4Orbits

IA Dynamics and Relativity

4.2 Motion in a central force field

Now let’s put in our central force. Since V = V (r), we have

F = −∇V =

dV

dr

ˆ

r.

So Newton’s 2nd law in polar coordinates is

m(¨r − r

˙

θ

2

)

ˆ

r + m(r

¨

θ + 2 ˙r

˙

θ)

ˆ

θ = −

dV

dr

ˆ

r.

The θ component of this equation is

m(r

¨

θ − 2 ˙r

˙

θ) = 0.

We can rewrite it as

1

r

d

dt

(mr

2

˙

θ) = 0.

Let

L

=

mr

2

˙

θ

. This is the

z

component (and the only component) of the

conserved angular momentum L:

L = mr ×

˙

r

= mr

ˆ

r × ( ˙r

ˆ

r + r

˙

θ

ˆ

θ)

= mr

2

˙

θ

ˆ

r ×

ˆ

θ

= mr

2

˙

θ

ˆ

z.

So the angular component tells us that

L

is constant, which is the conservation

of angular momentum.

However, a more convenient quantity is the angular momentum per unit

mass:

Notation

(Angular momentum per unit mass)

.

The angular momentum per

unit mass is

h =

L

m

= r

2

˙

θ = const.

Now the radial (r) component of the equation of motion is

m(¨r − r

˙

θ

2

) = −

dV

dr

.

We eliminate

˙

θ using r

2

˙

θ = h to obtain

m¨r = −

dV

dr

+

mh

2

r

3

= −

dV

eff

dr

,

where

V

eff

(r) = V (r) +

mh

2

2r

2

.

We have now reduced the problem to 1D motion in an (effective) potential — as

studied previously.

The total energy of the particle is

E =

1

2

m|

˙

r|

2

+ V (r)

=

1

2

m( ˙r

2

+ r

2

˙

θ

2

) + V (r)

(since

˙

r = ˙r

ˆ

r + r

˙

θ

ˆ

θ, and

ˆ

r and

ˆ

θ are orthogonal)

=

1

2

m ˙r

2

+

mh

2

2r

2

+ V (r)

=

1

2

m ˙r

2

+ V

eff

(r).

Example.

Consider an attractive force following the inverse-square law (e.g.

gravity). Here

V = −

mk

r

,

for some constant k. So

V

eff

= −

mk

r

+

mh

2

2r

2

.

We have two terms of opposite signs and different dependencies on

r

. For small

r

,

the second term dominates and

V

eff

is large. For large

r

, the first term dominates.

Then V

eff

asymptotically approaches 0 from below.

r

V

eff

E

min

r

∗

The minimum of V

eff

is at

r

∗

=

h

2

k

, E

min

= −

mk

2

2h

2

.

We have a few possible types of motion:

–

If

E

=

E

min

, then

r

remains at

r

∗

and

˙

θh/r

2

is constant. So we have a

uniform motion in a circle.

–

If

E

min

< E <

0, then

r

oscillates and

˙r

=

h/r

2

does also. This is a

non-circular, bounded orbit.

We’ll now introduce a lot of (pointless) definitions:

Definition

(Periapsis, apoapsis and apsides)

.

The points of minimum and

maximum

r

in such an orbit are called the periapsis and apoapsis. They

are collectively known as the apsides.

Definition

(Perihelion and aphelion)

.

For an orbit around the Sun, the

periapsis and apoapsis are known as the perihelion and aphelion.

In particular

Definition

(Perigee and apogee)

.

The perihelion and aphelion of the

Earth are known as the perigee and apogee.

–

If

E ≥

0, then

r

comes in from

∞

, reaches a minimum, and returns to

infinity. This is an unbounded orbit.

We will later show that in the case of motion in an inverse square force, the

trajectories are conic sections (circles, ellipses, parabolae and hyperbolae).

Stability of circular orbits

We’ll now look at circular orbits, since circles are nice. Consider a general

potential energy V (r). We have to answer two questions:

– Do circular orbits exist?

– If they do, are they stable?

The conditions for existence and stability are rather straightforward. For a

circular orbit,

r

=

r

∗

= const for some value of

h 6

= 0 (if

h

= 0, then the object

is just standing still!). Since ¨r = 0 for constant r, we require

V

0

eff

(r

∗

) = 0.

The orbit is stable if r

∗

is a minimum of V

eff

, i.e.

V

00

eff

(r

∗

) > 0.

In terms of V (r), circular orbit requires

V

0

(r

∗

) =

mh

2

r

3

∗

and stability further requires

V

00

(r

∗

) +

3mh

2

r

4

∗

= V

00

(r

∗

) +

3

r

∗

V

0

(r

∗

) > 0.

In terms of the radial force F (r) = −V

0

(r), the orbit is stable if

F

0

(r

∗

) +

3

r

F (r

∗

) < 0.

Example. Consider a central force with

V (r) = −

mk

r

p

for some k, p > 0. Then

V

00

(r) +

3

r

V

0

(r) =

− p(p + 1) + 3p

mk

r

p+2

= p(2 − p)

mk

r

p+2

.

So circular orbits are stable for

p <

2. This is illustrated by the graphs of

V

eff

(

r

)

for p = 1 and p = 3.

V

eff

p = 1

p = 3