Celestial Mechanics. Basic Orbits. Why circles? Tycho Brahe. PHY celestialmechanics  J. Hedberg


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1 PHY celestalmechancs  J. Hedberg Celestal Mechancs. Basc Orbts. Why crcles? 2. Tycho Brahe 3. Kepler 4. 3 laws of orbtng bodes 2. Newtonan Mechancs 3. Newton's Laws. Law of Gravtaton 2. The shell theorem. 3. Work and Energy 4. Potental Energy. Potental Plot 2. Escape Speed 5. Kepler's Laws. Cartesan & Polar Coordnates 2. st Law  Ellpses 6. The Vral Theorem Basc Orbts Why crcles? Coperncus was rght: the earth and the other planets do orbt around the sun. However, he was stll operatng under the assumpton that all the celestal bodes had to move n perfect crcles. Ths turned out to be a rather major flaw n the framework. Other astronomers of the era were unable to match the predctons of the Coperncan system wth observatons. More 'tweaks' were added to the Coperncan systems of crcular helocentrc orbts n order to 'save the apearances'. In order to make the model match wth observatons, they stll had to employ epcycles and so forth. However, the seed was planted. The sun should be at the center! Tycho Brahe Page
2 PHY celestalmechancs  J. Hedberg Late 6th century. Hybrd system: geohelocentrc model had the planets gong around the sun, but the sun stll went around the Earth. Close... Fg. Dansh astronomer Tycho Brahe [54660]. Made many very good measurements of the stars and planets. Kepler Kepler played a major role n the 7thcentury scentfc revoluton. He s best known for hs laws of planetary moton, based on hs works Astronoma nova, Harmonces Mund, and Eptome of Coperncan Astronomy. These works also provded one of the foundatons for Newton's theory of unversal gravtaton. Fg. 2 German mathematcan, astronomer, and astrologer. [57630] Kepler was Tycho's assstant and was beleved n the Coperncan system. 3 laws of orbtng bodes. A planet orbts the sun n an ellpse. The Sun s at one focus of that ellpse. 2. A lne connectng a planet to the Sun sweeps out equal areas n equal tmes 3. The sqaure of a planet's orbtal perod s proportonal to the cube of the average dstance between the planet and the sun: P 2 a 3. Ellpses An ellpse satsfes ths equaton: Page 2
3 F r r ae a F b r + a s the semmajor axs r and r are the dstances to the ellpse from the two focal ponts, F and F. e represents the eccentrcty of the ellpse ( 0 e ) r PHY celestalmechancs  J. Hedberg = 2a The semmajor axs s onehalf the length of the long (.e. major) axs of the ellpse. Conc Sectons The crcle s one of several conc sectons. (There's really nothng all that specal about crcles.) See appolonus  on concs. Ellptcal Orbts orbtng body aphelon central body prncple focus perhelon The frst law for Kepler holds that the orbtng body (.e. planet) wll trace out an ellptcal path as t orbts the central body (.e. the sun). The central body s located at the prncple focus. When the dstance between the planet and the sun s the shortest, the planet s sad to be at perhelon, or the pont of closes approach. When that dstance s greatest, the planet s located the aphelon. Polar Coordnates r = a ( e2) + e cos θ (2) Page 3
4 PHY celestalmechancs  J. Hedberg r r F The polar coordnate equaton can be obtan by startng wth the basc ellpse equaton Eq. () Take the pont at the top of the semmnor axs, where r = r. Snce r + r = 2a we know that r = a. Usng the Pythagorean theorem, we can say that: F ae b a r 2 = b 2 + a 2 e 2. Puttng a n for r yelds: Fg. 3 Ellpse  the pont where r = r b 2 = a 2 ( e 2 ) For polar coordnates, we'll use r and θ, r beng the dstance from the prncple focus and θ the angle measured counterclockwse from major axs of the ellpse. Agan, from Pythagorus: r F r θ r 2 = r 2 sn 2 θ + (2ae + r cos θ) 2 ae Fg. 4 Ellpse  and polar coordnates r and θ. F Expandng the 2nd RHS term: r 2 = r 2 sn 2 θ + 4 a 2 e 2 + 2aer cos θ + r 2 cos 2 θ = r 2 + 4ae (ae + r cos θ) and usng the fact that r 2 = (2a r) 2, we can obtan the polar equaton for the ellpse shown above n eq (2). Devaton from a Crcles.0 e = e = Mars' orbt [ e = ] looks a lot lke a crcle. Page 4
5 PHY celestalmechancs  J. Hedberg Equal Areas n Equal Tmes t orbtng body area t 2 focus focus t 2 area t Fg. 5 The blue areas n these fgures wll be the same f t 2 t s the same. The other concs Parabolas: e = 2p r = + cos θ (5) Hyperbolas: e > a( ) r = e2 + e cos θ (6) Newtonan Mechancs Newton's Laws In words: In symbols:. No change n velocty f there s no net force actng. 2. The net force actng on a body wll cause a proportonal acceleraton nversely proportonal to the mass of that body. 3. Forces come n equal and opposte pars. Δv = 0 f F net = 0 2. F net = n = F = ma 3. = F AB F BA Law of Gravtaton 2 Fg. 6 Two masses and a Page 5
6 PHY celestalmechancs  J. Hedberg The scalar form: gravtatonal nteracton F = G Mm r 2 (7) And n vector form: F 2 = G Mm r (8) r 2 Ths s Newton's Law of Unversal Gravtaton. It consders two masses: M and m and the dstance between them r. Also ncluded s G, the Unversal Gravtatonal Constant. ( G = m 3 kg  s 2 From nst The drecton of the force s always attractve and s along a lne connectng the centers of the two masses. F 2 s the force appled on object 2 by object. r s the dstance between the two objects ( r 2 2 ) And s the unt vector pontng from object to object to 2. r 2 r 2 = r 2 r r 2 r Example Problem #: The shell theorem. Show that at sphercal mass (of densty ρ(r) ) acts as a pont mass for objects outsde. Fg. 7 Consder a mass m a dstance r away from the center of a sphercal mass M. The radus of the sphercal mass s. Frst, the force from the rng whch a mass dm s: R 0 m dm rng d = G cos ϕ F rng We can reexpress the mass of the rng n terms of ts densty ρ(r). (The densty s at most only a functon of the radus,.e. sphercally symmetrc) d M rng The volume of a rng s just t's thckness: dr, tmes t's wh: Rdθ, tmes t's crcumference: 2πR sn θ whch equals: s 2 = ρ(r) d V rng 2 d rng = ρ(r) 2π sn θ dr dθ Page 6
7 PHY celestalmechancs  J. Hedberg d M rng = ρ(r) 2πR 2 sn θ dr dθ The cos ϕ can be expressed as: cos ϕ = r R cos θ s and s can be wrtten as the followng (from Pythagorus) s = (r R cos θ) 2 + R 2 sn 2 θ = r 2 2rR cos θ + R 2 Let's put all these back nto the force from the rng: m dm rng df rng = G cos ϕ s 2 = Gm [ρ(r) 2πR 2 r R cos θ sn θ dr dθ] [ ] [ ] s Then, we'll have to ntegrate over both varables. dr wll range from 0 to R 0 and dθ from 0 to π to get all the rngs that make up the entre sphere. s 2 F = 2πGm [ Integratng over θ yelds: 0 R 0 0 π ρ(r) (rr 2 sn θ R 3 sn θ cos θ) ( r 2 + R 2 3/2 2rR cos θ) dθ dr] (9) However, the mass of a shell of thckness dr s just: whch s just quanty nsde the ntegrand. Thus the force from on shell wll be, Ths force acts as f t s located at the center of the shell. We can then ntegrate over all the mass shells and the above becomes: whch s just the equaton for the force between two pont masses. R0 Gm F = 4πR 2 ρ(r) dr (0) r 2 O d M shell = 4πR 2 ρ(r)dr d F shell = Gm dm shell r 2 F = G mm r 2 () lttle g M F = G m (2) ( R + h) 2 Let's consder the gravtatonal force a dstance h above the surface of the earth. Snce F = ma, we can see that the acceleraton a wll be equal to: a = G M = g Page 7
8 PHY celestalmechancs  J. Hedberg a = G We generally call ths value 'lttle g'. Work and Energy M R 2 = g (3) z r m dr U f U = ΔU = F dr r f r ( M F r f y x Fg. 8 The change n potental energy s gven by the negatve of the work done by gravty. Expressng t n the ntegral above allows for changng force as a functon of poston (n contrast to the elementary defnton: W = F d) Potental Energy Fg. 9 Fnd the work done on an object by gravty as t moves from pont P to nfnty, where the potental energy s defned as 0. W = The force s always aganst the dsplacement, so R F(r) dr GM m F(r) dr = E dr r 2 We can pull the constants out of the ntegral, and we are left wth somethng very smple to ntegrate: W = G Em dr = GM m [ E ] Page 8
9 PHY celestalmechancs  J. Hedberg GM W = GM E m dr = [ E m ] r 2 r Evaluatng at the lmts of and R, we see that the work done s: From the orgnal defnton of work and potental energy: R GM W = [ E m GM ] = 0 E m GM = E m r R R R ΔU = U f U = W thus, snce U = 0 we can say: U = W And now we have a potental energy functon for masses at a dstance r from the center of the Earth: R U = G M e r m Potental Plot r U Escape Speed How fast to get somethng to reach nfnty, (and stop there)? Mm E = m G 2 v2 r (5) Ths s the total energy of a partcle: knetc plus potental. Snce the partcle wll be at r = and wll have stopped then the total energy must be zero. However, due to conservaton of energy, the total mechancal energy wll be the same. Thus, mv 2 Mm G = 0 at all tmes. Or: 2 Mm m = G 2 v2 r r Page 9
10 PHY celestalmechancs  J. Hedberg whch leads to v escape 2GM = (6) r For Earth, v escape = 82.4 m/s. Apollo traveled at 0423 m/s [nasa ref]. It got a lttle help from the moons gravty too. Kepler's Laws Cartesan & Polar Coordnates z y θ r m x cartesan y θ polar x Fg. 0 The standard Cartesan Coordnates. In polar coordnates, we have dfferent unt vectors: θ and r. and r = cos θ + j sn θ (7) θ = sn θ + j cos θ (8) We can also express some dervates: dr^ = ^ sn θ + j^ cos θ = θ^ (9) dθ dθ^ = ^ cos θ j^ sn θ = r^ (20) dθ And usng the chan rule, we can also express tme dervatves: These wll be useful later. dr^ dr^ dθ = = dθ dθ^ θ^ dθ dθ^ dθ = = r^dθ dθ (2) (22) Page 0
11 PHY celestalmechancs  J. Hedberg st Law  Ellpses The angular momentum per unt mass wll be gven by: L = m r2 dθ and the centrally drected gravtatonal force s gven by: F GMm dv = r = m t r 2 (23) (24) The acceleraton of the orbtng body s therefore: but dv GM a = = t r r (25) 2 dθ r = ( ) dθ (26) We can put eq (26) nto the acceleraton equaton (25) dv GM = t r ( dθ dθ^ ) 2 (27) Combnng ths wth (23) Integratng both sdes of ths equaton yelds: L dv = GMm dθ^ (28) L = + GMm v θ^ e (29) The e s a constant of ntegraton that depends on the ntal conons, whch we may choose. Page
12 PHY celestalmechancs  J. Hedberg y sun x Fg. We'll say that at t = 0, the orbter s at perhelon and we can orent the axes so that t's moton s entrely n the +y drecton, n other words: e = ej where e s a constant. Now: L v = θ + ej (30) GMm We can take the dot product wth θ of both sdes: L v θ = θ θ + ej θ (3) GMm Smplfyng, usng the defntons above (.e. (8)): L = + e cos θ GMm v t (32) ( ) s the tangental velocty of the orbter. From the defnton of angular momentum: L = r p v t mr v t = L (33) whch when put back nto (32), yelds: or solvng for r: L 2 = + e cos θ (34) GMm 2 r r = L 2 (35) GM m 2 ( + e cos θ) Ths s just the polar coordnates expresson for a conc secton. The Vral Theorem Page 2
13 PHY celestalmechancs  J. Hedberg The Vral theorem shows that for a gravtatonally bound system n equlbrum, the total energy s onehalf the tme averaged potental energy. Consder the quantty Q. Q p r (36) where p s the lnear momentum and r the poston of some partcle. The product rule gve the followng f we take the tme dervatve of the rght hand sde. dq dp = ( dr r + p ) (37) We could also say: dq d = = ( ) = dr m d r d 2 m r 2 d 2 I (38) 2 2 where I s the moment of nerta of the total number of partcles: I = m r 2 (39) d 2 I 2 2 dr p dp = r (40) Now, dr p = m v v = 2 = 2K 2 m v 2 (4) Thus: d 2 I 2 2 2K = F r (42) where we have used the 2nd law of Newton ( F = dp ) Let F j represent the force of nteracton between two partcles n the system. The force on due to j. If you have a partcle, then we need to add up the force contrbutons from every other partcle j where j. We can rewrte the poston of the th partcle as: F r = F j r (43) j j whch leads to r = ( + ) + ( ) 2 r r j 2 r r j (44) = j ( + j ) + j ( j ) Page 3
14 PHY celestalmechancs  J. Hedberg Snce the 3rd law of Newton says: zero. Thus: F r = ( + ) + ( ) 2 F j r r j 2 F j r r j (45) j j F j j = F j j, the frst term on the R.H.S. n the above equaton wll be F r = ( ) 2 F j r r j (46) j j Assumng only gravtatonal nteractons: we can wrte F j m m j = G r 2 j r^j (47) F r = 2 G m m j 2 j j j j r 3 j m m j G r j 2 ( r j r ) (48) (49) The potental energy U j between the and j partcles s just the term n the sums: Thus, U j m m j = G (50) r j F r = = U 2 U j (5) j j Now we can fll out eq: (42) d 2 I 2 K = U (52) 2 2 where we have replaced all the quanttes wth the tme averaged values. The average of d 2 I 2 s gong to be: d2i τ = 2 τ d 2 I 0 2 di di = ( ) τ τ 0 (53) (54) but for perodc motons lke orbts: di τ di = 0 (55) So 2 I = 0 Page 4
15 PHY celestalmechancs  J. Hedberg d 2 I = 0 (56) 2 whch leads fnally to 2 K = U or n terms of the total energy E E = U 2 The Vral Theorem: Uses It allows for statstcal results for many body systems. Page 5
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