Gravitation (Symon Chapter Six)

Transcription

1 Grvittion (Symon Chpter Six) Physics A31 Summer 6 Contents Preliminries 1 Course Outline Composite Properties in Curviliner Coördintes 1 The Volume Element in Sphericl Coördintes 3 3 Clculting the Center of Mss in Sphericl Coördintes 5 1 The Grvittionl Force 7 11 Force Between Two Point Msses 7 1 Force Due to Distribution of Msses 9 Grvittionl Field nd Potentil 1 1 Grvittionl Field 1 Grvittionl Potentil 1 1 Wrnings bout Symon 11 3 Exmple: Grvittionl Field of Sphericl Shell 1 3 Guss s Lw 15 A Appendix: Correspondence to Clss Lectures 17 Copyright 6, John T Wheln, nd ll tht 1

2 Tuesdy, My 9, 6 Preliminries 1 Course Outline 1 Grvity nd Moving Coördinte Systems Ch 6 Grvittion Ch 7 Moving Coördinte Systems Lgrngin nd Hmlionin Mechnics (Ch 9) 3 Tensor Anlysis nd Rigid Body Motion Ch 1 Tensor Algebr Ch 11 Rottion of Rigid Body Subject to modifiction, s Crl Brns tkes over fter chpter 7 Composite Properties in Curviliner Coördintes As precursor to our nlysis of the grvittionl influence of extended bodies, let s return to the clcultion of quntities such s totl mss nd center-of-mss position vector, for composite or extended body Recll from lst semester the definitions of totl mss M nd the center-of-mss position vector R for either collection of point msses or n extended solid body: Point Msses Mss Distribution N M = m k M= ρ( r) d 3 V k=1 R = 1 M N k=1 m k r k R = 1 M Reclling tht the position vector cn be written r ρ( r) d 3 V r = xˆx + yŷ + zẑ (1) nd defining the Crtesin coördintes of the center of mss s (X, Y, Z), so tht R = X ˆx + Y ŷ + Zẑ () it s esy to see tht X = 1 M Y = 1 M Z = 1 M x ρ(x, y, z) dx dy dz y ρ(x, y, z) dx dy dz z ρ(x, y, z) dx dy dz (3) (3b) (3c)

3 (A crucil step in the demonstrtion is the bility to pull the Crtesin bsis vector ˆx, ŷ, or ẑ out of ech integrl, which is oky becuse the Crtesin bsis vectors do not depend on loction in spce) The volume integrls in ech cse cover the solid in question Lst semester we looked t solids like prisms nd pyrmids which were esily descriubed in Crtesin coördintes Now let s consider sphere of uniform density ρ nd rdius centered t the origin This cn be defined in Crtesin coördintes by x + y + z (4) We could use the techniques from lst semester to find the limits of integrtion needed in Crtesin coördintes to clculte the mss, nd the result would be: z y z M = ρ dx dy dz (5) z y z The limits on the x nd y integrls mke the y nd z integrls messy to evlute However, the limits of integrtion re simple if we work in sphericl coördintes (r, θ, φ) defined implicitly by Then the sphere is defined by x = r sin θ cos φ y = r sin θ sin φ z = r cos θ (6) (6b) (6c) r (7) θ π (7b) φ π (7c) However, the volume element d 3 V is no longer s simple s it is in Crtesin coördintes 1 The Volume Element in Sphericl Coördintes The esiest wy to work out, or remember, the form of d 3!V in curviliner coördintes is to recll the infinitesiml chnge d r in the position vector r ssocited with infinitesiml chnges in the coördintes The position vector is nd its differentil, which we worked out lst semester, is r = xˆx + yŷ + zẑ = rˆr (8) d r = ˆx dx + ŷ dy + ẑ dz = ˆr dr + ˆθ r dθ + ˆφ r sin θ dφ (9) We derived this lgebriclly, but it lso hs geometric interprettion This interprettion is somewht simpler (nd esier to drw) if we look t it in two dimensions, where the forms of r nd d r in Crtesin nd polr coördintes re nd r = xˆx + yŷ = rˆr (1) d r = ˆx dx + ŷ dy = ˆr dr + ˆφ r dφ (11) 3

4 The re of the cell with (x, y) nd (x + dx, y + dy) on the corners is d A = dx dy (1) dx The re of the cell with (r, φ) nd (r + dr, φ + dφ) on the corners is So likewise, in 3 dimensions where the volume element is dy d A = (dr)(r dφ) = r dr dφ (13) r dφ dr d r = ˆr dr + ˆθ dθ + ˆφ r sinφ (14) d 3 V = (dr)(r dθ)(r sin θ dφ) = r sin θ dr dθ dφ (15) so the mss of sphere with uniform density ρ nd rdius is M = π π The θ integrl is done with the substitution so tht ρ r sin θ dr dθ dφ = ρ r dr sin θ dθ dφ = 4π 3 ρ3 (16) }{{}}{{}}{{} π r 3 3 = 3 3 π ; see below π µ = cos θ dµ = sin θ dθ (17) µ : cos cos π = 1 1 (17b) sin θ dθ = 1 1 ( dµ) = 1 1 π dµ = (18) The substitution µ = cos θ is usully the best wy to do the θ integrl in sphericl coördintes 4

5 3 Clculting the Center of Mss in Sphericl Coördintes How bout the center of mss R = 1 ρ( r) r d 3 V (19) M of sphere? Should we lso use r = r ˆr to find the sphericl coördintes of the center of mss? There s problem with this! ˆr depends on θ nd φ so it cn t be pulled out of the integrl like ˆx, ŷ, nd ẑ Insted find the Crtesin coördintes X, Y, nd Z, even if the integrls re done in sphericl (or cylindricl) coördintes: X = 1 M Y = 1 M Z = 1 M ρ(r, θ, φ) x r sin θ dr dθ dφ = 1 M ρ(r, θ, φ) y r sin θ dr dθ dφ = 1 M ρ(r, θ, φ) z r sin θ dr dθ dφ = 1 M ρ(r, θ, φ) (r sin θ cos φ) (r sin θ) dr dθ dφ () ρ(r, θ, φ) (r sin θ sin φ) (r sin θ) dr dθ dφ ρ(r, θ, φ) (r cos θ) (r sin θ) dr dθ dφ (b) (c) etc Now, for the sphere, X =, Y =, nd Z = (becuse of the symmetry of the problem with uniform density sphere centered t the origin) (Exercise: show this!) Consider insted different shpe Imgine sphere, gin of constnt density ρ nd rdius nd centered t the origin, but with code of opening ngle 45 tken out long the negtive z xis Here s the y = cross-section: Wht re its mss nd center-of-mss position vector? Well, the rnge of r nd φ vlues is the sme, but θ only runs from 3π, so, using 4 3π/4 sin θ dθ = 1 1/ dµ = = 1 + = + (1) 5

6 we hve M = π 3π/4 ( ) ( ρ r 3 + ) sin θ dr dθ dφ = ρ 3 (π) = ( + )πρ 3 3 () Agin X = = Y (exercise!) but Z = 1 M = π 3π/4 3 ( + πρ 4 )πρ ρ r cos θ r sin θ dr dθ dφ = ρ 4 π M 4 3π = ( + )8 = 3π( ) (4 )8 µ 1 1/ = 1 (1 1 )= 1 4 { }}{ 1 1/ µ dµ = 3( )π 16 (3) so R 345 ẑ, which puts the center of mss bout 1/3 of the wy up the z xis 6

7 Thursdy, My 11, 6 1 The Grvittionl Force 11 Force Between Two Point Msses We ve lredy introduced the grvittionl force between two objects smll enough to be idelized s point msses, which is: 1 Proportionl to the product of the msses Inversely proportionl to the squre of the distnce between them 3 Attrctive nd directed on line from one mss to the other To represent this mthemticlly in vector eqution, we consider the grvittionl force on mss m with position vector r due to mss m 1 with position vector r 1 The key geometricl quntity is the vector r 1 = r r 1 (11) which points from one to the other: m 1 r 1 m r 1 r O The distnce between the two msses is nd the unit vector pointing from mss 1 to mss is We this write the force on mss due to mss 1 s Some importnt fetures of this force: 1 It stisfies Newton s third lw F 1 = G m m 1 r 1 r 1 = r 1 (1) ˆr 1 = r 1 r 1 (13) F 1 = G m 1m ˆr r1 1 (14) ˆr 1 = G m 1m ˆr r1 1 = F 1 (15) 7

8 Becuse the grvittionl chrge is just the sme s the inertil mss, the ccelertion experienced by prticle is F = 1 = G m 1 ˆr m r1 1 (16) which depends on prticle s loction but not ny properties of the prticle itself This is clled the Equivlence Principle nd ws instrumentl in the development of Einstein s Generl Theory of Reltivity, which describes grvity in terms of the geometry of spcetime 3 The coupling constnt hs been numericlly determined s G = m3 kg s (17) We cn check the units on this by considering two one-kilogrm msses locted one meter wy from ech other: F = m3 1 kg 1 kg 11 kg m = = N (18) kg s (1 m) s We see not only tht Newton s constnt G hs the right units, but tht the grvittionl force between everydy objects is very smll This is why it is still difficult to determine Newton s constnt to the sme ccurcy s other constnts of nture Most things which re big enough to exert n pprecible grvittionl force (like moons, plnets, nd strs) re too big to hve their msses directly compred to everydy objects whose msses we know in kilogrms So we know the combintion GM pretty well for things like the Erth nd the Sun, from eg, Kepler s third lw, but to get n independent mesure of G (nd hence of the msses of plnet-sized things) required sophisticted experiments to mesure the grvittionl forces exerted by lbortory-sized objects While the form (14) emphsizes tht the mgnitude of the grvittionl force is inversely proportionl to the squre of the distnce between the msses, for prcticl clcultions, it s more useful to replce ˆr 1 using (13) nd sy F 1 = Gm 1 m r 1 r 3 1 = Gm 1 m r r 1 r r 1 3 (19) which is the form of Symon s eqution (63) 1 The extr fctor of r 1 in the denomintor serves to cncel out the fctor of distnce in the vector r 1 For the ske of completeness, we spell out the form of the grvittionl interction in terms of the x, y, nd z coördintes of the msses involved The position vectors re r 1 = x 1ˆx + y 1 ŷ + z 1 ẑ r = x ˆx + y ŷ + z ẑ (11) (11b) 1 Symon puts the minus sign in different plce by tlking bout r 1 r, but the two expressions re equivlent 8

9 which mkes the displcement vector from object 1 to object r 1 = r r 1 = (x x 1 )ˆx + (y y 1 )ŷ + (z z 1 )ẑ (111) The distnce r 1 between the objects is the length of this vector r 1 = r 1 = (x x 1 ) + (y y 1 ) + (z z 1 ) (11) So we cn write the grvittionl force in gory detil s F 1 = Gm 1 m r r 1 r r 1 3 = Gm 1m (x x 1 )ˆx + (y y 1 )ŷ + (z z 1 )ẑ [(x x 1 ) + (y y 1 ) + (z z 1 ) ] 3/ (113) 1 Force Due to Distribution of Msses Becuse Newtonin grvity obeys the principle of superposition, we cn build up the grvittionl force on point mss due to collection of other point msses, or due to continuous mss distribution, by dding up the effects due to ll of the individul source msses We re interested in the grvittionl force on point mss, locted t loction often referred to s the field point When the source of the grvittionl force ws lso point mss, we clled the mss t the field point m nd its position vector r Now we refer to it simply s m, nd its position vector s r Insted of single source mss m 1 t loction r 1, we now hve set of N source msses {m i i = 1 N} locted t positions { r i i = 1 N}, respectively The force on our mss m due to the ith source mss is, by nlogy with (19), which mkes the totl force F = r r i F due to i = Gmm i r r i 3 (114) N F due to i = i=1 N r r i Gmm i r r i 3 (115) which is equivlent to Symon s eqution (66) As we did when clculting centers of mss in Chpter 5, we cn replce the sum over point msses with n integrl over mss distribution of density ρ( r ), with the infinitesiml mss ρ( r )d 3 V : F = i=1 Gmρ( r r r ) r r 3 d3 V (116) The integrl is over the volume occupied by the mss distribution, lthough s Symon points out, it cn be thought of s n integrl over ll spce if we define ρ( r ) = for points r outside the mss distribution Note tht there re two position vectors in the expression (116): the source point r = ˆxx + ŷy + ẑz (117) which is integrted over (in Crtesin coördintes, d 3 V = dx dy dz ), nd the field point r = ˆxx + ŷy + ẑz (118) 9

10 which is not When evluting (116) for prticulr mss distributions, one should note tht the nswer cnnot depend on the primed coördintes x, y, nd z, but only on the unprimed ones x, y, nd z For given mss distribution, we cn often find the force which would be experienced by point prticle t n rbitrry position r nd consider this to be function of r nd the mss m of the prticle: F ( r) = { N i=1 Gmm i r r i r r i 3 Gmρ( r ) r r r r 3 d 3 V collection of point msses continuous mss distribution (119) This is force field, just like we considered lst semester Grvittionl Field nd Potentil 1 Grvittionl Field As we noted in the cse of single point source, the ccelertion experienced by point prticle due to grvittionl forces is ctully independent of the prticle s mss It is useful to think of ccelertion, dependent on the position but not the mss of the prticle in question, s vector field, which we cll the grvittionl field g( r) = F ( r) m = { N i=1 Gm i r r i r r i 3 Gρ( r ) r r r r 3 d 3 V collection of point msses continuous mss distribution (1) Grvittionl Potentil The force field F ( r) defined by (119) is conservtive, s one could demonstrte by explicitly clculting the curl F in Crtesin coördintes However, it s esier just to write down potentil energy V ( r) which stisfies F = V () In fct, we ve lredy worked out the grvittionl potentil energy ssocited with point source (see our study of centrl force motion, in prticulr Symon s eqution (39), or problem on Problem Set 1 from lst semester) In the lnguge of Section 11 of these notes, it s which generlizes to the superposition cse s V ( r ) = Gm 1m r 1 (3) V ( r) = { N Gmm i i=1 r r i Gmρ( r )d 3 V r r collection of point msses continuous mss distribution (4) The one cvet is tht the grvittionl force of the test prticle bck on the source msses might cuse them to move But we cn typiclly ssume they re held in plce by some other forces nd consider their loctions to be fixed 1

11 Just s it s useful to divide out the mss m nd produce vector field g( r) which depends only on the source msses, one cn similrly define grvittionl potentil ϕ( r) = V ( r) m = { N Gm i i=1 r r i Gρ( r )d 3 V r r collection of point msses continuous mss distribution (5) In prctice, it is often esier to clculte the potentil ϕ with (5) nd then differentite it to get the grvittionl field g = ϕ (6) thn it is to clculte the field g( r) directly using (1) 1 Wrnings bout Symon 1 There is typo in eqution (611) The left-hnd side should red V mmi rther thn V mm i Symon defines grvittionl field G( r) = V ( r)/m with the opposite sign to ϕ( r) nd clims this sign convention is stndrd Perhps it ws in 1971, but I ve never seen it before We ll work with ϕ( r) insted, since the sign convention preserves the symmetricl reltionship mong F, g, V, nd ϕ 11

12 Fridy, My 1, 6 3 Exmple: Grvittionl Field of Sphericl Shell Let S be sphericl shell of constnt density nd totl mss M with inner rdius nd outer rdius b, centered on the origin Find the grvittionl field g( r) resulting from this shell, t n rbitrry position r This is slightly different thn the exmple given in Symon, since we llow the shell to hve finite thickness, nd we solve the problem by slightly different method, integrting for the grvittionl field directly rther thn finding the potentil first For economy of nottion, we ll cll the density ρ We cn relte ρ to M by performing volume integrl: M = ρ d 3 V = 4π b3 3 ρ (7) 3 S In the clcultions, we ll just work with constnt ρ nd then relte it to M t the end Now, the grvittionl field is given by the integrl g( r) = Gρ( r r r ) r r 3 d3 V (8) In generl, we d need to perform triple integrl over r for ech of the three components g x, g y, nd g z, but in this cse we cn tke dvntge of the sphericl symmetry of the problem to restrict the form of g( r) Becuse no direction is preferred over ny other direction, the mgnitude of the field must depend only on the rdil coördinte r = r nd not on the direction of the position vector r Likewise, the field must point in the rdil direction, becuse there s nothing to pick out one direction over nother (When we do the integrl, there will be non-rdil components of the grvittionl field due to little pieces of the shell, but they will integrte to zero) So the field must hve the form g( r) = g r (r)ˆr (9) nd we just need to find the rdil component ( ) g r (r) = ˆr Gρ( r r r ) r r 3 d3 V (1) If we write r = rˆr, this becomes g r (r) = where the mgnitude in the denomintor is Gρ( r r ˆr r ) rˆr r 3 d3 V (11) rˆr r = r + r rˆr r (1) So the crucil piece of geometry is the dot product ˆr r Due to the geometry, we wnt to do the integrl d 3 V in sphericl coördintes, but the choice of xis with which to define the 1

13 θ nd φ coördintes is bsiclly rbitrry So we choose the xis to lie long ˆr (which is fine becuse it s r nd not r which is vrying in the integrl) which mens ˆr r = r cos θ This mkes the integrl π g r (r) = Gρ = πgρ π b b π (r r cos θ ) [r + r rr cos θ ] 3/ r dr sin θ dθ dφ (r r cos θ ) [r + r rr cos θ ] 3/ sin θ dθ r dr (13) The quntity inside the squre brckets is r + r r r cos θ (14) Note tht this is lwys positive number, which is equl to (r r) t θ =, incresing with θ ll the wy to (r + r) t θ = π So we cn chnge vribles, replcing θ with u = r + r r r cos θ (15) to get the differentil, we note tht so We lso note tht which tells us u du = d(u ) = d(r + r r r cos θ ) = r r sin θ dθ (16) g r (r) = πgρ = πgρ r sin θ dθ = u du r r r cos θ = r + r u r b r+r b (r r + r u r r r ( r+r r r + u du r r u }{{} I(r ) (17) (18) ) 3 u du u r r r dr ) r dr (19) Looking t the integrl I(r ) = 1 r+r r r ( r r ) + 1 du = 1 [ r r u u ] r+r + u r r () we see the vlue depends on whether r is greter or less thn r Considering ech cse seprtely, we find I(r > r) = 1 ( ) (r + r)(r r) + (r + r) (r + r)(r r) (r r) = (1) r + r r r 13

14 nd I(r < r) = 1 Armed with the result tht ( (r + r )(r r ) + (r + r ) + (r + ) r )(r r ) (r r ) r + r r r = r + r + r + r + r + r r + r I(r ) = { r = r () r < r r > r we need to think bout how the possible vlues of r compre to r in the integrl g r (r) = πgρ r b There re three cses, depending on the vlue of r (3) I(r )r dr (4) < r < In this cse, ll of the vlues r b in the integrl re lrger thn r, nd therefore g r (r) = πgρ b ()r dr = when < r < (5) r r > b Here, ll the possible vlues of r re smller thn r nd thus g r (r) = πgρ r b (r )r dr = 4πGρ b 3 3 r 3 = GM r when r > b (6) < r < b In this cse, r cn be lrger or smller thn r, but the only non-zero contributions re from r < r so the integrl becomes g r (r) = πgρ r r (r )r dr = 4πGρ r 3 3 r 3 = GM r 3 3 r b 3 3 Putting it ll together, we get r g( r) = GM r 3 3 ˆr r b r b 3 3 GM ˆr r b r when < r < b (7) (8) where r = r nd ˆr = r/r s usul Note tht this mens tht outside sphericl shell of mtter, the grvittionl field is the sme s if the whole mss were concentrted t the center, while inside sphericl shell, there is no grvittionl field due to the shell And since ny sphericlly symmetric distribution cn be described s superposition of sphericl shells, it mens the grvittionl field due 14

15 to such sphericlly symmetric distribution, distnce r from the center, is just the field due to the mss closer to the center: g(r) = GM(r) r ˆr (9) where r M(r) = 4π ρ(r ) r dr (3) is the mss inside sphere of rdius r 3 Guss s Lw Section 63 of Symon concerns grvittionl field theory which is nlogous to electrosttic field theory Here we derive one of the key results, which concerns the flux of the grvittionl field through closed surfce The nottion we ll use refers to the surfce s V, nd the volume it encloses s V The flux through V is thus where the vector-vlued re element for the surfce is V g d A (31) d A = n d A (3) with n being n outwrd-directed unit vector norml (perpendiculr) to the surfce V Symon uses geometricl rguments to evlute the integrl, but we ll rely on result from vector clculus, the divergence theorem, which sys g d A = ( g)d 3 V (33) V which we cn pply whenever the divergence g is well-defined First, we consider the flux of the grvittionl field due to point mss M If we define our coördintes so this point source is t the origin, the field is V g( r) = GM ˆr (34) r At ny point other thn the origin (t which r = ) we cn clculte the divergence: ( g = r + 1 r θ + 1 ) (g r (r)ˆr) (35) r sin θ φ Now, we could look up the form of the divergence in sphericl coördintes, but the reltively simple form of g mens we cn lso just work with the derivtive opertors, if we recll how 15

16 the unit vector ˆr vries from point to point: Using this nd the product rule, we hve g = dg r dr (ˆr ˆr) + g r = if r ˆr r = ˆr θ = ˆθ ˆr = sin θ ˆφ (36c) φ ( 1 r (ˆθ ˆθ) + 1 r sin θ ( ˆφ sin θ ˆφ) ) (36) (36b) = dg r dr + r g r = GM ( r 3 + r ) r (37) So, if the volume V enclosed by the surfce V does not include the origin (where the point source is locted), we cn use the divergence theorem to show tht the grvittionl flux through V vnishes If V does include the origin, there must be some minimum rdius such tht bll of rdius centered on the origin (which we cll B ) is entirely inside V Then we cn split up V into B nd V, which is the volume V with bll of rdius removed from it If we re creful bout the geometry, we cn show tht V g d A = B g d A + g d A (38) V since the contribution from the inner boundry of V cncels out tht through B At ny rte, it s cler tht V = B V (39) nd so, ppeling to the divergence theorem, g d A = ( g)d 3 V = ( g)d 3 V + V B V V ( g) d 3 V = }{{} = for r V B g d A (31) The flux through sphere of rdius we cn clculte directly, since r = nd ˆn = ˆr everywhere on the sphere, nd thus π π ( g d A GM ) B ˆr (ˆr sin θ dθ dφ ) = 4πGM (311) So in generl the flux through closed surfce due to point source is zero if the source is not enclosed in the surfce nd 4πG times the mss, if it is But ny mss distribution cn be built up out of point sources, so the generl result is g d A = 4πGM enc (31) V 16

17 where M enc is the enclosed mss M enc = V ρ( r )d 3 V (313) This cn be written in differentil form s which is nlogous to Guss s lw in electrosttics g = 4πGρ (314) A Appendix: Correspondence to Clss Lectures Dte Sections Pges Topics 6 My 9 6 Outline; Volume in Sphericl Coördintes 6 My Grv Force, Field & Potentil 6 My Sphericl Shell; Guss s Lw 17