Newton s Laws, Keple s Laws, and Planetay Obits PROBLEM SET 4 DUE TUESDAY AT START OF LECTURE 28 Septembe 2017 ASTRONOMY 111 FALL 2017 1 Newton s & Keple s laws and planetay obits Unifom cicula motion Cente of mass Elliptical obits and thei consistency with Newtonian mechanics Keple s laws (Pe-)validation of Newtonian dynamics by Keple and Tycho 28 Septembe 2017 ASTRONOMY 111 FALL 2017 2 1
Newton s Laws Newton s fist law: It takes foce to change a body s velocity eithe in magnitude o diection. Newton s second law: The ate at which a body changes its velocity (i.e. acceleates) is popotional to the foce: F = ma (Note that the foce F and the acceleation a ae vectos.) Newton s thid law: Evey foce is opposed by an equal and opposite foce in eaction. 28 Septembe 2017 ASTRONOMY 111 FALL 2017 3 Newton s Law of gavity, foce, potential F & ( = Gm &m ( ( & ( U &,( = Gm &m ( & ( & ( whee G = 6.674215 ± 0.000092 10 <= cm @ g <& s <( 10 <&& m @ kg <& s <( m 1 F 1 2 1 2 m 2 ˆ 1 2 Note: spheical bodies act gavitationally on objects outside them as if all thei mass is concentated at thei centes. F =-F 2 1 1 2 28 Septembe 2017 ASTRONOMY 111 FALL 2017 4 2
Unifom cicula motion Suppose an object (mass m) moves in a cicle at constant speed v. (Speed = magnitude of velocity) It is then acceleating, because its velocity is constantly changing diection. The acceleation is always pependicula to the velocity, because the speed does not change. What is the acceleation? Beak down the velocity into its components in Catesian coodinates. y v y v x h v h = wt = v t m x 28 Septembe 2017 ASTRONOMY 111 FALL 2017 5 Unifom cicula motion (cont.) The angle v makes with the x-axis is η + F (, so y Thus, v I = v cos ωt + π 2 v P = v sin ωt + π 2 a I = dv I dt a P = dv P dt = v sin ωt = v cos ωt = ωv cos ωt = ωv sin ωt v y v x h v a m x a = a ( I + a ( P = ωv = v( at = xt cos ωt yt sin ωt = X 28 Septembe 2017 ASTRONOMY 111 FALL 2017 6 3
Unifom cicula motion (cont.) A body in unifom cicula motion at speed v and adius acceleates with a constant magnitude a = YZ [, diected towad the cente of the cicle: a = v( X This equies a foce in the X diection. Note that because v and thus ω ae constant in a cicula obit, thee is a simple elation with the peiod of the obit, P: P = 2π v = 2π ω y v a h = wt m x 28 Septembe 2017 ASTRONOMY 111 FALL 2017 7 Unifom cicula motion and gavity Suppose a lage mass M lies a distance fom a mass m. At what speed will the small mass obit in a cicle about the lage mass? F = GMm ( X = ma = m v( X v = GM What is the coesponding total enegy and the angula momentum of the small mass elative to the cente of its obit? E = K + U = 1 2 mv( GMm = GMm 2 GMm = GMm 2 L = mv = zxmv sin π = zxm GM 2 28 Septembe 2017 ASTRONOMY 111 FALL 2017 8 4
Cente of mass What if the masses ae not vey diffeent? Then both move in an obit about the system s cente of mass. The cente of mass is defined accoding to momentum consevation. In a fame of efeence in which the bodies have zeo momentum, the position of the cente of mass is consistent in time: m & + m ( d cd dt d & = m & dt + m d ( ( dt = 0 m & + m ( cd = m & & + m ( ( + constant cd = m & & + m ( ( m & + m ( 0, fo the most convenient coodinate oigin choice. 28 Septembe 2017 ASTRONOMY 111 FALL 2017 9 COM (cont.) Suppose the two masses ae sepaated by a displacement, and we place the cente of mass at the oigin of the coodinates: m & & + m ( ( m & + m ( = 0 = ( & m & & + m ( + & m & + m ( = 0 Usually one also defines the educed mass: & = m ( m & m & + m ( = ( m & + m ( μ = m &m ( m & + m ( 28 Septembe 2017 ASTRONOMY 111 FALL 2017 10 5
COM (cont.) Then Thus, & = μ m & v & = μ m & d dt = μ m & v ( = μ m ( v ( = μ m ( v E = 1 2 m &v & ( + 1 2 m (v ( ( Gm &m ( ( & Vey simila to the lage-mass case. = 1 2 μv( Gμ m & + m ( The motion of one of the two bodies, in a efeence fame at which the othe is at est, is the same as the motion of the educed mass. 28 Septembe 2017 ASTRONOMY 111 FALL 2017 11 COM Example 1 Descibe the obits of two masses, m 1 and m 2, about thei common cente of mass if thei sepaation is. In a coodinate system centeed on m 1, the foce on m 2 appeas as F ( = Gm h ( &m ( ( X = m ( a h v ( ( = m ( X v ( h = Gm & That is, cicula motion about m 1. But in a coodinate system with its oigin at the cente of mass, & = μ m & ( = μ m ( v & = μ m & d dt = μ m & v ( h = μ m & Gm & v ( = μ m ( Gm & 28 Septembe 2017 ASTRONOMY 111 FALL 2017 12 6
COM Example 1 (cont.) So each mass tavels in a cicula obit about the cente of mass, one with adius & = j k l and speed v & = j mk l k l [ and the othe with adius ( = j k Z and speed v ( =. j kz mk l [ Note that if m 1 = m 2 = m, the masses ae equidistant fom the cente of mass and obit at the same speed: n = m m + m = 2 v n = m m + m Gm = 1 2 Gm 28 Septembe 2017 ASTRONOMY 111 FALL 2017 13 Escape speed & example If an object is gavitationally bound to anothe, with total enegy E (a negative numbe), and an extenal agent adds a kinetic enegy to the object equal to E, then the object can (just baely) escape. Escape fom the suface of an isolated planet with mass M, adius R, initially at est but given an impulse: E o = K o + U o = GMm R E q = K q + U q = 1 2 mv ( st GMm R = 0 v st = 2GM R 2 times the obital speed at distance R 28 Septembe 2017 ASTRONOMY 111 FALL 2017 14 7
Elliptical obits Equation x ( a ( + y( b ( = 1 Foci at x = ±c c = a ( b ( Eccenticity ε = c a = a( b ( Focal lengths (aphelion and peihelion distances) a f = a ± c = a 1 ± ε Do not confuse the semimajo axis with the acceleation. 28 Septembe 2017 ASTRONOMY 111 FALL 2017 15 Elliptical obits (cont.) It is a simple algebaic execise to show that by substituting x h = cos η and y h = sin η into That is, pola coodinates with the oigin on one focus and angle measued fom the coesponding majo axis, the equation fo the ellipse becomes = x h + c ( a ( + y( b ( = 1 a 1 ε( 1 + ε cos η x y h 28 Septembe 2017 ASTRONOMY 111 FALL 2017 16 8
Algeba that we will not discuss Fist eliminate b and c in favo of a and ε: c = εa b ( = a ( c ( = a ( 1 ε ( Then the equation of the ellipse becomes x h + c ( a ( + y ( b ( = xh + εa ( a ( + y ( a ( 1 ε ( = 1 ( cos η ( + ε ( a ( + 2εa cos η a ( + ( sin η ( a ( 1 ε ( = 1 ( cos η ( + ε ( a ( + 2εa cos η 1 ε ( + ( sin η ( = a ( 1 ε ( ( 1 ε ( cos η ( + 2εa cos η 1 ε ( a ( 1 ε ( ( = 0 28 Septembe 2017 ASTRONOMY 111 FALL 2017 17 Algeba (cont.) This is a quadatic equation in, with the solution = 1 2 1 ε ( cos η ( } 2εa cos η 1 ε( ± 4ε ( a ( cos η ( 1 ε ( ( + 4 1 ε ( cos η ( a ( 1 ε ( ( ~ = 1 2 1 ε ( cos η ( εa cos η 1 ε ( ± a 1 ε ( = a 1 ε ( ±1 ε cos η 1 + ε cos η 1 ε cos η Nomally, is consideed to be a positive numbe, so we choose the uppe sign and find that a 1 ε( = 1 + ε cos η 28 Septembe 2017 ASTRONOMY 111 FALL 2017 18 9
Elliptical obits (cont.) Suppose a vey lage mass M occupies the oigin, and a small mass m is in a (not necessaily cicula) obit about the lage one. In these pola coodinates, we can wite the total enegy pe unit (obite) mass as E m = v( 2 GM = v ( [ ( 2 + v 2 GM The velocity has been expessed as components in the adial diection and the diection pependicula to this ( η ). Similaly, the angula momentum pe unit mass is L m h = v = ( dη dt Since in a shot time dt the angle η changes by dη = Y ƒ [. 28 Septembe 2017 ASTRONOMY 111 FALL 2017 19 Elliptical obits (cont.) You will late (in Phys 235) lean that this fomula fo the enegy yields the Hamiltonian equation of motion fo the obiting paticle with mass m. Both enegy and angula momentum ae conseved (thus ae constant, whateve the obit), so E m = v ( [ 2 + h( 2 ( GM But So v [ = d dt = d dη dη dt = d dη E m = d dη A fist-ode, nonlinea diffeential equation. ( h ( h ( 2 + h( 2 ( GM 28 Septembe 2017 ASTRONOMY 111 FALL 2017 20 10
Elliptical obits (cont.) You will also late lean (in Phys 235) that this diffeential equation has the unique solution a 1 ε( = 1 + ε cos η That is, the equation of an ellipse whee the semimajo axis a and eccenticity ε ae given in tems of the angula momentum pe unit mass h and total enegy E by h = GMa 1 ε ( E m = GM 2a a = GMm 2E ε = 1 + 2 h GM ( E m 28 Septembe 2017 ASTRONOMY 111 FALL 2017 21 Moe algeba that we will not discuss But we can demonstate that this solution woks. It is not difficult to do, but it does take a lot of witing. Fist, the deivative: = d dη = a 1 ε( 1 + ε cos η a 1 ε( d 1 + ε cos η ( dη a 1 ε( = ε sin η 1 + ε cos η ( = ( ε sin η a 1 ε ( cos η = a 1 ε( ε 1 + ε cos η 28 Septembe 2017 ASTRONOMY 111 FALL 2017 22 11
Algeba Then the squae: d dη ( ε ( = a ( 1 ε ( ( sin η ( ε ( = a ( 1 ε ( ( 1 cos η ( ε ( = a ( 1 ε ( ( 1 a 1 ε( ( ε ( ( = a ( 1 ε ( ( 1 + 2a a( 1 ε ( ( 28 Septembe 2017 ASTRONOMY 111 FALL 2017 23 Algeba E m = = d dη ( h ( 2 + h( 2 ( GM h ( 2a ( 1 ε ( 1 + 2a a( 1 ε ( ( + h( 2 ( GM h ( 0 = 2a ( 1 ε ( E m + h ( a 1 ε ( GM 1 + h( 2 h( 2 This will be tue fo all if and only if the squae-backet tems ae all zeo. The last one we get fo fee. The othe two detemine a and ε in tems of M and h: h ( a 1 ε ( = GM and h ( 2a ( 1 ε ( = E m 1 ( 28 Septembe 2017 ASTRONOMY 111 FALL 2017 24 12
Algeba Divide the fist by the second: Substitute back: h ( 2a ( 1 ε ( a 1 ε ( h ( GMm 2E = 2a = GMm E a = GMm 2E 1 ε ( h ( = 1 GM ε = 1 + 2Eh( G ( M ( m 28 Septembe 2017 ASTRONOMY 111 FALL 2017 25 Elliptical obits Example 3 A body of mass m is in an elliptical obit with eccenticity ε. What is the atio of obital speeds at peihelion and aphelion? Consult the popeties of ellipses: ˆ = a 1 ε Use consevation of angula momentum: mˆvˆ = m v = a 1 + ε ma 1 ε vˆ = ma 1 + ε v vˆ = 1 + ε v 1 ε 28 Septembe 2017 ASTRONOMY 111 FALL 2017 26 13
Elliptical obits Example 4 What is the speed of this object at a point fo which its distance fom the lage mass is? Use consevation of enegy: E = 1 2 mv( GMm = GMm 2a v = GM 2 1 a This useful esult is often called the vis-viva equation. 28 Septembe 2017 ASTRONOMY 111 FALL 2017 27 Keple s Laws Befoe Newton discoveed the laws of motion and gavity, Johannes Keple noticed the following facts about planetay obits, which had been measued with geat accuacy by his mento Tycho Bahe: Each planet follows and elliptical obit with the Sun at one focus. The line between the Sun and each planet sweeps out equal aeas in equal times, wheeve the planet is in its obit. The squae of a planet s obital peiod is popotional to the cube of its obital semimajo axis. All of these empiical facts can be deived fom Newton s laws; that is, Newton s theoies ae validated by Tycho s obsevations, as Newton himself noted. 28 Septembe 2017 ASTRONOMY 111 FALL 2017 28 14
Keple s fist and second laws Keple s fist law: we just showed, above, that elliptical obits ae consistent with gavity and the laws of motion, and the elationship between paametes of the ellipse and the conseved mechanical quantities. Keple s second law (equal aeas): da h = d h h dη [ da = dη h d h = ( Œ 2 dη da dt = ( dη 2 dt = 1 2 v = h 2 = 1 2 GMa 1 ε( da dh da 28 Septembe 2017 ASTRONOMY 111 FALL 2017 29 Keple s thid law (peiod and semimajo axis) Integate ove one obital peiod: da Œ da dt = h 2 Ž = h 2 dt Œ A = h P = πab 2 P ( = 4π( a ( b ( h ( = 4π( a 1 ε ( GMa 1 ε ( P ( = 4π( GM a@ 28 Septembe 2017 ASTRONOMY 111 FALL 2017 30 15
Tycho & Keple, Newton & Copenicus The Keple/Tycho esults ae consistent with the pedictions of Newtonian dynamics and can be egaded as the fist expeimental (pe-)validation of Newton s theoies. They also supply a cucial missing piece of the Copenican model of the Sola System: Copenicus hypothesized cicula obits fo the planets. Measuements like Tycho s wee accuate enough to ule out cicula heliocentic obits. The Ptolemaic geocentic theoy, with ~15 epicycles pe planet, was thus in much bette ageement with the obsevations. Copenicus had only one fee paamete pe planet: the obital adius. A theoy with moe fee paametes can always fit expeimental esults bette than one with fewe fee paametes. Since neithe theoy had an explanation fo why anything was obiting anything else, a contempoay of Copenicus could maintain not just asset that the Ptolemaic theoy was bette, as it gave bette ageement with obsevations. 28 Septembe 2017 ASTRONOMY 111 FALL 2017 31 16