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 2 2. Newton’s law of universal gravitation § 6.1 The gravitational force 12 12 21 21 ? ? 2 21 21 2 21 12 r r mm GF r r mm GF ?= ?= r r 1 The vector force 2The principle of linear superposition Gravitational force is always attractive force. 12 F r 1221 FF rr ?= 12 ? r 2112 rr = 21 ? r § 6.1 The gravitational force 3 The value of G G is a fundamental constant of nature. Determined by experiment(Henry Cavendish): 2211 /kgmN1067.6 ?×≈ ? G 3 § 6.1 The gravitational force 4Gravitational mass and inertial mass r r Mm GrF ? )( 2 ?= r m –gravitational mass )()( ramrF r r ′ = m ′ --inertial mass AAIA amF = 2 r Mm GF AG A = 2 r GM m m a AI AG A ?= Free-fall body experiment: § 6.1 The gravitational force BBIB amF = 2 0 r Mm GF BG B = 2 r GM m m a BI BG B ?= Experimental result: a A =a B constantmeansthat ==== k m m m m BI BG AI AG L choose suitable unit, k=1, m G =m I 4 The experimental results for proving m G =m I experimenter Galilei Newton Eotvos Dicke Braginsky time 1610 1680 1890-1915 1964 1971 IGI mmm /)( ? 3 102 ? ×< 3 101 ? ×< 9 103 ? ×< 11 101 ? ×< 13 109 ? ×< § 6.1 The gravitational force § 6.1 The gravitational force 5Principle of equivalence Einstein’s thinking experiment 1: 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 ① If the elevator accelerates with an acceleration a relative to an inertial frame outside the elevator: mamgT mamgTF y += =?= ∑ T r gm r mgTamgTa <<>> ,0;,0 § 6.1 The gravitational force 6 ③ If the cable breaks, a= -g ② relative to the elevator mamgT mamgTF y += =??= ∑ 0 a r T r gm r am r 0=T 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. 7 § 6.2 Gravitational force of a uniform sphere on a particle 3. Proof of the shell theorems αcos d d 2 x mm GF A A = The axial component of the force: )dd(d cos d ddd 2 L L ++= = ++= BA BA mmM x Mm G FFF α ?d =M § 6.2 Gravitational force of a uniform sphere on a particle θθπρρθθπ dsin2dsin2d 2 tRtRRM =???= x rR x r xRr R x Rr ddsin 2 cos cos cos 222 = ?+ = ? = θθ θ θ α x x Rr r mRGt F d)1(d 2 22 2 + ? = ρπ The force exerted by the circular ring dM on m: 8 § 6.2 Gravitational force of a uniform sphere on a particle The total force on m due to the entire shell: ∫∫ + ? + ? == Rr Rr x x Rr r mRGt FF d)1(d 2 22 2 ρπ 22 )4( r mM GR r mRGt F == ρπ Inside the shell: 0 d)1( d 2 22 2 = + ? == ∫ ∫ + ? Rr rR x x Rr r mRGt FF ρπ § 6.2 Gravitational force of a uniform sphere on a particle Exercise 1: Find the gravitational force between The small ball of mass m 1 and the thin staff of mass m and length L. Solution: r L m rm i r mm GF ddd ? d d 2 1 == = λ r i LddL mm Gi r r L mm GFF Ld d ? ) 11 ( ? d d 1 2 1 + ?=== ∫∫ + rr i LddL mm GF ? ) 11 ( 1 + ?= r x F r d 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. xO m M R § 6.2 Gravitational force of a uniform sphere on a particle Solution: x O m M R F r d α Md αcosdd FF x = 2 d d r Mm GF = 2322 220 22 )( )( d cosd xR GmMx xR x xR MGm FF M x + = + + == ∫∫ α iFF x ? ?= r F r d 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 RMrGm F 3 2 33 )( = = 3 3 3 3 34 34 R r R r = π π r M r M § 6.2 Gravitational force of a uniform sphere on a particle R r F rF ∝ 2 /1 rF∝ 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 11 ? cc O § 6.3 Kepler’s laws of planetary motion 1. Kepler’s first law The path of a planet orbiting the sun is an ellipse with the Sun at one focus of the ellipse; the other focus is empty. ε ε θε ε ? + = ? ? = 1 1 cos1 )1( peri aph 2 r r a r s asr 2=+ a c ≡ε r —polar coordinate ε --eccentricity a—semimajor axis peri aph Newton found that the elliptical paths described by Kepler are the consequence of a central inverse-square force law. More generally, Newton showed that the paths of the mass under the influence of this force had to be conic sections. § 6.3 Kepler’s laws of planetary motion 12 § 6.3 Kepler’s laws of planetary motion 2. Kepler’s second law During equal time intervals the radius vector from the sun to a planet sweeps out equal areas. M § 6.3 Kepler’s laws of planetary motion Central force—force that acts along the line between two particles 0sin ==× πrFFr r r 0=× amr rr 00 ) d d ()( d d )( d d )( d d =×+= ×+×= ×=× amr t v mrvm t r vmr t pr t rr r rr r rrrr 13 § 6.3 Kepler’s laws of planetary motion Define orbital angular momentum: prL rr r ×= The orbital angular momentum of a particle (or planet) under the influence of a central force is conserved. 0)( d d d d =×= pr tt L rr r Then § 6.3 Kepler’s laws of planetary motion θd Ad θdr r r We can prove that the conservation of the angular momentum is equivalent to the Kepler’s law of area. rrArrA rr r d 2 1 d)d( 2 1 d ×== θ m L vr t r r t rr t A 22 1 d d 2 1 d d 2 1 d d r rr r r rr r =×= ×= × = 14 § 6.3 Kepler’s laws of planetary motion 3. Kepler’s third law If T is the time that it takes for a planet to make one full revolution around the sun, and if a is half major axis of the ellipse, then 3 2 2 4 a GM T π = M a m P252~P254 § 6.4 The gravitational field and Gauss’s law for the the gravitational field 1. The gravitational field The existence of M sets up or establishes the gravitational field at all points in space whether the other mass m is presented or not. r r GM m F rg ? )( 2 ?== r r The gravitational field can be imagined as the force per unit mass at a given point in space. It is the same as the acceleration due to gravity at that location. 15 § 6.4 The gravitational field and Gauss’s law for the the gravitational field Gravitational force obey the principle of linear superposition, the total gravitational field established at some point in space by a number of masses is the vector sum of the individual gravitational fields at that point by each mass acting individually. 2. Principle of linear superposition 1individual masses Ni N i i gggg gg r L r L rr rr ++++= = ∑ 21 1 g r 2 g r 3 g r g r g′ r 1 m 2 m 3 m § 6.4 The gravitational field and Gauss’s law for the the gravitational field r r M Grg ? d )( 2∫ ?= r 2extended mass r M g ? r d -Gd 2 = r r ? g r d Md r P The influence of a mass M on another mass m is thought of as being conveyed by the field. The idea of action-at-a-distance take placed by idea of field . 16 § 6.4 The gravitational field and Gauss’s law for the the gravitational field Example 1: x O P λ R g r d g r d α Md 2322 22 2 0 22 2 )( 2 d )( dcosd xR GxR s xR x xR G s r x r G gg R x + = + + = == ∫ ∫∫ λπ λ λ α π αcosdd gg x = s r G r M Gg d d d 2 2 λ = = g r igxg x ? )( ?= r 3. Gravitational field lines A useful alternative geometric representation of the field. r r GM m F rg ? )( 2 ?== r r 1The direction of the gravitational field at any point is tangent to the field line passing through that point and in the direction indicated by arrows on the field line. The properties of the field lines: § 6.4 The gravitational field and Gauss’s law for the the gravitational field 17 2The gravitational field is strong where field lines are close together and weak where they are far apart. The number of lines passing through a square meter oriented perpendicular to the lines is proportional to the magnitude of the gravitational field. 3Field lines are never cross. The field at any point has a unique direction. § 6.4 The gravitational field and Gauss’s law for the the gravitational field SvSv r r ?== )cos( θΦ § 16.5 Gauss’s law for electric field and its applications S r Uniform vector field 4. Gauss’ law for the the gravitational field 1the flux of a vector 18 § 6.4 The gravitational field and Gauss’s law for the the gravitational field The differential flux dΦ of the gravitational field vector through the differential area dS is defined to be the scalar product of with . g r S r d g r Sg v r dd ?=Φ ? ? ? ? ? ?= = = ==?= SgSg SgSg Sg SgSg dcosd d0cosd 02cosd cosddd π π θΦ v r S r d g r θ Sv r r dd ?=Φ ∫∫ ?=?= SareaSarea SdSd r r r r vv ΦΦ 1 S 2 S Nonuniform vector field § 6.4 The gravitational field and Gauss’s law for the the gravitational field ∫∫ ∫∫ ?== ?== SS Sg Sg r r r r dd dd SS ΦΦ ΦΦ The total flux through the area S or a closed surface S: Choose the outward direction as the direction of the surface element dS for a closed surface. Example 2: find the flux of the local gravitational field near the surface of the earth through each of the five surfaces of the inclined plane and the total flux of the entire closed surface. 19 § 6.4 The gravitational field and Gauss’s law for the the gravitational field Solution: For three vertical surfaces 0d ver =?= ∫ Sg r r Φ For the inclined surface θθ θθΦ coscosd ) ? cos ? sin(d ? d incl gSSg jiSjgSg S S ?=?= +???=?= ∫ ∫∫ r r For the bottom surface θcosd ) ? d() ? (d btm gSSgSg jSjgSg S S = ′ = ′ = ′ ???= ′ ?=Φ ∫ ∫∫ r r x y θ S r d g r § 6.4 The gravitational field and Gauss’s law for the the gravitational field 5. Gauss’s law for the gravitational field 1solid angle 222 cosd ? dd d r S r rS r A φ ? = ? == r 2 Gauss’s law The flux of the gravitational field of M through a closed surface ∫ ?= S clsd dSg r r Φ S r d AS dcosd =φ r ? r r M ?d φ 20 Because φθ coscosdd 2 r GM SgSg ?==? r r ∫∫∫ ?=?=?= ? φ Φ d cosd d 2 S clsd GM r S GMSg r r GMSg πΦ 4d S clsd ?=?= ∫ r r § 6.4 The gravitational field and Gauss’s law for the the gravitational field S r d AS dcosd =φ r ? r r M ?d φ S θ g r GM S r GM rSr r GM Sg SSS π Φ 4 d ? d ? d 22 ?= ?=??=?= ∫∫∫ r r The flux through the spherical surface which surrounded the M The flux through the arbitrary surface which surrounded the M GMSg S πΦ 4d ?=?= ∫ r r § 6.4 The gravitational field and Gauss’s law for the the gravitational field 21 The flux through the arbitrary surface which don’t surrounded M 0d =?= ∫ S Sg r r Φ The flux through the arbitrary surface which surrounded several particles ∑ ∫ ?=?= N i Si S MGSg )within( 4d πΦ r r § 6.4 The gravitational field and Gauss’s law for the the gravitational field