Problem Goldstein 3-3 (3rd ed. 3.11)

We can think of this two particle system as a single particle of reduced massµ in a circular orbit of period τ , radius R, and angular momentum l. We canrelate these quantities using Kepler’s Law of equal areas in equal time

dA

dt=

l

2µ, (1)

which, for a circular orbit of area πR2, integrates to

πR2 =lτ

2µ. (2)

Because we have no information about l, we would like to eliminate it fromconsideration. We can do this using another property of a circular orbit, namely,that at r = R, dV 0/dr = 0, where V 0 is the effective potential for the one-dimensional radial motion,

V 0 = −kr+

l2

2µr2. (3)

Setting the first derivative of V 0 equal to zero results in

R =l2

µk, (4)

and when combined with Eq.(2) gives us

R3 =k

µ

³ τ

2π

´2. (5)

At the instant that the circular motion is halted, the angular momentumbecomes zero (l = 0) and the radial EOM simplifies to

µ··r = − k

r2. (6)

A first integral of this EOM results in a statement of conservation of energy

1

2µ·r2= −k

µ1

R− 1r

¶, (7)

where·r = 0 at r = R. Now solve this equation for

·r, and rearrange it into the

form s2k

µR3dt = − dr

R3/2q

1r − 1

R

= − dxq1x − 1

, (8)

1

where

x = r/R , (9)

and we choose the minus sign because t will increase as r decreases. If we makethe variable substitution,

x = sin2 α , dx = 2sinα cosαdα , (10)

the integration of Eq.(7) becomes straightforward. We integrate the left sidefrom t = 0 to t = tc and the right side from α = π/2 (x = 1) to α = 0 (x = 0)to obtain s

2k

µR3tc = −2

Z 0

π/2

sin2 αdα =

Z π/2

0

(1− cos 2α)dα = π

2. (11)

Thus, the time until the particles collide is

tc =π

2

rµR3

2k=

τ

4√2, (12)

where the right hand side is found after we substitute Eq.(5) for R.

2