(1) Te motion is planar Notes on Planetary Motion Use 3-dimensional coordinates wit te sun at te origin. Since F = ma and te gravitational pull is in towards te sun, te acceleration A is parallel to te position vector R. Now d(r V ) = dr V + R dv = V V + R A. But V V = 0 and R A = 0, since R and A are parallel vectors. Tus d(r V )/ = 0, so R V is a constant, say C. Since R (R V ), we see tat te position vector R is in te plane containing te origin O and perpendicular to C. Note tat we only used te fact tat F is parallel to R ere (a central force field). We did not use te magnitude of F. (2) Polar coordinates in te plane Assume tat te motion is in te (x, y)-plane and use polar coordinates (r, θ). Tus, bot r and θ are functions of t. Define te unit vectors U r = i cos θ + j sin θ, U θ = i sin θ + j cos θ. Note tat U r U θ, du r /dθ = U θ and du θ /dθ = U r, so du r = dθ U θ, du θ = dθ U r. Since x = r cos θ and y = r sin θ, we ave R = x i + y j = ru r and ence V = dr = d(ru r) = dr U r + r du r = dr U r + r dθ U θ. Differentiating again, we get A = dv = [ d 2 ] r 2 r( dθ ) 2 U r + Observe tat te U θ component of A is equal to r d2 θ 2 + 2dr dθ = 1 r 1 d [ r d2 θ 2 + 2dr ( r 2 dθ ). ] dθ U θ.
(3) Central force fields and Kepler s second law Assume tat te force F is directed in or out from te origin O. Tis is a central force field. As in part (1), we know tat te motion is planar. Use polar coordinates. Since F = ma, te acceleration A is in te same direction as R and ence te same direction as U r. Tus te U θ -component of A is zero. Tis yields ( d r 2 dθ ) = 0, so r 2 (dθ/) =, a constant. Recall tat te area A(t) swept out by te moving position vector R is so A(t) = t 0 1 2 r2 dθ, da = 1 dθ r2 2 = 1 2 and A = (t)/2. Tis is Kepler s second law of motion wic says Te radius vector from te sun to a planet sweeps out area at a constant rate. (4) Initial conditions Assume tat R 0, te initial position vector of a planet (at time t = 0) is in te positive x-direction, so tat θ = 0, and tat it represents te closest approac of te planet to te sun. Ten dr/ = 0 at tis point, so V = (dr/)u r + r(dθ/)u θ implies tat V 0 is in te U θ direction. But U r = i and U θ = j wen θ = 0, so V 0 = v 0 j, were v 0 is te initial speed. Since v 0 = r(dθ/) at t = 0, we get r 0 v 0 = r 2 (dθ/) =. (5) Conic sections and Kepler s first law We now use Newton s law of gravitational attraction, namely F = (GMm/r 2 )U r, were M is te mass of te sun, m is te mass of te planet, and G is te universal gravitational constant. Since F = ma, we get A = (µ/r 2 )U r, were µ = GM. Furtermore, r 2 (dθ/) =, so dv = A = µ r 2 U r = µ dθ U r = µ du θ. 2
Taking antiderivatives of bot sides, we get V = (µ/)u θ + C, were C is a constant. Now at t = 0, we ave V = v 0 j and U θ = j, so C = (v 0 µ/)j and V = (µ/)u θ + (v 0 µ/)j. Next, we take te dot product of tis expression wit U θ using te facts tat V U θ = r(dθ/) (for a general V ), j U θ = cos θ and U θ U θ = 1. We get r dθ = µ (v + 0 µ ) cos θ. But r(dθ/) = /r, so and solving for r yields r = µ (v + 0 µ ) cos θ r = Since r 0 v 0 = and µ = GM, we get 2 /µ 1 + [(v 0 /µ) 1] cos θ. r = 2 /µ 1 + e cos θ were e = (r 0 v0 2 GM)/(GM) and 2 /µ > 0. We recognize te above as te equation of a conic section wit one focus at te origin and wit eccentricity equal to e. Note tat e 0 since r is minimal wen θ = 0, and ence r 0 v0 2 GM. Tis is Kepler s first law, namely Te orbit of eac planet is an ellipse wit te sun at one focus. (6) Comets and comments For more general objects like comets, tere are additional options. If r 0 v 2 0 = GM, we get e = 0 and te orbit is a circle. If r 0 v 2 0 < 2GM, ten e < 1 and te orbit is an ellipse. If r 0 v 2 0 = 2GM, ten e = 1 and we ave a parabola. Finally, if r 0 v 2 0 > 2GM, ten te orbit is a yperbola. Do tere exist comets wit circular, yperbolic or parabolic orbits? 3
Remember tat Isaac Newton invented calculus, introduced te basic laws of motion and gravitational force, and ten used all of tis macinery to derive Kepler s laws. (7) Periods of revolution Kepler s tird law relates te period (year) of a planet to te lengt of its semimajor axis. To understand tis, recall tat if x 2 a 2 + y2 b 2 = 1 is an ellipse wit semimajor axis a (so tat a b) and wit eccentricity e, ten we know tat te area of te ellipse is equal to πab and tat b 2 = a 2 c 2 = a 2 (ae) 2 = a 2 (1 e 2 ). Now suppose tat te planet of interest as te above equation if te axes are suitably sifted. If τ is te period (year) of te planet, ten te area formulas yield πab = τ/2, so = 2πab/τ and 2 = 4π2 a 2 b 2 τ 2 = 4π2 a 4 (1 e 2 ) τ 2. Also, using te polar equation for r, we know tat 2a is equal to r at θ = 0 plus r at θ = π. Tus so 2a = 2 /µ 1 + e + 2 /µ 1 e = 22 /µ 1 e 2, 2 = aµ(1 e 2 ). Setting te two formulas for 2 equal to eac oter, canceling a factor of a and 1 e 2, we get GM = µ = 4π 2 a 3 /τ 2, so τ 2 a 3 = 4π2 GM = γ. Note tat te rigt and term γ depends only on te mass M of te sun. Tus it is te same for all planets and we obtain Kepler s tird law, namely 4
Te square of te period of revolution of a planet is proportional to te cube of te semimajor axis of its orbit. Exercises 1. A particle moves in 2-space wit polar coordinates given by r = 1 + sin θ, θ = 2t Describe R, V and A in terms of t, U r and U θ. 2. A particle moves troug 2-space under te action of a central force field (centered at te origin). If θ = e 2t and r(0) = 5, find r as a function of t and ten as a function of θ. 5