University Physics Al No. 4 The gravitational force and the gravitational field Class Number ame I. Choose the Correct Answer 1. The magnitude of the force of gravity between two identical objects is given by Fo. If the mass of each object is doubled but the distance between them is halved, then the new force of between the objects will be (A) A)16F0 (B)4F0 (C)F0 (D)Fo/ Solution GMm The magnitude of the gravitational force is F according to the problem, we get 16—,=16F (r/2)2 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 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 (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 (C)can increase or 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 g·ds=4G∑M1=4ad whenp=A/r2, the acceleration g will increase as you go deeper, when p=Ar the
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 2 0 2 2 2 16 16 ( / 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: 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 ∫ ∑ ∫ g ⋅ s = G Mi = G V s d 4π 4π ρd r r when 2 ρ = A r , the acceleration g r will increase as you go deeper, when ρ = Ar the acceleration g r will decrease as you go deeper
I. Filling the blanks Two masses m, 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 mI- m2( Fill or= or>) Solution The magnitude of the gravitational force is F_ gm, m, gm,(M-m) dF G(M-2m) The 2. A mass m is inside a uniform spherical shell of mass Mand a mass m is outside the shell as shown in Figure 1. The magnitude of the total gravitational force on m is Mm (s+R)2-d R Solution on nl Fig1 0 F=F F GMm GMm A certain neutron star has a radius of 10.0 km and a mass of 4.00 x 10 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 2.67x10 m/s2 The ratio of the magnitude of this acceleration and g is 2. 72x10 If the student is in a circular orbit of radius 100 km about the neutron star, the orbital period is 3.84x10-s Solution (a) The magnitude of the force is F GMm =(x10)y=26×10m
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 1 1 2 1 2 ( ) r Gm M m r Gm m Fgrav − = = , We get 2 0 ( 2 ) 0 d d 2 1 1 1 M m r G M m m F = ⇒ = − = ⇒ Then m1 = m2 . 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 2 2 (s R) d Mm F G + − = . Solution: 2 2 2 on ' on total 'on on ( ) 0 s R d GMm r GMm F F F F F F F total total M m M m M m M m + − ∴ = = ∴ = = = + r r r Q r r r 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 10 2 5 2 11 30 2 2.67 10 m/s (1 10 ) 6.67 10 4 10 = × × × × × = = − r GM a M s m M′ R Fig.1 d
267×10 (b) The ratio of the magnitude of this acceleration and g is 81=272x109 (c)Since the acceleration is a=or=/2T 4r2 The orbital period is 4zr_4z23_|4×3142×(1×105)2 GMV667×10-×4×10x≈3.84×10-(s) 4. The magnitude of the total gravitational field at the point 1.60×10°m Moon P in Figure 2 is_ 2.37x10- m/s-_, the magnitude of the P9597.36×102kg acceleration experienced by a 4.00 kg salt lick at point P 2.37x10 m/s-, the magnitude of the total gravitational force on the salt lick if it is placed at P 9.48 N 4.16×108m Solution (a)The total gravitational field at the point P is gota =8M+ge=2Mi+2E(cos0i+sin]) ③o Fig 2 667×101×736×102:667×10-11×598×102416×108:3.84×103 (1.6×10 (416×103 4.16×1084.16×103 192×10-i+23×10(0.38+0.92j) (1.07×10-)+(2.12×10-)j The magnitude of the total gravitational field at the point P is gmu=2.37×103m/s (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) a=gmu=237×10-3m2 (c)The magnitude of the total gravitational force on the salt lick if it is placed at P is F=ma=4×2.37=948N I. Give the solutions of the following problems
(b) The ratio of the magnitude of this acceleration and g is 9 10 2.72 10 9.81 2.67 10 = × × = g a (c) Since the acceleration is 2 2 2 2 2 4 T r r T a r π π ω ⎟ = ⎠ ⎞ ⎜ ⎝ ⎛ = = The orbital period is 3.84 10 ( ) 6.67 10 4 10 4 4 4 3.14 (1 10 ) 5 11 30 2 2 3 2 5 2 s GM r a r T − − = × × × × × × × = = = π π 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 )ˆ ˆ (cos ˆ sin 2 2 i j r GM i r GM g g g PE E PM M total = M + E = + θ + θ r r r i j i i j i i j ˆ (2.12 10 ) ˆ (1.07 10 ) ) 92 ˆ 0. 38ˆ 2.3 10 (0. 92 10 ˆ 1. )ˆ 4.16 10 3.84 10 ˆ 4.16 10 1.6 10 ( (4.16 10 ) 6.67 10 5.98 10 ˆ (1.6 10 ) 6.67 10 7.36 10 3 3 4 3 8 8 8 8 8 2 11 24 8 2 11 22 − − − − − − = × + × = × + × + × × + × × × × × × + × × × × = The magnitude of the total gravitational field at the point P is 3 2 2.37 10 m/s − gtotal = × (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). 3 2 2.37 10 m/s − a = gtotal = × (c)The magnitude of the total gravitational force on the salt lick if it is placed at P is F = ma = 4× 2.37 = 9.48 N III. Give the Solutions of the Following Problems Moon Earth 7.36×1022kg 5.98×1024kg 1.60×108 m 4.16×108 m 90° P Fig.2 i ˆ j ˆ
1. Several planets(the gas giants Jupiter, Saturn, Uranus and Neptune) possess nearly circular surrounding ring erhaps composed of material that failed to form a satellite. dM In addition, many galaxies contain ring-like structures Consider a homogeneous ring of mass M and radius R(a) Find an expression for the gra the ring on a particle of mass m located a distance x from e center of the ring along its axis. See Fig 3.(b) Suppose that the particle falls from rest as a result of the attraction Fig 3 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 GdM GdM d g =-dg cose cos0 dg R2+x2 dg, -dgsin =GdM sine According to the symmetry of the ring-like structures 0 So the gravitational force is g= dg, i According to the graph, we get cos0=r GxdM R+x (R2+y2)3/2 Thus the gravitational force is g=dg i ('+r)/adMi=-GrM (b) Apply Newtons second law of motion, we have MMx GMx dy dv dx d Due to g (R2+x2)2 dt dx dt dx We can get the speed GMx 2GM( R√R
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 2 2 d d R x G M g + = ⎪ ⎪ ⎩ ⎪ ⎪ ⎨ ⎧ + = = + = − = − ⇒ θ θ θ θ sin d d d sin cos d d d cos 2 2 2 2 R x G M g g R x G M g g y x According to the symmetry of the ring-like structures, d = 0 ∫ g y . So the gravitational force is g g i x d ˆ ∫ = v . According to the graph, we get 2 2 cos R x x + θ = , so 2 2 3 / 2 ( ) d d R x Gx M gx + = − Thus the gravitational force is i R x GxM Mi R x Gx g g ix ˆ ( ) d ˆ ( ) d ˆ 2 2 3 / 2 2 2 3 / 2 + = − + = = − ∫ ∫ v (b) Apply Newton’s second law of motion, we have i R x GMmx F mg ˆ ( ) 2 2 3 / 2 + = = − v v Due to x v v t x x v t v R x GMx g d d d d d d d d ( ) 2 2 3 / 2 = = = + = , We can get the speed ) 1 1 d 2 ( ( ) d 2 2 0 2 2 3 / 2 0 R R x x v GM R x GMx v v x v + ⇒ = − + = ∫ ∫ m R M Fig.3 x θ x dM y
2. A mass M is in the shape of a thin uniform disk of radius R. Let the =-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 de (a)Find the gravitational field of the disk at a coordinate along the symmetry axis of the disk? Fig 4 (b) What is the expression for g for >>R in part(a)? (c)Let o be the surface mass density of the disk(the number of kilograms per square meter of the disk, so that as↓ zr. what is the gravitational field in part (a)as :>0 along the positive c-axis? olution (a)The mass Mis a uniform disk, so the surface mass density of the disk ISoM Choose a circular differential mass dM shown in figure. dM= 2mrdr.o The gravitational field established by the ring at a coordinate z is GzdM Gzo·2mdr Thus the gravitational field of the disk at a coordinate along the symmetry axis of the disk is -k =>0 g==-=+7)yk M -1-,k:>R, the disk can be treated as a point mass, so 8= (c)In part(a) 2GM R When 20, the gravitational field is 8=-2GMK R
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 z → 0 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, dM = 2πrdr ⋅σ The gravitational field established by the ring at a coordinate z is k z r Gz r r k z r Gz M g ˆ ( ) 2 d ˆ ( ) d d 2 2 3 / 2 2 2 3 / 2 + ⋅ = − + = − v σ π Thus the gravitational field of the disk at a coordinate z along the symmetry axis of the disk is ⎪ ⎪ ⎩ ⎪ ⎪ ⎨ ⎧ + − − = + ⋅ = = − ∫ ∫ 0 ˆ] ( ) 2 [ 1 0 ˆ] ( ) 2 [1 ˆ ( ) 2 d g d 2 2 2 1/ 2 2 2 2 1/ 2 0 2 2 3 / 2 k z z R z R M G k z z R z R M G k z r Gz r r g v v R σ π (b) For z >> R , the disk can be treated as a point mass, so k z GM g ˆ 2 = − r (c) In part (a) ( ) k R z z R GM g ˆ 1 2 2 1 2 2 2 ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ + = − − r , When z → 0 , the gravitational field is k R GM g ˆ 2 2 = − r 0 z R M k ˆ Fig.4 dθ dr r
The surface mass density of the disk 0= ap2, so the gravitational field is 8=-2nGok=constant 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 2p and radius 3R/4 is surrounded by a mantle of density p 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. Mantle density p (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 Core density 2p 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 丌(R)3=zRp (b) The mass of the mantle is m=p Mante=p l mp34 R]=0TR'p (c)The total mass of the planet is M+m=rRp+.Rp=zrR'p (d) The magnitude of the gravitational field at the surface of the planet is G(M+m)91 丌GRp=1.896xGR R 48 (e) The magnitude of the gravitational field at the interface between the core and the mantle GM 16 g =G·mR`p·=2zGRp
The surface mass density of the disk 2 R M π σ = , so the gravitational field is constant ˆ g = −2πGσ k = 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 ρ ρ π π ρ 3 3 core 8 9 ) 4 3 ( 3 4 M = V ⇒ 2 ⋅ R = R (b) The mass of the mantle is ρ ρ π π π ρ 3 3 3 mantle 48 37 ) ] 4 3 ( 3 4 3 4 m = ⋅V = ⋅[ R − R = R (c) The total mass of the planet is π ρ π ρ π ρ 3 3 3 48 91 48 37 8 9 M + m = R + R = R (d) The magnitude of the gravitational field at the surface of the planet is πGRρ πGRρ R G M m g 1.896 48 ( ) 91 surface 2 = = + = (e) The magnitude of the gravitational field at the interface between the core and the mantle is π ρ πGRρ R G R 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? The surfaces are shown in figure (a)The flux of the total gravitational field of the Earth and Moon through a closed surface S, that encloses the Moon is =+8ds=-47GMM (b) The flux of the total gravitational field of the Earth and Moon through a closed surface S2 encloses both the earth and moon 02=5g:d=-x(M+ME) The flux of the gravitational field changed ④=中2-1=-4mGMg
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 S1 that encloses the Moon is M S Φ g ds 4πGM 1 1 = ⋅ = − ∫ v v (b) The flux of the total gravitational field of the Earth and Moon through a closed surface S2 encloses both the earth and Moon is d 4 ( ) 2 2 M E S Φ = ∫ g ⋅ s = − πG M + M v v The flux of the gravitational field changed is Φ Φ Φ πGM E 4 ∆ = 2 − 1 = − MM ME S2 S1