Kepler s Laws JWR October 13, 1 Kepler s first law: sun at one focus. A planet moves in a plane in an ellipse wit te Kepler s second law: Te position vector from te sun to a planet sweeps out area at a constant rate. Kepler s tird law: Te square of te period of a planet is proportional to te cube of its mean distance from te sun. (Te mean distance is te average of te closest distance and te furtest distance. Te period is te time required to go once around te sun.) 1. Let (x, y, z) be te position of a planet in space were x, y, and z are all function of time t. Assume te sun is at te origin (,,). We define te position vector r, te velocity vector v, and te acceleration vector a by r = xi + yj + z, Newton s law of motion is v = dx i + dy j + dz, F = ma a = d x i + d y j + d z. were F is te force on te planet and m is te mass of te planet. Newton s inverse square law of gravity is F = GMm r r 3 Tese notes are adapted from: Milnor, J: On te geometry of te Kepler problem. Amer. Mat. Montly 9 (1983) 353-365. 1
were G is a universal gravitational constant and M is te mass of te sun. (Te inverse square law is so called because te magnitude F = GMm r of te force F is proportional to te reciprocal of te square of te distance r from te planet to te sun.) Newton s laws imply a = d r = GMr r 3. (1) Note tat m cancelled. Tis means tat te mass of te planet does not affect its motion. (We are assuming tat te sun is motionless. In more advanced treatments, tis assumption is not made.). First we sow tat te planet moves in a plane. By te product rule for differentiation d (r v) = v v + r a () By (1) and te fact tat te cross product of parallel vectors is te rigt and side of () is. It follows tat te vector = r v is constant. We conclude tat bot te position and velocity vector lie in a plane normal to. Coose coordinates so tat tis plane is te xy-plane. Ten = for some constant scaler. (We ave assumed ; in case = it can be sown tat te planet moves on a straigt line.) 3. Since te planet moves in te xy plane we ave r = xi + yj = r cos(θ)i + r sin(θ)j (3) were te polar coordinates r and θ are functions of t. Te derivative of r is [ ] [ ] dr dr v = cos(θ) r sin(θ)dθ i + sin(θ) + r cos(θ)dθ j (4) From r v = we get [ ] dr r cos(θ) sin(θ) + r cos(θ)dθ [ ] dr r sin(θ) cos(θ) r sin(θ)dθ =.
After multiplying out and simplifying tis reduces to r dθ = (5) Te area A swept out from time t to time t 1 by a curve in polar coordinates is A = 1 t1 r dθ so A = (t 1 t )/ by (5). Tis is Kepler s second law. Teorem 4 (Hamilton). Te velocity vector v moves on a circle. Proof. Since r = r equation (1) can be written dv = GM r Divide (6) by (5) and use te cain rule: Now integrate to obtain dv dθ = GM v = GM t [ cos(θ)i + sin(θ)j [ ] cos(θ)i + sin(θ)j. ]. (6) [ ] sin(θ)i + cos(θ)j + c (7) were c is te constant of integration. Hence v c = GM/ so v moves on a circle centered at c wit radius GM/. 5. Now we can prove Kepler s first law. Coose coordinates (in te xy plane) so tat c is parallel to j (and in te same direction) and let e = (c j)/gm so tat (7) taes te form v = GM [ ] sin(θ)i + (cos(θ) + e)j. (8) Te cross product of (3) and (8) is = = r v = GMr (1 + e cos(θ)) so r = (9) 1 + e cos θ were = /GM. Equation (9) is te polar equation of a conic section wit eccentricity e. 3
6. Assume tat te conic section (9) is an ellipse, i.e. tat e < 1. Te ellipse (9) as one focus at te origin and te oter on te negative x-axis so te closest and fartest te planet comes to te sun are given by r min = r = θ= 1 + e, r max = r = θ=π 1 e. Tese are te values of x = r cos(θ) were te i component dx/ of v vanises, i.e. by equation (7) were sin(θ) =. Te quantity a = r min + r max = 1 e (1) is te major semi-axis of te ellipse. Te minor semi-axis b can be found by maximizing y on te orbit. Tis maximum value of y = r sin(θ) occurs wen dy/ =, i.e. wen te j component of v vanises, i.e. by equation (7) wen cos(θ) + e =. Tus b = y θ=cos 1 ( e) = sin(cos 1 ( e)) 1 + e cos(cos 1 ( e)) = 1 e = 1 e Te equation for te ellipse on rectangular coordinates is (x + ea) a 1 e. (11) + y = 1. (1) b 7. We now prove Kepler s tird law in te form ( ) 3 T = 4π rmin + r max (13) GM were T is te period of te planet, i.e. te time it taes te planet to go around te sun one time. Te area of te ellipse is πab so by te second law we get πab = 1 π r dθ = 1 T r dθ = 1 T = T (14) Tus ( ) πab T = = 4π 4 (1 e ) 3 = 4π a 3 = 4π a 3 GM. (15) Te first equality in (15) comes from (14), te second from (1) and (11), te tird from (1), and te fourt from te definition = /GM of given after equation (9). Equation (13) results by substituting (1) in (15). 4
8. Historical Remar. Kepler publised is laws in 169. Newton s Principia was publised in 1687. Te proof of Hamilton was publised in 1846. Exercise 15 below ass you to prove Newton s inverse square law assuming (a) Kepler s first law (9) and (b) tat te force on te planet is directed towards te sun. Peraps tis is ow Newton discovered te inverse square law of gravitation. 9. In te special case were te orbit of te planet is a circle Kepler s tird law is muc easier to prove. Sow tat T = (4π /GM)a 3 under te assumption tat d r = GMr, r = a cos(ωt)i + a sin(ωt)j a 3 were a and ω are constants. 1. Te proof of Kepler s second law did not use te full force of (1). Prove Kepler s second law under te ypotesis tat m d r = gr (16) were g = g(x, y, z) is any function. Te proof in te text is te special case g = GMm r 3. 11. Te planet eart is 93 million miles from te sun and orbits te sun in one year. Te planet Pluto taes 48 years to orbit te sun. How far is Pluto from te sun? 1. Halley s comet goes once around te sun every 77 years. Its closest approac is 53 million miles. Wat is its furtest distance from te sun? Wat is te maximum speed of te comet and wat is te minimum speed? 13. Calling te quantity a defined in equation (1) te mean distance is a misnomer. Sow tat 1 T T r = a(1 + e /). Hint: By equation (1) te ellipse as parametric equations x = ea + a cos φ, y = b sin φ. Express te integral first in terms of θ and ten in terms of φ. 5
14. Sow tat 1 T r 1 = a 1. T 15. Assume tat te motion of a particle satisfies Kepler s first law and tat te force is directed toward te origin; i.e. assume Equations (9) and (16). Sow tat f = cr (along te orbit) were c is a constant. (See Exercise 7 on page 566 of Tomas & Finney fift edition.) 16. Te quantity W = GMm r is called te potential energy of te planet, te quantity K = m v is called te inetic energy, and te quantity is called te energy. Sow tat E = K + W = m v E = m mgm r ( ) GM (e 1). Conclude tat E is constant along solutions of (1) and tat te orbit (9) is an ellipse, a parabola, or a yperbola according as te energy E is negative, zero, or positive. Hint: Use (7). (9), and te definition of. Also sow tat E is negative, zero, or positive according as te origin lies inside, on, or outside te velocity circle of Teorem 4. 17. Sow tat te force F = GMm r r 3 in te Kepler problem is te negative gradient F = grad W of te potential energy. Conclude tat de = along any solution of (1). Tis confirms te result of Exercise 16. (Do tis exercise after you ave learned partial differentiation.) 6
Answers to te Exercises 18. Answer to Exercise 9. Te formula r = a cos(ωt)i + a sin(ωt)j says tat te planet moves on a circle of radius a wit angular velocity ω. Differentiating twice gives tat te acceleration vector is a = ω r. From Newton s law te acceleration vector is a = GM r 3 r = GMa 3 r so ω = GMa 3. Te period is te time required to go around once so ωt = π and ence ω = 4π /T. Equate te two formulas for ω to get T = GMa 3 /4π wic is Kepler s tird law. 19. Answer to Exercise 1. Part I. Suppose tat ma = gr; we sow tat r lies in a plane. By te product rule d dr (r v) = v + r dv = v v + r a = v v + g m r r = as te cross product of a vector wit itself is. Hence te vector (r v) is constant; we coose coordinates so tat it is a multiple of say r v =. Since r r v we conclude tat (if ) r, i.e. tat r lies in te (x, y)-plane.. Answer to Exercise 1. Part II. Suppose tat r v = were is constant so tat r lies in te (x, y)-plane. We sow tat te radius vector r sweeps out area at a constant rate. Te radius vector r is given in polar coordinates by te formula r = xi + yj = r cos θi + r sin θj = r(cos θi + sin θj) were x, y, r, θ depend on t. Tus by te product rule for differentiation v = dr dθ (cos θi + sin θj) + r ( sin θi + cos θj) Te first term on te left is proportional to r and te second is perpendicular to r so = r v = r(cos θi + sin θj) r dθ dθ ( sin θi + cos θj) = r. so = r dθ. Te rate at wic te planet sweeps out area is i.e. da/ is constant. da = r dθ =, 7
1. Answer to Exercise 11. Te period of te eart is T 1 = one year and its mean distance to te sun is a 1 = 93 1 6 miles. Te period of Pluto is T = 48 years; we are to find its mean distance a to te sun. Using Kepler s tird law T1 = 4π a 3 1 /(GM) and te data for te eart we can find te constant GM: GM = 4π a 3 1 T 1 = 4π 93 3 1 18 (17) so using Kepler s law wit te data for Pluto determines its mean distance ( ) T 1/ a = GM 4π = 48 /3 93 1 6 miles.. Answer to Exercise 1. As in te previous exercise te period of Halley s comet is T = 77 years; let a be its mean distance to te sun. Ten 77 = T = 4π a 3 GM = a 3 93 3 1 18 so a 3 = 77 93 3 1 18 so a = 77 /3 93 1 6 miles. Now a = (r min + r max )/ so r max = a r min = ( 77 /3 93 53) 1 6. By Equation (7) te speed is v = (GM/) sin θ + (cos(θ) + e) = (GM/) 1 + e cos θ + e (18) and is largest wen θ = and smallest wen θ = π. From te values r min and r max we can determine te constants e and in Equation (9); we determined GM in Equation (17), and we can determine from te definition = /GM of wic appears after Equation (9). Specifically, a = r min + ea so e = 1 (r min /a) and = a(1 e ) so te largest and smallest values of v are GM 1 ± e + e = GM(1 ± e) were GM = 4π 93 3 1 18, e = 1 (53/(77 /3 93)), and = 77 /3 93 1 6 (1 e ). 3. Answer to Exercises 13 and 14. Exercise 13 ass us to evaluate te integral I n = 1 T T r n for n = 1; Exercise 14 ass us to evaluate tis integral for n = 1. Te planet lies on te ellipse r = 1 + e cos θ 8
were = /GM and is te constant in Kepler s second law r dθ/ =. Tus /dθ = r / so I n = 1 π r n+ dθ. (19) T Te semimajor axis a = (r min + r max )/ of te ellipse, te semiminor axis b, and te distance c between te foci are given by a = 1 e, b = a c, c = ea. In rectangular coordinates x = r cos θ and y = r sin θ te equation of te ellipse is (x + c) a so te ellipse as parametric equations + y b = 1 x = c + a cos φ, y = b sin φ, φ π. We calculate dθ/dφ. Differentiate te equation to get c + a cos φ = x = r cos θ = cos θ 1 + e cos θ a sin φ dφ dθ = sin θ (1 + e cos θ) = yr. substitute y = b sin φ and divide by a sin φ to get and ence from Equation (19) we get I n = 1 T π dφ dθ = br a n+ dθ a r dφ = dφ T b π r n+1 dφ. () We express te constant a/t b in terms of a and e using te formulas a = 1 e, T = πa3/ GM, = GM. Ten = a(1 e ) and = GM = a(1 e )GM so a T b = a a(1 1 GM e ) a(1 e )GM πa 3/ 1 a 1 e = 1 πa 9
so Equation () simplifies to Taing n = 1 in Equation (1) gives I n = 1 π r n+1 dφ. (1) πa I 1 = 1 π dφ = 1 πa a wic completes Exercise 14. For n = 1 we use r = x + y, a = b + c, and c = ea so from te parametric equations for te ellipse and te alf angle formula we get r = c ac cos φ + a cos φ + b sin φ = a (1 e cos φ + e cos φ) ( ) = a 1 e cos φ + e (1 cos φ). Now π cos φ dφ = π cos φ dφ = so by Equation (1) we ave Tis completes Exercise 13. I 1 = a (1 + e 4. Answer to Exercise 15. We assume tat te force is directed toward te origin ). ma = m d r and tat te planet moves in an ellipse = gr (16) r = 1 + e cos θ (9) We are to prove tat ma is proportional to 1/r. We introduce te unit vector u = cos θi + sin θj in te direction r. Ten r = ru and te unit vector du dθ = sin θi + cos θj 1
is perpendicular to u, and a calculation (see Tomas Finney) sows tat ( d ( ) ) r dθ ( d ) a = r θ u + + dr dθ du dθ From Equation (16) we get d r r ( ) dθ = gr () d θ + dr dθ =. (3) As is sown in Tomas Finney, Equation (3) leads to anoter proof of Kepler s second law r dθ/ =. Differentiate Equation (9) and use (9) and Kepler s second law to get dr = and ence e sin θ (1 + e cos θ) dθ = dθ e sin θ (1 + e cos θ) By Kepler s second law again = r dθ e sin θ = e sin θ d r e cos θ = dθ. (4) r ( ) dθ = r r ( dθ ) = r dθ = (1 + e cos θ) dθ (5) so subtracting (5) from (4) gives d r r ( dθ ) = dθ = r. (6) By () and (3) te left and side of (6) is a so a is inversely proportional to r as claimed. 5. Answer to Exercise 16. By Equation (8) K = m v = m ( GM ) (sin θ + (cos θ + e) ) = m ( ) GM (1 + e cos θ + e ). Recall tat = /GM [see te definition after Equation (9)] so by te definition of W and and Equation (9) we ave W = mgm r ( mgm(1 + e cos θ) GM = = m ) (1 + e cos θ) 11
so te total energy ( ) GM (e 1) E = K + W = m is constant along te orbit. Wen te orbit is an ellipse (i.e. e < 1) te energy is negative and taing θ =, π/, π, 3π/ in Equation (8) sows tat v passes troug all four quadrants so te origin lies inside te velocity circle. 6. Answer to Exercise 17. Were r = r = x + y + z we ave so by te cain rule ( ) mgm W = r so de = dk + dw r = r x i + r y j + r z = r r = mgm r r = mgm r 3 r = F = ma = mv a + W v = mv a ma v =. Possible Examination Questions 7. A particle moves in space according to a central force law m d r/ = gr were g = g(x, y, z) is any function and r = xi + yj + z is te position vector of te particle. Prove tat te particle stays in a plane. Answer: Write wat is in section 19 above. 8. Suppose tat te position vector r = xi + yj of a particle in te (x, y)-plane satisfies r v = were is constant and v is te velocity vector. Sow tat te radius vector r sweeps out area at a constant rate Answer: Write wat is in section above. 9. A planet moves in space according to Newton s law m d r/ = mgm r 3 r were r = xi + yj + z is te position vector of te planet. Let v be te velocity vector, K = m v /, and W = mgm/ r. Sow tat te quantity E = K + W is constant. Answer: Write wat is in section 6 above. 1