yields m 1 m 2 q 2 = (m 1 + m 2 )(m 1 q m 2 q 2 2 ). Thus the total kinetic energy is T 1 +T 2 = 1 m 1m 2

Transcription

1 1 I iediately have 1 q 1 = f( q )q/ q and q = f( q )q/ q. Multiplying these equations by and 1 (respectively) and then subtracting, I get 1 ( q 1 q ) = ( + 1 )f( q )q/ q. The desired equation follows after dividing by 1 +, which is licit since I assue 1, > 0. The kinetic energies of the particles are T 1 = 1 1 q 1 and T = 1 q. Since 1 q 1 + q = 0, I have 0 = 1 q 1 + q = 1 q q 1 q + q. Adding this to 1 q = 1 q 1 1 q 1 q + 1 q yields 1 q = ( 1 + )( 1 q 1 + q ). Thus the total kinetic energy is T 1 +T = q, the first ter of the desired forula. Meanwhile, the particles potential energies are V 1 = 1 V ( q 1 q ) and V = 1 V ( q q 1 ). Each of these is siply 1 V ( q ), so the total potential energy is the other ter of the desired forula. 3 The angular oenta of the particles are J 1 = 1 q 1 q 1 and J = q q. Since 1 q 1 + q = 0, I have 0 = ( 1 q 1 + q ) ( 1 q 1 + q ) = 1 q 1 q (q 1 q + q q 1 ) + q q. Adding this to 1 q q = 1 q 1 q 1 1 (q 1 q + q q 1 ) + 1 q q yields 1 q q = ( 1 + )( 1 q 1 q 1 + q q ). Thus the total angular oentu is J 1 + J = 1 1+ q q, which is the desired forula. Since the z coponents of q 1 and q are zero, so the z coponents of q and q are also zero. Thus, the x and y coponents of q q are zero. 4 I have r = q and r cos θ = q ˆx, where ˆx is the unit vector along the positive x axis. Differentiating these forulas, I get rṙ = q q and ṙ cos θ r θ sin θ = q ˆx. Multiplying the first of these forulæ by 1 ˆx and subtracting the other forula ultiplied by q, I get q (ˆx q) = rṙˆx ṙq cos θ + r θq sin θ, where I ve used the vector identity that q (ˆx q) = ( q q)ˆx ( q ˆx)q. Now, ˆx q = rẑ sin θ, where ẑ is the unit vector along the positive z axis, and q ẑ = q since q has zero z coponent. Thus I have r q sin θ = rṙˆx ṙq cos θ + r θq sin θ = r ṙ + ṙ q cos θ + r θ q sin θ rṙ ˆx q cos θ + r ṙ θˆx q sin θ rṙ θ q cos θ sin θ, (1) 1

2 where I ve used that ˆx = 1. Substituting r for q and r cos θ for ˆx q, this becoes r q sin θ = r ṙ + r ṙ cos θ + r 4 θ sin θ r ṙ cos θ + r 3 ṙ θ cos θ sin θ r 3 ṙ θ cos θ sin θ = r ṙ sin θ + r 4 θ sin θ. () Thus, q = ṙ + r θ, and the forula for the total energy is proved. (The conclusion is valid even when sin θ = 0, by continuity, since sin θ can t be constantly 0 in the physically relevant situation where an orbit exists. Siilarly, I ve been assuing r > 0 all along.) As for the angular oentu, q q = ˆx (q q) = (ˆx q)q (ˆx q) q = ṙq cos θ r θq sin θ r q cos θ = ṙ q cos θ + r θ q sin θ + r q cos θ rṙ θ q cos θ sin θ rṙ(q q) cos θ + r θ(q q) sin θ cos θ = r ṙ cos θ + r 4 θ sin θ + r (ṙ + r θ ) cos θ r 3 ṙ θ cos θ sin θ r ṙ cos θ + r 3 ṙ θ sin θ cos θ = r 4 θ, (3) so = q q = r θ. I in fact have = r θ, since both sides are positive for counterclockwise otion, although ultiately I won t need the sign. 5 Note that the sign of θ always equals the sign of the constant in this forula, and the physical application requires that θ not be constantly 0, so I can conclude that 0, which will be needed when I divide by later on. To continue, E = 1 (r θ + ṙ )+V (r) = 1 (r r 4 + ṙ )+V (r) = 1 ṙ +(V (r) + r ), as desired. By the way, note that the centrifugal force is (d/dr)( /r ) = /r 3, not /r as stated. 6 Solving for ṙ in the forula for E, I get E U(r) = 1 ṙ, or ṙ = (E U(r)), so ṙ = ± (E U(r)). However, ṙ ay or ay not be nonnegative. 7 If ṙ 0, θ/ṙ = ± r / (E U(r)), or θ = ± r ṙ/ ṙ 0. Integrating this with respect to tie, θ = θ(0)± 0 ( r ṙ dt/ (E U(r)), when (E U(r))).

3 Since dr = ṙ dt, this the desired forula except for the sign abiguity, if θ 0 := θ(0). This forula won t work, however, for an orbit of constant r, where the analysis can only be saved by arguing by continuity of the dependence of the solution on the paraeters. My analysis, however, will avoid this proble entirely by not relying on any division by ṙ. 8 U(r) = V (r)+ /r = /r k/r. As r nears zero, U(r) nears / =. As r nears infinity, U(r) nears 0. U(r) = 0 when /r = k, or r = /k. U (r) = 0 when k/r /r 3 = 0, or k = /r, which is r = /k; U( /k) = k /. Finally, U (r) = 0 when 3 /r 4 k/r 3 = 0, or 3 /r = k, which is r = 3 /k; U(3 /k) = 4k /9. Now I have plenty of inforation to sketch a graph: At an energy of E > 0, the effective particle will coe in fro infinity and then go back out. At an energy of E = 0, the effective particle will do this sae thing, but extreally. At an energy of k / < E < 0, the effective particle will follow a bound path. At an energy of E = k /, the effective particle will reain otionless at r = /k; of course, this doesn t ean that the original syste is otionless then, only that the value of r will be constant. Finally, an energy of E < k / is ipossible; the potential energy alone is enough to rule that out, since the effective kinetic energy 1 ṙ can never be negative. 3

4 9 To avoid picking branch cuts for and arccos (or even for θ!), let e revert to a differential equation and look at θ. Since θ 0 but I can t be so sure about ṙ, I ll use ṙ θ = (E U(r)) / r 4 = r4 (E V (r) /r ) = r (Er + kr ). Let e siplify the constants by introducing p := /k and e := 1 + E /k now, so = kp and k e = k +E = k +Ekp, or Ep = ke k. Note that since E k /, or E /k 1, the square root defining e is real. Then ṙ = r (Er + kr kp) = r (Epr + kpr kp ) θ kp kp = r (ke r kr + kpr kp ) kp = r (e r r + pr p ) p. (4) The nuerator now contains the subtraction of a square, which suggests the use of trigonoetry. Since the left side of this equation ust be nonnegative, I can conclude that e r r pr + p = (p r). Thus, there ust always be an angle α such that p r = er cos α. In ters of this angle, then, I have ṙ / θ = r (e r e r cos α)/p = e r 4 sin α/p. If ṙ 0, then, I can now write this as θ/ṙ = ±p/er sin α, or θ = ±pṙ/er sin α. In order to integrate this, I ll want to understand α, so differentiate the equation p r = er cos α to get ṙ = eṙ cos α er α sin α; since e cos α = p/r 1, this becoes ṙ = pṙ/r ṙ er α sin α, or α = pṙ/er sin α. I need analyse this no further, for now I see that θ = ± α, so θ = ±α + θ 0 for soe constant θ 0. In particular, cos(θ θ 0 ) = cos(±α) = cos α = ( p r 1)/e, so I ay write θ = θ 0 + arccos ( ( p r 1)/e) for soe branch of θ and arccos. In preparation for writing this in ters of the original paraeters, let e ultiply both sides of the fraction in the arccosine by k/. Then θ = kp r θ 0 + arccos( ke = θ 0 + arccos( r k k + E ) = θ 0 + arccos( k k kr k ) 1 + E k ), (5) which is the desired forula. Note that this θ 0 ay not be the θ(0) fro above, but that could be fixed by adusting the indefinite integral. 10 I already reduced the clutter in the previous proble. However, to ake sure of the forula for r, I should check also the situation when ṙ = 0. I know fro 4

5 the qualitative analysis that this occurs only when E = k / and that r = /k then. In ters of e and p, these are equations are, irrespectively, r = p and k (e 1)/ = k /, or e = 0. Thus, the equation r = p/(1 + e cos(θ θ 0 )) holds then as well. 11 You used the translation and Galilean syetries by setting the origin at the centre of ass for all tie, and you used soe of the rotational syetry by placing the proble within the x, y plane. But there reains a rotational syetry within that plane, so I use it to set θ 0 to 0. (There is also a syetry of reflection through the z axis, which could be set by requiring > 0, and a scaling syetry, which could be set by requiring p = 1. But I will not do this yet.) Then the equation for the orbit becoes p = r + er cos θ. Now, r cos θ = x, so I get r = ex p, or r = e x epx + p. Since r = x + y, this becoes x e x +epx+y = p, as desired. If e = 0, then this equation is x +y = p, a circle of radius p centred at the origin. If 0 < e < 1, then prepare to coplete the square by writing (1 e ) x + ep(1 e )x + (1 e )y = p (1 e ), or (1 e ) x +ep(1 e )x+e p +(1 e )y = p (1 e )+e p, so ( (1 e )x + ep ) + (1 e )y = p, which becoes (x + ep 1 e ) p / (1 e ) + y / p 1 e = 1, which is an ellipse centred at ( ep/(1 e ), 0) with inor radius p/(1 e ) along the x axis and aor radius p/ 1 e along the y axis. If e = 1, then the equation is px + y = p, or x = p/ y /p, which is a parabola centred along the x axis, opening to the left, and with a vertex at (p/, 0). Finally, if e > 1, then I can coplete the square as in the elliptic case, only now the equation should be written (x ep e 1 ) p / (e 1) y / p e 1 = 1, which is a hyperbola centred at (ep/(e 1), 0) with inial radius p/(e 1) along the x axis and asyptotes x = ±(e 1)y. 5