Kepler s problem gravitational attraction

Transcription

1 Kele s oblem gavitational attaction Summay of fomulas deived fo two-body motion Let the two masses be m and m. The total mass is M = m + m, the educed mass is µ = m m /(m + m ). The gavitational otential is U() = Gm m The effective otential is then: = α, whee α = Gm m > 0 U eff () = l l + U() = µ µ α The conseved angula momentum l = µ φ is equal to its initial value l = [ 0 µ v 0 ] e z, whee by definition 0 = (0) (0) and v 0 = v (0) v (0). In geneal l 0. (The case l = 0 is athe tivial: l = 0 φ = 0 i.e. φ = const, and theefoe the two bodies move eithe towads one anothe, o away fom one anothe, along the line uniting them. This haens if 0 and v 0 ae aallel. I ll obably give this to you fo homewok, o we ll discuss it in the tutoial). Fo what follows, I assume l 0. U eff () min max U 0 E el Fig. Effective otential fo l 0, and tuning oints fo a value E el < 0. The effective otential eaches its minimum value at = given by whee it takes the minimum value (see Fig. ) du eff d = = 0 = l µα () U 0 = U eff () U 0 = µα l = α To find out the allowed ange fo the elative distance, we need to find the elative enegy E el = µ v 0/ + U( 0 ) fom the initial conditions. We know that at any time E el = µṙ + U eff() U eff () () ()

2 since µṙ / 0. This means that motion is only ossible if E el > U 0, since that s the minimum value of U eff. It is vey convenient to intoduce the eccenticity e defined by e = + E el E el = (e )U 0 (4) U 0 It follows that if 0 e < U 0 E el < 0; if e = E el = 0; and if e > E el > 0. We ae now eady to find the tajectoy of the elative coodinate. The tajectoy The equation fo the tajectoy (obtained fom Eq. + the fact that l = µ φ) is ld φ = µ(eel U eff ()) Fo simlicity, hee I assume that the constant of integation is zeo (that can always be aanged by oienting the system of coodinates oely). We efom the integal in the following way: we intoduce the new vaiable Also, following the definitions u = du = d d = du E el U eff () = (e )U 0 l µ + α We now elace = /u = l /(µαu) and U 0 = µα l, and afte some stuggle we find that E el U eff () = µα [ e (u ) ] l We now substitute eveything in the integal to find: du φ = l µ µα [e l (u ) ] = du e (u ) if one uses the fact that = l /(µα). To do the last integal we equest that (u ) = e cos x, such that the condition 0 (u ) e is always obeyed (this is necessay to insue that the quantity unde squae oot is ositive). Then, u = + e cos x du = e sin xdx and e (u ) = e sin x In tems of x, the integal is tivial: e sin xdx φ = = x e sin x Since u = / and u = + e cos x, the equation of the tajectoy is: = + e cos φ (5) Let us analyze this equation fo vaious values of the eccenticity.

3 . e = 0 : the cicle If e = 0, i.e. E el = U 0, the tajectoy is =, i.e. a cicle of adius. This makes efect sense if one looks at Fig., and emembes that at all times we must have E el U eff ().. 0 < e < : the ellise In this case, U 0 < E el < 0, and we exect the tajectoy to be finite (see Fig. ). Let us go back to x = cosφ and y = sin φ (see Fig. ) to undestand what tajectoy is descibed by this equation: = + e cos φ = + ex = ex = x + y = ex + e x We gou tems togethe to get: ( y + ( e ) x + e ) = e e (exand the squae and show that the two exessions ae equivalent). Let us define: a = e ;b = ;x e 0 = ea (6) (note: we can only do this fo e <!!) It follows that the equation of the tajectoy is: (x + x 0 ) a + y b = which is an ellise of lage semi-axis a, small semi-axis b, and shifted by x 0 fom its cente (see Fig. ). Note: if e = 0 a = b =,x 0 = 0, i.e. the ellise becomes a cicle, as it should. y b x 0 ae φ x max min Fig. Ellitic tajectoy fo the elative coodinate. Some inteesting (and useful) elations about these quantities: a = e = U 0 E el E el = α a a

4 (I used Eq. 4 and ). So the lage semiaxis is staightfowad to find fom E el. Anothe way to find the lage semiaxis is the following: fom Fig. we see that the minimum/maximum distance between objects is min = a ae = a( e), max = a + ae = a( + e) (7) and theefoe a = ( min + max )/. The values min and max ae the etun oints, so they ae given by the solutions of the equation E el = U eff (). Anothe way to find them is fom the tajectoy equation: since cosφ e + e /( + e) /( e) min = /( + e); max = /( e). These distances have secial names, namely eihelion ( min ) and ahelion ( max ). Now it would be inteesting to know what each of the two objects is actually doing, when the elative tajectoy is an ellise. Let us assume that the CM is at est (we can fix this by choosing an inetial efeence system in which the CM seed is zeo). Remembe how and R whee linked to and (see Fig. ): = R + m, M = R m. M If the CM is at est, we can take R = 0. We know that otates in time, descibing an ellise. So uting the two togethe, the motion of the bodies must be like in Fig. 4, with each object descibing an ellise, ootional to m, esectively M m. M The CM is in the focus of both ellises, and the objects ae always oosite to one anothe. m R CM Fig. Relation between vaious vectos. m CM Fig 4. The motion of the two objects, when the CM (the focus of both ellises) is at est. If the CM is not at est, just imagine moving the focus unifomly in some diection, as the masses otate on ellises about it. That s a bit too difficult fo me to daw. Finally, you can see what haens if one object is much much heavie than the othe one (fo instance, we have a Sun () and a lanet ()). The ellise descibed by the lanet is given by M S /(M S + m ), while the Sun descibes an ellise given by m /(M S + m ) 0. In othe wods, the CM is basically in the Sun, which is so vey much heavie. Theefoe, the Sun stays in the focus, and the lanet obits aound it at the elative distance. This is the confimation fo: 4

5 Kele s st law: All lanetay obits ae ellises, with the Sun at the focus. We have aleady confimed: Kele s nd law: The aea swet e unit time by the line joining the lanet to its sun, is constant. This is just equivalent with the consevation of the angula momentum, as discussed fo geneal motion in a cental field, whee we showed that da dt = φ = l µ = const. This allows us to find the eiod of the obital motion ight away, since this constant must be the total aea of the ellise πab divided by the total eiod T. Theefoe: πab T = l µα µ µ = µ T = πab (8) α (see Eq. ). Now let s ut all lengths in tems of a. We have (Eq. 6) b = a e ; = a( e ) ab = a So we get the beautiful fomula a T = π α = π GM In othe wods, Kele s d law: The squae of the eiod is ootional to the cube of the lage semiaxis. If you emembe, we showed that this must be the ight elationshi using scaling laws (mechanical similaity). This should convince you that those scaling laws ae vey useful (although they don t tell us what the ootionality constant is!) Let s now look at the othe ossible eccenticities:. e =, e > : the aabola and the hyebola We oceed in the same way as fo the ellise, and find that fo e = we have = + x = ( x) y = x x = ( y ) As you know, this is the equation of a otated aabola (see Fig. 5). The minimum distance is / (when y = 0). Indeed, looking at Fig. we see that this is the tuning oint when E el = 0. Finally, the case e > is teated similaly, and the esulting cuve is called a hyebola. The shotest distance thee (the tuning oint) is min = /( + e), since cos φ. The shae is somewhat like a distoted aabola, so I won t daw anothe figue. Again, if you want to think about the motion of the two objects, just ut the CM in the focus (oigin), daw two tajectoies scaled in ations of m /M and m /M etc. The ight-side ictue in Fig. 5 should hel visualize this when m m. I will ost on the website a link to a PhET simulation that will allow you to look at the tajectoies of both objects, to bette visualize how this eally woks. 5

6 y y / x x Fig 5. The aabola fo e =. The hyebola fo e > looks somewhat simila, excet that the shotest distance is not /, but /( + e). Time deendence of the motion on the ellise To find (t), we need to integate the equation: d t = [E µ el U eff ()] Hee it is convenient to exess eveything in tems of a; emembe that E el = α/a. The tem ootional to l /µ in the effective otential can be witten in tems of and theefoe in tems of a = /( e ). Afte some stuggle, the exession becomes: d t = α e a ( a) Suise suise... we will ask that ( a) = e a cos ξ a = ea cos ξ = a( e cos ξ) (the minus sign and the ξ notation ae just conventions fo this aticula oblem). Then d = ae sin ξdξ, e a ( a) = ea sin ξ and the integal becomes quite tivial: t = a( e cos ξ)dξ t = (ξ e sin ξ) α α This is the aametic vesion: fo any time t we can find the coesonding ξ and fom that find the value of = a( e cos ξ), so we have (t). The eiod can be easily found now: a full evolution is descibed by ξ = 0 π, since in this case = min max min. But if ξ vaies by π, you can see that this coesonds to an incease in time: T = π α The oblem can be solved similaly fo the aabola and hyebola -tye of tajectoies. The textbook gives you thefinal esults. You should ty deiving them, it s good math actice. 6