A1 Dynamical Astronomy 1

Quick facts #2: The two-body problem

Introduction

The most straightforward orbit calculations oc-cur when the central body is much more massivethan the orbiting body, as is the case for the orbitsof man-made satellites around the Earth. We as-sumed that this is also the case for planetary orbitsabout the Sun – a good approximation, especiallyfor the minor planets. However it isnot appro-priate in a binary star system where the masses ofthe two stars orbiting each other are similar. Evenfor planetary motion there is a small but importantcorrection to be made once the orbital motion ofthe Sun is taken into account. The good news isthat we can apply all our old results, with suitablemodifications.

There are two ways to approach the problem.The first, covered in the notes, is to realise that ingeneral both bodies perform elliptical orbits abouta fixed, common, centre of mass. For example,the massm1 moves as if in motion about a fixedmass of magnitudem0

2D m3

2=.m1 C m2/2. All

our ‘one-body’ results can now be applied, usingm0

2fixed at the focus, in place of a movingm2.

Sometimes however, it is useful to think in termsof the relative motion of the two masses ratherthan their individual motions about the centre ofmass.

Relative motion

The general elliptical analysis isjust beyond thescope of this course (though not by much). In-stead we will analyse the case of two stars of sim-ilar mass performing circular orbits about a com-mon centre (Figure 1).

Let the stars have massesm1 andm2, at distancesr1 andr2 from their common centre. We will de-fine theradius vector of the system as

r D r1 C r2:

m1

m2

r1r2

r

v1

v2

Figure 1: Massesm1 and m2 in circular orbits(m1 > m2).

r is therefore the distance between the stars. Itis constant in our example, but would vary if theorbits were elliptical. The two stars must havethe sameangular speed! (otherwise one masswould catch up the other, and the gravitationalforce would not be directed towards the centre ofthe circles). The orbital speeds of the two massesare therefore

v1 D !r1;

v2 D !r2:

The gravitational force between the stars dependson the inverse-square of their separation,r , andsupplies the centripetal force that keeps themmoving in circles of radiir1 andr2 respectively.So

Gm1m2

r2D

m1v2

1

r1

D m1!2r1 (1)

Dm2v2

2

r2

D m2!2r2: (2)

The equality of the two right-hand terms gives

r1

r2

Dm2

m1

;

which is just the condition that the common cen-tre of the circles is the centre of mass of the sys-tem. Using Equations 1 and 2 we can also write

!2D

Gm1

r2r2

DGm2

r2r1

: (3)

2 A1 Dynamical Astronomy - two-body problem

Also, since

r D r1 C r2 D r1 Cm1

m2

r1;

we get

r1 Dm2r

m1 C m2

:

Combining this with Equation 3 we get

!2D

G.m1 C m2/

r3:

We can write this in terms of the period of theorbit, T D 2�=!, as

T 2D

4�2r3

G.m1 C m2/;

i.e.,

r3

T 2D

G.m1 C m2/

4�2:

This is Kepler’s third law for circular orbits.It is always true, even when one mass is muchgreater than the other, but is especially usefulwhen m1 ' m2. If m1 � m2, then r2 ' r

and the equation reduces to the ‘one-body’ ver-sion derived earlier in the course.

In fact all the earlier results can be applied tothe two-body problem with the following trick(you will be shown the proof in honours astron-omy!). We can define

M D m1 C m2

� Dm1m2

m1 C m2

v D v1 C v2:

The quantity� is called thereduced mass of thesystem. Note that ifm1 � m2 thenM ' m1 and� ' m2.

In general, the two-body problem can betreated as an equivalent one-body problem inwhich the reduced mass� is orbiting about afixed massM at a distancer .

The reduced mass will describe an imaginedellipse aboutM with semi-major axisa D a1 C

a2 and eccentricitye (which is the same as theeccentricities of the two real orbits). The radiusvector,r , tells us the relative separation of the twomasses. Here are some results (derived using theabove rule) for the two-body system. Note theirsimilarity to the one-body results derived earlierin the course, but remember the definitions ofM ,�, v, a andr in terms of the parameters of the twostars:

Energy: E D1

2�v2

�GM�

rAngular momentum: L D �rv sin�

Speed: v2D GM.2=r � 1=a/

Period: T 2D

4�2a3

GM.

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