Math Notes on Kepler s first law 1. r(t) kp(t)

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1 Math 7 - Notes on Keple s fist law Planetay motion and Keple s Laws We conside the motion of a single planet about the sun; fo simplicity, we assign coodinates in R 3 so that the position of the sun is fixed at the oigin. Let pt) denote the position of the planet at time t, let v = p be its velocity, a = v = p its acceleation, and = p its distance fom the sun. Keple s Fist Law states that the obit of the planet is an ellipse with the sun as a focus of the ellipse; Keple s Second Law states that the aea of an elliptical secto swept out by the obiting planet depends only on the length of time the planet takes to sweep it out i.e., it is independent of whee the planet is along its obit). Newton s invese squae law fo the foce of gavity states that the gavitational attaction acting on the planet has the fom F = mmg pt) t) t) whee m and M ae the masses of the planet and the sun, and g is the gavitational constant; note that the fist faction is the magnitude of the foce, the second faction is a unit vecto in the diection of the planet, and the minus sign indicates that the foce is diected towad the sun. Using anothe of Newton s laws, a = F/m, so that by using the fomula above fo F we obtain a = Mg t) pt) t) kpt) t) 3 ) whee k = Mg is a stictly positive constant. Lemma 0. pt) at) = 0 fo all t Poof: By equation ) a is paallel to p, so thei coss poduct is 0. Lemma 0. pt) vt) is a constant vecto that we will denote by J. This can be intepeted as consevation of angula momentum, but we won t get into this hee.) Poof: By the poduct ule, d dt p v) = p v + p v = v v + p a = = 0. Since J is the coss poduct of p with anothe vecto, J and p ae pependicula. This means that if J = 0 then the position of the planet is always in the plane though the oigin that is pependicula to J. On the othe hand, if J = 0 then lemma 0. means that p and v ae paallel. in this case we know fom equation ) that p and a ae paallel, so we conclude that a and v ae paallel, meaning that the nomal component of acceleation is 0. This means that the path of the planet lies along a staight line. Why would this not be a good situation?) Fom now on we will assume that J = 0, so that the obit of the planet lies in a plane though the oigin. By choosing coodinates appopiately, we can assume this is the plane z = 0 in these coodinates J is vetical). Discoveing potential enegy One of Newton s basic physical insights was the idea of inetia, o moe geneally, the idea that some aspects of a system ae peseved unless thee is a net gain o loss of enegy in the system. Most of us ae familia with the idea of kinetic enegy fom high school physics as a quantity popotional to the squae of the speed. Ou model of the obit of a planet should have a total enegy H that is conseved, H should be the sum of kinetic enegy vt) and a potential enegy ft). We expect that H should be constant, so that It follows that 0 = d dt H = d dt vt) + ft)) = v a + f t). f = v a = v a) = v kp/ 3 ) = k v p) 3 = k 3 p p) = k ) 3 ) = k k = k = so if we set the potential enegy to ft) = k/, then we have the following esult. )

2 Math 7 - Notes on Keple s fist law Theoem 0.3 Consevation of enegy) The quantity Hv, p) = vt) k t) is constant. The fist tem epesents the kinetic enegy with the constant m/ factoed out), similaly, the second epesents the potential enegy. These ae not expessed in physical units.) Constants of motion To finish ou analysis of the motion of the planet, we need anothe constant of motion. Since we ae assuming that J = p v = 0, p and v ae not paallel, and so they span the plane of motion the plane pependicula to J the plane z = 0). The vecto J a is pependicula to J, so we can wite J a = αp + βv fo some scala functions of t, α and β. Since J a is pependicula to a which is paallel to p, we see that 0 = p J a) = αp p + βp v so that α = β p v p p and we can wite J a = β v p v ) p p v. ) Now we need to identify β; to do this, note that J a = J a because a and J ae pependicula), so that J a = β v p v ) p p v. 3) The tem in the paentheses is analogous to the nomal vecto) component of acceleation; it is actually the component velocity that is pependicula to position. You may ecall the calculation that we did ealie showing that the length of the nomal component of acceleation is the length of v a divided by v ; the analogous fomula hee is that v p v p p v = p v p = p v You can check this by diffeentiating p/.) Combined with 3) and the assumption that J is nonzeo, this gives This implies βt) = ± a = ± kp 3 = ± k. We will show below that β < 0, so we have Now etun to ). Since J is constant, it can be ephased as J v) = β v p v ) p p v = k = J. a = βt) /. 4) βt) = k. 5) v p v ) p p v. 6) Continuing the analogy between the tem in paenthesis and the nomal component of acceleation, note that ) d pt) = v p v ) dt t) p p p

3 Math 7 - Notes on Keple s fist law 3 you can check this by diffeentiating). Combining this with 6) shows that J v) = k ) pt). t) Note that both sides ae deivatives; since the diffeence in these two deivatives is 0, the diffeence in the two quantities without the deivatives) is constant, and we have found anothe constant of motion. This is summaized in the fist pat of the following the second pat is obvious). Poposition 0.4 The vecto Et) = v J k p the plane of motion. is a constant in t. E is pependicula to J, so when J = 0, E lies in Actually, we have not quite completed the poof of the poposition; it emains to veify that β < 0. Lemma 0.5 The function βt) defined above is always negative. Poof: Let v i = v p v p p p so that by ) J a = βv i. It follows that J a) v i = β v i and so β has the same sign as J a) v i = J a) v J a) p ) p v = J a) v p p the second tem is zeo because a and p ae paallel, so that J a is pependicula to p). Since J = p v, the tiple p, v, J) has the standad oientation, and since a points in the opposite diection to p, we see that a, v, J) has the nonstandad oientation; two paiwise switches take us fom a, v, J) to J, a, v), so this last tiple also has the nonstandad oientation. This means that the tiple poduct J a) v is negative, so β < 0 as well. Lemma 0.6 pt) E + kt) = J J Poof: Fom the definition of E poposition 0.4) p E = p v J kp ) = p v) J k p p = p v) J k = J J k whee in the last step we have used the defintion of J lemma 0.). Now we ae eady to finish. Thee ae two cases, depending upon whethe o not E = 0. If E = 0 the last lemma says that t) = J J)/k; i.e., t) is constant. This means that the planet stays at a constant distance fom the sun, so its obit is cicula. In this situation the velocity vecto is tangent to the sphee containing the obit of the planet, which means that v is pependicula to a ecall that a is paallel to p). Hence the tangential component of acceleation is 0, which, as we saw ealie, means that speed is constant. Hee is anothe veification that speed is constant: Howeve which is 0 because t) is constant. d v v = v a = v dt ) kp 3 v p = d dt p p) = d dt = k v p 3.

4 Math 7 - Notes on Keple s fist law 4 Now assume E = 0; choose coodinates in R 3 so that E = ε, 0, 0) with ε > 0 and J is vetical this means that the plane of motion is the x-y plane and that the positive x-axis is in the diection of E; ecall fom poposition 0.4 that E is in the plane of motion. Let t), θt), 0) denote the position of the planet at time t as given in pola coodinates. The choice of coodinate system means that θt) is the angle between pt) and E, so we have Combining this with lemma 0.6 gives E pt) = E pt) cosθt)) = ε cosθt)). ε cosθt)) = E pt) = J J k, [ε cosθ) + k] = J J. This gives an equation fo the path of the planet in pola coodinates: = J J k + ε cosθ). Pola equations fo conics The equation above fo the obit of ou planet can be ewitten in the fom = a/ + b cosθ)), whee a and b ae positive constants. To identify this cuve we convet it to ectangula coodinates. Rewite it as If 0 < b < this is an ellipse. If b = this is a paabola. If b > this is a hypebola. + b cosθ) = a = a b cosθ) = a bx x + y = = a bx) = b x abx + a x b ) + abx + y = a. The case b = 0 gives a cicle; this is the E = 0 case that we took cae of ealie. ) Hee ae some pictues that wee geneated by the following Mathematica commands: [a_, b_, t_] = a/ + b Cos[t]); plotconic[a_, b]:= PolaPlot[[a, b,t], {t, 0, Pi}]

5 Math 7 - Notes on Keple s fist law Gaph of plotconic[3,.] -3 Gaph of plotconic[3,.5] Gaph of plotconic[3,.8] -4 Gaph of plotconic[3,.9] Gaph of plotconic[., ] escaled) Gaph of plotconic[3,.] Finally, we show that if the obit of the planet is one of these conics then the velocity is neve 0; this means eithe that the planet goes aound an elliptical obit in one diection, o else it moves in one diection along a paabolic o hypebolic path. In the latte cases the planet would eventually move off towads infinity ; since this doesn t happen fo actual planets, we can conclude that the obit of a planet like eath is oughly elliptical we say oughly because ou analysis above ignoes, among othe things, the influence of one planet upon anothe). Suppose that at time t 0 velocity was 0: vt 0 ) = 0. Since J = p v is constant, it follows that J = 0 and so as noted ealie) the path of the planet lies on a staight line, not one of ou conic sections. In fact, moe can be said. Since the path lies on a line, by a suitable choice of coodinates we can identify the vecto pt) with its magnitude t); the govening equation ) then simplifies to = k/ < 0. Since acceleation is always negative and velocity is 0 at t 0 we see that velocity will be negative fo t > t 0, meaning that the planet is appoaching the sun on a staight line path, with eve-inceasing velocity. Boom.)