24.2. PLANETARY MOTION 451

Now the other part of 24.7 and 24.8 implies

r′′ (t)− r (t)θ′ (t)2 = r′′ (t)− r (t)

(c2

r4

)=−k

(1r2

). (24.9)

It is only r as a function of θ which is of interest. Using the chain rule,

r′ =drdθ

dt=

drdθ

( cr2

)(24.10)

and so also

r′′ =d2rdθ

2

(dθ

dt

)( cr2

)+

drdθ

(−2)(c)(r−3) dr

dt

=d2rdθ

2

( cr2

)2−2(

drdθ

)2(c2

r5

)(24.11)

Using 24.11 and 24.10 in 24.9 yields

d2rdθ

2

( cr2

)2−2(

drdθ

)2(c2

r5

)− r (t)

(c2

r4

)=−k

(1r2

).

Now multiply both sides of this equation by r4/c2 to obtain

d2rdθ

2 −2(

drdθ

)2 1r− r =

−kr2

c2 . (24.12)

This is a nice differential equation for r as a function of θ but its solution is not clear. Itturns out to be convenient to define a new dependent variable, ρ ≡ r−1 so r = ρ−1. Then

drdθ

= (−1)ρ−2 dρ

dθ,

d2rdθ

2 = 2ρ−3(

)2

+(−1)ρ−2 d2ρ

dθ2 .

Substituting this in to 24.12 yields

2ρ−3(

)2

+(−1)ρ−2 d2ρ

dθ2 −2

(ρ−2 dρ

)2

ρ−ρ−1 =

−kρ−2

c2

which simplifies to

(−1)ρ−2 d2ρ

dθ2 −ρ

−1 =−kρ−2

c2

since those two terms which involve(

)2cancel. Now multiply both sides by −ρ2 and

this yieldsd2ρ

dθ2 +ρ =

kc2 , (24.13)

which is a much nicer differential equation. Let R = ρ − kc2 . Then in terms of R, this

differential equation isd2Rdθ

2 +R = 0.

24.2. PLANETARY MOTION 451Now the other part of 24.7 and 24.8 impliesrine’? =") re (G) =-*(4). (24-9)rIt is only r as a function of @ which is of interest. Using the chain rule,F wee 7 (5) (24.10)"= 6 a d0and so alsodr (d@\c\. dr dr dopeo EI —_ (2 —3\) MEMJ (ae) (3) + MO) a6 atrpc? dr\? (ec_ adrcy?_,(dr\ (ee 24.11aa (7) (is) (s) (24.11)Using 24.11 and 24.10 in 24.9 yieldsa? 2 dr\? (¢? 2 1(5) -2(“) (<)--pm( 5) =-«( =).de? \r? do r lad rNow multiply both sides of this equation by r4/c* to obtain@r dr\7 1 —kr?~-_9( 2) ---= . 24.12do" (5) , c ( )This is a nice differential equation for r as a function of 0 but its solution is not clear. Itturns out to be convenient to define a new dependent variable, p = r~' sor = p~!. Thendr dp @r _3;(dp\’ ap“=(-1)p? =, — =2 ae ~1)p 2.Substituting this in to 24.12 yields2 2 _2p? (3) + (1p 28 - (075) p-pt= <5which simplifies to ; ;(-1)p 28 —pl= sssince those two terms which involve (%) ° cancel. Now multiply both sides by —p? andthis yields ;cota, (24.13)which is a much nicer differential equation. Let R = p — 5. Then in terms of R, thisdifferential equation isaR—,+R=0.do*