Using the Runge-Lenz vector to derive the orbit in a 1/r potential

Richard Low <richard@wentnet.com> June 2004

Here we find a conserved quantity, the Runge-Lenz Vector, in any orbit in a 1/r potential and use this to show the orbit is elliptical or, in general, any conic section. First define the specific angular momentum h

h = r × ṙ

and the force law for an inverse square central force

r̈ = −\frac{kr}{r^{3}}

where k is a constant e.g. for gravity, k = GM. Now prove h is conserved

ḣ = ṙ × ṙ + r × r̈
= 0 − \frac{k}{r^{3}} r × r
= 0

Secondly, find the time derivative of r/r:

\frac{d}{dt} (\frac{r}{r} ) = \frac{1}{r^{2}} (r ṙ − ṙ r)

but

\frac{d}{dt} (r.r) = \frac{d}{dt} (r^{2})

so

2ṙ.r = 2rṙ

giving

ṙ = \frac{ṙ.r}{r}

so

\frac{d}{dt} (\frac{r}{r} ) = \frac{1}{r^{3}} (r^{2}ṙ − ṙ.r r)

Now consider \frac{d}{dt} (ṙ × h):

\frac{d}{dt} (ṙ × h) = r̈ × h + ṙ × ḣ
= r̈ × (r × ṙ)

= −\frac{k}{r^{3}} r × (r × ṙ)
= −\frac{k}{r^{3}} (r.ṙ r − r^{2}ṙ)
= \frac{k}{r^{3}} (r^{2}ṙ − r.ṙ r)
= k \frac{d}{dt} (\frac{r}{r} )

Therefore

\frac{d}{dt} (ṙ × h − k r/r) = 0

This gives the Runge-Lentz vector A

ṙ × h − k r/r ≡ A is conserved.

Now consider A.r:

A.r = Ar cos θ
= r. (ṙ × (r × ṙ)) − kr
= (r × ṙ) . (r × ṙ) − kr
= h^{2} − kr

So

r = \frac{h^{2}}{k + A cos θ}
= \frac{\frac{h^{2}}{k}}{1 + \frac{A}{k} cos θ}

This is a conic section because h and A are constant. The semi-latus rectum is h^{2}/k and eccentricity A/k with the ‘sun’ at one focus. A special case is when A/k = 0, a circular orbit. For bound orbits, 0 ≤ A/k < 1, which gives an ellipse.