UNIVERSITY PHYSICS I CHAPTER 6 Gravitational force and Gravitational Field 86.1 The gravitational force 1. How did Newton deduce the gravitational force lay Every particle in the <Y Moon universe attracts every 中a other particle with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them The direction Earth of the force is along the line joining the particles
1 1. How did Newton deduce the gravitational force law Every particle in the universe attracts every other particle with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them.The direction of the force is along the line joining the particles. §6.1 The gravitational force
86.1 The gravitational force 2. Newtons law of universal gravitation ① The vector force F12=-G ,m , m2 I Gravitational force is always attractive force FE- FES FE ② The principle of linear superposition 86.1 The gravitational force ③ The value of g G is a fundamental constant of nature Quartz fiber Determined by experiment(Henry Cavendish): Mirror G≈667×101N.m2/kg
2 2. Newton’s law of universal gravitation §6.1 The gravitational force 12 12 21 21 ˆ ˆ 2 1 2 21 2 1 2 12 r r m m F G r r m m F G = − = − r r 1 The vector force 2The principle of linear superposition Gravitational force is always attractive force. F12 r F21 F12 r r = − 12 rˆ 12 21 r =r 21 rˆ §6.1 The gravitational force 3 The value of G G is a fundamental constant of nature. Determined by experiment(Henry Cavendish): 11 2 2 ≈ 6.67×10 N⋅m /kg − G
86.1 The gravitational force @Gravitational mass and inertial mass F(r)=-G Mm r m-gravitational mass F(r)=ma(r) nertial mass Free-fall body experiment G GM =a Al 86.1 The gravitational force m GM = f=m BIB Experimental result: a=a that means AG=BG=.=k=constant nt BI choose suitable unit, k=1, mg=m
3 §6.1 The gravitational force 4Gravitational mass and inertial mass r r Mm F(r) G ˆ 2 = − r m –gravitational mass F(r) m a(r) r r = ′ m ′--inertial mass A AI A F = m a 2 r m M F G AG A = 2 r GM m m a AI AG A = ⋅ Free-fall body experiment: §6.1 The gravitational force B BI B F = m a 0 2 r m M F G BG B = 2 r GM m m a BI BG B = ⋅ Experimental result: aA=aB that means = = = k = constant m m m m BI BG AI AG L choose suitable unit, k=1, mG=mI
86.1 The gravitational force The experimental results for proving mg=m, experimenter time (m, -mG)/m, Galilei 1610 < Newton 1680 <1x10- Eotvos 1890-1915 <3×10-9 Dicke 1964 <1x10 Braginsky 1971 <9×10-13 86.1 The gravitational force 5 Principle of equivalence Einstein's thin king experiment 1
4 The experimental results for proving mG=mI experimenter Galilei Newton Eotvos Dicke Braginsky time 1610 1680 1890-1915 1964 1971 m I mG m I ( − ) / 3 2 10 − < × 3 1 10 − < × 9 3 10 − < × 11 1 10 − < × 13 9 10 − < × §6.1 The gravitational force §6.1 The gravitational force 5Principle of equivalence Einstein’s thinking experiment 1:
86.1 The gravitational force Einstein's thinking experiment 2 Inertial force and gravitational force are equivalent. No experiments can distinguish a physical system uniformly accelerating in the absence of gravity and a physical system at rest, subject to a uniform gravitational force. 86.1 The gravitational force Experiment: A person weighs a fish on a spring scale attached to the ceiling of an elevator, as shown in Figure Show that if the elevator accelerates, the spring scale reads a weight different from the true weight of the fish OIf the elevator accelerates with an acceleration a relative to an inertial frame outside the elevator: ∑F=T-mg=ma T mg t ma >0, T>mg; a<0, T<mg
5 Einstein’s thinking experiment 2: Inertial force and gravitational force are equivalent. No experiments can distinguish a physical system uniformly accelerating in the absence of gravity and a physical system at rest, subject to a uniform gravitational force. §6.1 The gravitational force Experiment: A person weighs a fish on a spring scale attached to the ceiling of an elevator, as shown in Figure. Show that if the elevator accelerates, the spring scale reads a weight different from the true weight of the fish. a r ①If the elevator accelerates with an acceleration a relative to an inertial frame outside the elevator: T mg ma Fy T mg ma = + ∑ = − = T r mg r a > 0, T > mg; a < 0, T < mg §6.1 The gravitational force
86.1 The gravitational force relative to the elevator ∑F,=T-mg-ma=0 T=mg+ma @If the cable breaks, a n T=0 The fish is weightless 86.2 Gravitational force of a uniform sphere on a particle 1. Shell theorem #1 A uniformly dense spherical shell attracts an external particle as if all the mass of the shell were concentrated at its center 2. Shell theorem #2 a uniformly dense spherical shell exerts no gravitational force on a particle located anywhere inside it 6
6 ③If the cable breaks, a= -g ②relative to the elevator T mg ma F T mg ma y = + ∑ = − − = 0 a r T r mg r ma r T = 0 The fish is weightless. §6.1 The gravitational force §6.2 Gravitational force of a uniform sphere on a particle 1. Shell theorem #1 A uniformly dense spherical shell attracts an external particle as if all the mass of the shell were concentrated at its center. 2. Shell theorem #2 A uniformly dense spherical shell exerts no gravitational force on a particle located anywhere inside it
86.2 Gravitational force of a uniform sphere on a particle 3. Proof of the shell theorems The axial component of the force dF=g mdmA cos al dF=dF4+dFB+… mdM =G-cosa dM=(dm4+dmg+…) dM=? 86.2 Gravitational force of a uniform sphere on a particle dM=2rRsin 8Rd8..p=2xptR sin d8 r-Rcos 8 cos a r+R Rcos e sin 0de dx rR The force exerted by the circular ring dm on m: dPs rGtpmR2-P2 +1)dx
7 §6.2 Gravitational force of a uniform sphere on a particle 3. Proof of the shell theorems cosα d d 2 x m m F G A A = The axial component of the force: d (d d ) cos d d d d 2 L L = + + = = + + A B A B M m m x m M G F F F α dM = ? §6.2 Gravitational force of a uniform sphere on a particle d 2π sinθ dθ ρ 2πρ sinθdθ 2 M = R ⋅ R ⋅ t ⋅ = tR x rR x r r R x R x r R sin d d 2 cos cos cos 2 2 2 = + − = − = θ θ θ θ α x x r R r Gt mR dF ( 1)d 2 2 2 2 + − = π ρ The force exerted by the circular ring dM on m:
86.2 Gravitational force of a uniform sphere on a particle The total force on m due to the entire shell R 2 f=dF nGtpmR rr+R +1)dx mM F nGtpmR (4R)=G Inside the shella F=dF nGtpmR R +1)dx 0 86.2 Gravitational force of a uniform sphere on a particle Exercise 1: Find the gravitational force between The small ball of mass m, and the thin staff of m and length l Solutio mdm dm dF=g dm= adr=-dr L F=[dF=( Gmm可:=cm L l dd+l M117 F=G d+L 8
8 §6.2 Gravitational force of a uniform sphere on a particle The total force on m due to the entire shell: ∫ ∫ + − + − = = r R r R x x r R r Gt mR F dF ( 1)d 2 2 2 2 π ρ 2 2 (4 ) r mM R G r Gt mR F = = π ρ Inside the shell: 0 ( 1)d d 2 2 2 2 = + − = = ∫ ∫ + − r R R r x x r R r Gt mR F F π ρ §6.2 Gravitational force of a uniform sphere on a particle Exercise 1: Find the gravitational force between The small ball of mass m1 and the thin staff of mass m and length L. Solution: r L m m r i r m m F G d d d ˆ d d 2 1 = = = λ r i L d d L m m i G r r L m m F F G d L d ˆ) 1 1 ( ˆ d d 1 2 1 + = = = − ∫ ∫ r r + i L d d L m m F G ˆ) 1 1 ( 1 + = − r F x r d
86.2 Gravitational force of a uniform sphere on a particle Exercise 2: a particle of mass m is placed on the axis of a circular ring of mass M and radius Find the gravitational force exerted by the ring on the particle located a distance from the center of the ring. R n 86.2 Gravitational force of a uniform sphere on a particle Solution: dm dF=dF cosa M dF dF mdM dF=g Gm dM F dF cos a (R2+x2)√R2+x MmX (R2+x 3/2 F=-F i
9 §6.2 Gravitational force of a uniform sphere on a particle Exercise 2: a particle of mass m is placed on the axis of a circular ring of mass M and radius R. Find the gravitational force exerted by the ring on the particle located a distance from the center of the ring. O x m M R §6.2 Gravitational force of a uniform sphere on a particle Solution: x O M m R F r d α dM dFx = dF cosα 2 d d r m M F = G 2 2 3 2 0 2 2 2 2 ( ) ( ) d d cos R x GmMx R x x R x Gm M F F M x + = + + = = ∫ ∫ α F F i x ˆ = − r F r d
86.2 Gravitational force of a uniform sphere on a particle Exercise 3: P240-241 (a)a point mass m located outside the sphere; (b)a point mass m located within the volume enclosed by the sphere itself. Solution: (a) F_GmM M 4m3/3 (b) 4丌R/3R gm(rM/R) GmM 86.2 Gravitational force of a uniform sphere on a particle Exercise 4: A spherical hollow is made in a lead sphere of radius R, such that its surface touches the outside surface of the lead sphere and passes through its center. The mass of the sphere before hollowing was m. what is the attracting force between a small sphere of mass m and the hollowing sphere. 10
10 Exercise 3: P240~241 Solution: (a) A point mass m located outside the sphere; (b) A point mass m located within the volume enclosed by the sphere itself. (a) 2 r GmM F = (b) r R GmM r Gm r M R F 3 2 3 3 ( ) = = 3 3 3 3 4 3 4 3 R r R r = π π r M r M §6.2 Gravitational force of a uniform sphere on a particle R r F F ∝ r 2 F∝1/r 2 R GmM §6.2 Gravitational force of a uniform sphere on a particle Exercise 4: A spherical hollow is made in a lead sphere of radius R, such that its surface touches the outside surface of the lead sphere and passes through its center. The mass of the sphere before hollowing was M. What is the attracting force between a small sphere of mass m and the hollowing sphere. d R m