17.3. PLANETARY MOTION 359

Multiply both sides by dRdθ

. Then using the chain rule,

12

ddθ

((dRdθ

)2

+R2

)= 0

and so( dR

dθ

)2+R2 = δ

2 for some δ > 0. Therefore, there exists an angle ψ = ψ (θ) suchthat

R = δ sin(ψ) ,dRdθ

= δ cos(ψ)

Since( dR

dθ

)2+R2 = δ

2,( 1

δ

dRdθ, 1

δR)

is a point on the unit circle. But differentiating, the firstof the above equations,

dRdθ

= δ cos(ψ)dψ

dθ= δ cos(ψ)

and so dψ

dθ= 1. Therefore, ψ = θ +φ . Choosing the coordinate system appropriately, you

can assume φ = 0. Therefore,

R = ρ − kc2 =

1r− k

c2 = δ sin(θ)

and so, solving for r,

r =1(

kc2

)+δ sinθ

=c2/k

1+(c2/k)δ sinθ=

pε

1+ ε sinθ

whereε =

(c2/k

)δ and p = c2/kε. (17.10)

Here all these constants are nonnegative.Thus

r+ εr sinθ = ε p

and so r = (ε p− εy). Then squaring both sides,

x2 + y2 = (ε p− εy)2 = ε2 p2 −2pε

2y+ ε2y2

And sox2 +

(1− ε

2)y2 = ε2 p2 −2pε

2y. (17.11)

In case ε = 1, this reduces to the equation of a parabola. If ε < 1, this reduces to theequation of an ellipse and if ε > 1, this is called a hyperbola. This proves that objectswhich are acted on only by a force of the form given in the above example move alonghyperbolas, ellipses or circles. The case where ε = 0 corresponds to a circle. The constantε is called the eccentricity. This is called Kepler’s first law in the case of a planet.

17.3.3 Kepler’s Third LawKepler’s third law involves the time it takes for the planet to orbit the sun. From 17.11 youcan complete the square and obtain

x2 +(1− ε

2)(y+pε2

1− ε2

)2

= ε2 p2 +

p2ε4

(1− ε2)=

ε2 p2

(1− ε2),