Question

just 18.3

In other words, the center of mass moves as a free particle (no external force) of mass m. The solution for R corre- sponds to uniform straight-line motion, and eliminates three of the six independent variables in the original equations of motion (three components each of, and r. Let us take, as the remaining three independent variables, the three components of r -. To ma- nipulate the original dynamical equations into a single vector equation for r. divide out m, in the first equation and my in the second. Subtract the results, and derive Newton showed that Kepler's laws are approximate to the extent that the planets and the Sun can be idealized as perfect spheres and that the mass of the Sun is much greater than that of any of the planets. Today, we further know that Kepler's laws are also approximate to the extent that relativistic and quantum-mechanical effects are negligible. All these approximations are valid in the actual solar system (some more than others), and our purposes here will be adequately served if we merely accept Newton's explanations for Kepler's laws (sce Problems 18.3 and 18.4). Indeed, the story of Kepler and Newton's discoveries provided such a cogent demon- stration of the explanatory power of the analytical method that it set the example par excellence for all subsequent scientific endeavors. After almost four cen- turies (Kepler's three laws were formulated in 1609- 1618). the formal results may seem ancient but the reasoning used to derive the results is as modern and as full of vitality as any other we have discussed in this book. Like rare wine, the great works of science do not suffer with age. Thus, the motion of body 2 relative to body 1 can be conceived as that of a test particle attracted to a center of gravitation of mass m m , + m2. To obtain the individual motions of bodies 1 and 2 with respect to the center of mass (which is an inertial frame) define the relative positions: R-r-R and R --R. Show by direct substitution that R, and R, can be written R, = r. R, r. m, + m2 Problem 18.31 for those who know vector calculus). lg. nore the perturbations of the other bodies in the solar system and consider only the mutual attraction between the Sun, mass m, at position r, relative to some inertial frame, and a planet, mass m, at position rg. In this approximation, we have a two-body problem, and we wish to show here how to reduce this two-body problem to an equivalent one-body problem in a field of external force. To begin. Newton's laws lead to the following dynamical equations for body I and body 2: Thus, R, and R, describe loci of the same shape as r, but they are smaller by the respective factors m/m, + m2) and m m, + ml. Compare this result with Figure 10.2. d r Problem 18.4. (also for those who know vector calculus). The nontrivial part of Problem 18.3 is the solution of the equivalent one-body problem: dr Gm, mn r . Gm, Pr. N dr dr = -mrr. where r = r-r, is the displacement vector from mass I to mass 2. Since the forces on the right-hand side are equal and opposite, we are motivated to add the above two vector equations. Show that this results in To simplify the problem, let us first derive the conserva- tion of angular momentum and energy. Take the cross product of equation (1) with r. The right-hand side is zero and the left-hand side is r x dr dr. Show that the latter can be expressed, with r = dr/dr, as (,, + m ) - 0. ir x ) = 0 (2) To interpret the above. define the total mass, m= m, + y. and center of mass position. R m , + m ) (m, + myl and show that the above equation may be written Show that equation (2) implies the conservation of angu- lar momentum per unit mass, and consult Figure 3.21 to see why this conservation principle implies Kepler's second law as well as the notion that the orbit takes place in a single plane. IR IN =0.

Answer 1

1 . 2 Gmi m2 p3 gr met - 9 monte Adding both equations, we get () + f (m) 20 min mobil 20

(6) R = mñ + ma Myt my And ma mi + m2 7 mini tom nie 7 min + my game dos (mm) 70 om zo ] 6) Dividing out an, in the first equation we get dão Gromar t2 = non Dividing out my in the 2nd Egn, we obtain Sutarting these two equations part 02

عام قوه و au tlue بوابة + - قابو - غلا و مقام بلا ة م - عام + عية ي Thethue عدة ( و ) The time Twthe الراية - دلي- " أي uthue ا = بلي بعدها با + - asing 3 d 12 - و ؟ a Gm مي [ په ته r »