ? ? ? ? ? ? == The orbital period is )(1084.3 1041067.6 )101(14.3444 5 3011 252322 s GM r a r T ? ? ×= ××× ××× === ππ 4. The magnitude of the total gravitational field at the point P in Figure 2 is 2.37×10 -3 m/s 2 ,the magnitude of the acceleration experienced by a 4.00 kg salt lick at point P is 2.37×10 -3 m/s 2 , 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 θθ ++=+= 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( 1098.51067.6 ? )106.1( 1036.71067.6 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 7.36×10 22 kg 5.98×10 24 kg 1.60×10 8 m 4.16×10 8 m 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. Choose a pointlike differential masses dM shown as in figure 3, the field produced by dM is 22 d d xR MG g + = ? ? ? ? ? ? ? + == + ?=?= ? θθ θθ 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 + =θ , so 2/322 )( d d xR MGx g x + ?= 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 θ 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 σ be the surface mass density of the disk (the number of kilograms per square meter of the disk), so that 2 R M π σ = . 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 π σ = . Choose a circular differential mass dM shown in figure, σπ ?= rrM d2d 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 + ? ?= + ?= πσv 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 πσvv (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 dθ dr r The surface mass density of the disk 2 R M π σ = , so the gravitational field is constant ? 2 =?= kGg σπ 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 2ρ and radius 3R/4 is surrounded by a mantle of density ρ 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 ρππρρ 33 core 8 9 ) 4 3 ( 3 4 2 RRVM =??= (b) The mass of the mantle is ρπππρρ 333 mantle 48 37 ]) 4 3 ( 3 4 3 4 [ RRRVm =??=?= (c) The total mass of the planet is ρπρπρπ 333 48 91 48 37 8 9 RRRmM =+=+ (d) The magnitude of the gravitational field at the surface of the planet is ρπρπ GRRG R mMG g 896.1 48 91)( 2 surface == + = (e) The magnitude of the gravitational field at the interface between the core and the mantle is ρπρπ GR R RG R GM g erface 2 9 16 8 9 ) 4 3 ( 2 3 2 int =??== 3R/4 R/4 Core density 2ρ Mantle density ρ Fig.5 4. What is the flux of the total gravitational field of the Earth and Moon through a closed surface that encloses the Moon? If another surface encloses both the earth and Moon, by how much is the flux of the gravitational field changed? Solution: The surfaces are shown in figure. (a) The flux of the total gravitational field of the Earth and Moon through a closed surface S 1 that encloses the Moon is M S GMsgΦ π4d 1 1 ?=?= ∫ vv (b) The flux of the total gravitational field of the Earth and Moon through a closed surface S 2 encloses both the earth and Moon is )(4d 2 2 EM S MMGsgΦ +?=?= ∫ π vv The flux of the gravitational field changed is E GMΦΦΦ π4 12 ?=?=? M M M E S 2 S 1