4 Classical Dynamics

4.1 Newtonian gravity

4.1.1 Basic law of attraction

Two p oint masses, with mass M and M , lying at r and r , attract one

1 2 1 2

another. M feels a force from M

1 2

GM M

1 2

F = force on M from M = r

12 1 2 12

where r = r r =vector from 2 to 1 and r = jr j

12 1 2 12 12

M1 F r1 12 F21

M2 Coordinate r2

Origin

Similarly M feels a force from M

2 1

GM M GM M

2 1 1 2

F = force on M from M = r =+ r = F

21 2 1 21 12 12

3 3

r r

21 12

where r = r r is the vector from 1 to 2. Wehave used r = r , and

21 2 1 21 12

jr j = jr j.Thus, the gravitational forces are equal and opposite. This is in

21 12

accordance with Newton's 3 law and ensures that the comp osite system of

(M + M ) do es not suddenly start moving as a whole and violating Newton's

1 2

1 law.

Notice also that the gravitational force at M from M is directed exactly

1 2

at M . It is a \central" force: F is parallel to r . This remark causes us

2 12 12

to digress and mention::: 60

4.1.2 The little known co dicil to Newton's 3 law

Many b o oks, e.g. Marion, state Newton's laws in something like this form:

1. A b o dy remains at rest or in uniform motion unless acted up on bya

force.

2. A b o dy acted up on by a force moves in such a manner that the time

rate of change of momentum equals the force.

3. If two b o dies exert forces on each other, these forces are equal in mag-

nitude and opp osite in direction.

This statement of 3 is WRONG! The correct statementis

3.Iftwo b o dies exert forces on each other, these forces are equal in mag-

nitude and opp osite in direction AND ALONG THE SAME LINE.

Newton knew this, but mo dern text b o ok writers have garbled it.

B B

OK NOT OK! A A

(a) (b)

If (b) were p ossible, then we could take the two particles A and B , nail

one to the hub of a wheel, the other to the center, and watch as the wheel

accelerated up to in nite angular velocity. 61

Thus, while one often hears ab out the \central force" nature of gravita-

tion, this is actually a prop ertyofany action-at-a-distance, classical force

law.

You mightwonder ab out the magnetic force b etween twomoving par-

ticles in electromagnetism. This do es not seem to b e a central force. The

explanation is that the electromagnetic eld itself carries momentum and

angular momentum. Classical electromagnetism cannot b e written as a pure

action-at-a-distance theory. That is, there is more to Maxwell's equations

than just the Coulomblaw.

4.1.3 Gravitational Potential

The force on M , due to M 's presence can b e written in terms of the gradient

1 2

of a gravitational potential:

F = force at M due to M

12 1 2

= force at p osition r due to mas at r

1 2

! ! !!

@ GM @ GM @ GM

2 2 2

= M ; ;

@x jr r j @y jr r j @z jr r j

1 1 2 1 1 2 1 1 2

= M r V (r )

1 1 1

@ @ @

where r is the gradient op erator at M ; . V (r) is the gravi- ; ;

1 1

@x @y @z

1 1 1

tational p otential at r ,

V = :

jr r j

1 2

The p otential at p oint r from a numb er of masses; at r ; r ; r ;:::;r is, in

1 2 3 n

general

V =

jr r j

i=1

where the sum excludes any mass exactly at r (do not include the in nite

self-energy!). 62

4.2 The 2-b o dy problem

Although every mass in the universe exerts a force on every other mass,

there is a useful idealization where the mutual interaction of just two masses

dominates their motion. The equations of motion are then

GM M

1 2

M r = (r r )

1 1 1 2

jr r j

1 2

GM M

1 2

M r = (r r ) :

2 2 2 1

jr r j

1 2

These apply, for example, to binary stars at large separations, but small

enough that the tidal e ects of other stars in the galaxy can b e neglected.

Or, the orbit of the Earth around the Sun, to the extent that the e ects of

the Mo on and other planets can b e ignored.

4.2.1 Conservation laws for 2-b o dy orbits

For a 2-b o dy system, wewant to solve for 6 functions of time r (t) and r (t),

1 2

and 12 constants will app ear in the solution (equivalentto2vectors of initial

p ositions and 2 vectors of initial velo cities). It is most ecient to approach

this problem by nding constants of the motion (to reduce the number of

equations to solve) and bycho osing a convenient set of co ordinates in which

to work.

Add the two equations of motion given ab ove. Then, since the forces are

equal and opp osite, the right hand side sums to zero, and we are left with

M r + M r =0:

1 1 2 2

Let the center of mass of the system b e at R ; then by the de nition of center

of mass as the mass-weighted average p osition,

M r + M r

1 1 2 2

R =

(M + M )

1 2 63

we get

(M + M )R =0 or R =0 :

1 2

Thus, the lo cation of the center of mass of the system is unaccelerated, since

there is no external force on the system.

It follows that the center of mass moves at a constantvelo city:

R = v = constant

R = R + v t:

0 0

This relation expresses the \Conservation of Linear Momentum." R and

V are the rst 6 \constants of motion."

The next logical thing to do is to change to center-of-mass co ordinates

0 0

r ; r , that is, to co ordinates relative to the center of mass.

1 2

M1 CoM ′ r1 r2′ M2 r 1 R r 2

The relevant relationships are:

r = R + r

R = R + v t:

0 0

Substituting these into the equations of motion, we get

GM M

1 2

0 0 0

M r = (r r )

1 1 2

0 0

jr r j

1 2 64

GM M

1 2

0 0 0

M r = (r r ) :

2 2 1

0 0

jr r j

1 2

Note that the equations preserve exactly the form they had b efore: We could

have just written down the answer without calculation. We can arbitrarily

cho ose co ordinates relative to the CoM, b ecause the CoM frame is a go o d

inertial frame.Anychange of origin (R ) and/or change of velo city(V )is

0 0

allowed by (so-called) \Gallilean invariance."

So let us cho ose to do this: co ordinates are, henceforth, measured relative

to the CoM; and we drop the primes.

In CoM system,

GM M

1 2

(r r ) M r =

1 2 1 1

jr r j

1 2

GM M

1 2

M r = (r r )

2 2 2 1

jr r j

1 2

with M r + M r =0 by de nition.

1 1 2 2

We can now construct further integrals of the motion. Consider the total

_ _

angular momentum ab out the CoM: L = M r r + M r r . Then

1 1 1 2 2 2

_ _ _ _

= M (r r + r r )+M (r r +r r )

1 1 1 1 1 2 2 2 2 2

= M r r +M r r :

1 1 1 2 2 2

Substitute the equations of motion for M r and M r , then

1 1 2 2

GM M dL

1 2

fr (r r )+r (r r )g =

1 1 2 2 2 1

dt jr r j

1 2

GM M

1 2

= (r r + r r )

1 2 2 1

jr r j

1 2

= 0

i.e. the total angular momentum of the system is conserved:

L = constant. 65

That is, the angular momentum vector is xed in direction and magnitude (b ecause no external torques are acting).

M1 r r 2 1 r1 CoM r 2 M

The angular momentum provides 3 more integrals of the motion, so we are

nowupto9.

We can also lo ok at the total energy of the system:

E = (kinetic energy) + (p otential energy)

1 GM M 1

1 2

_ _ _ _

M r r + M r r + : =

1 1 1 2 2 2

2 2 jr r j

1 2

To see that this is conserved, write

GM M d dE

1 2

_ _

= M r r + M r r + jr r j :

1 1 1 2 2 2 1 2

dt jr r j dt

1 2

Now

d d

1=2

jr r j = (r r + r r 2r r )

1 2 1 1 2 2 1 2

dt dt

______

r r + r r r r r r (r r )(r r )

1 1 2 2 1 2 1 2 1 2 1 2

= = :

jr r j jr r j

1 2 1 2

Therefore

! !

dE GM M GM M

1 2 1 2

_ _

(r r ) + r (r r ) = r M M r + r +

1 2 2 2 1 1 2 1 2 1

3 3

dt jr r j jr r j

1 2 1 2

= 0

using the equations of motion. Therefore the total energy of the system,

E = constant 66

(b ecause no work is b eing done on systems by an external force). The energy

is another integral of the motion (each one equivalent in counting to one

initial condition). These 10 integrals, R ; v ; L;E are general to N b o dies,

0 0

not just 2 (as we will see later). To completely sp ecify the system we need

12 constants of integration; 10 sp eci ed: 2remain. These turn out to b e

particular to the sp ecial case of 2-b o dies. That is, they are more like \initial

conditions" than \conservation laws."

4.2.2 Two-b o dy orbits

Finally we are ready to solve the 2-b o dy problem. Let us recap: Wehave

removed 6integrals of the motion byworking in the center of mass (= center

of momentum frame). Then the other integrals of the motion are total energy

and total angular momentum.

The p ositions of the two b o dies, r and r ,obey

1 2

GM M

1 2

M r = (r r ) (1)

1 1 1 2

jr r j

1 2

GM M

1 2

M r = (r r ) (2)

2 2 2 1

jr r j

1 2

and by de nition

M r + M r =0: (3)

1 1 2 2

We showed that

_ _

M r r + M r r = L L = constantvector

1 1 1 2 2 2

1 1 GM M

1 2

_ _ _ _

M M = E: E= constant scalar r r + r r

1 2 1 1 2 2

2 2 jr r j

1 2

Wenowwant to completely solve the system: for r (t) and r (t).

1 2 67

It is convenient to reduce the system to simpler form byworking in terms

of the separation vector of the two b o dies:

r = r r =vector from b o dy 2 to b o dy 1.

1 2

r1 − r2 = r

r1 r2

M1 CoM M2

Then consider equation (1) divided by M minus equation (2) divided by

M :

G(M +M )

1 2

r =(r r )= (r r )

1 2 1 2

jr r j

1 2

r r =

where M = M + M is the total mass of the system. This equation says

1 2

that b o dy 1 moves as if it was of very small mass b eing attracted by mass

(M + M ) at b o dy 2; the separation vector is accelerated by the total mass

1 2

of the system. The fact that the 2-b o dy problem reduces to the equation of

the \one-b o dy problem" (test mass in the central force eld of a ctitious

mass M ) is an amazing and nonobvious result!

With the relations

r = r r

1 2

M r + M r =0

1 1 2 2

we can nd how to calculate r and r from the solutions for r :

1 2

M M

2 1

r = r r = r :

1 2

M M 68

Note that the signs are opp osite (the masses must b e on opp osite sides of

the CoM).

Eliminating r and r in favor of r we nd that r (t) ob eys

1 2

r = r

r r = L

M M

1 2

1 GM M

_ _

E: r r =

2 r M M

1 2

Notice the app earance of a quantity M M =(M + M ) on the right

1 2 1 2

hand sides, where we exp ect (one over) a mass. The quantity is called the

reduced mass. Notice that is smaller than b oth M and M .

1 2

The angular momentum equation

r r = L

M M

1 2

_ _ _

has an imp ortant immediate consequence. Since r (r r )=r(rr)=0,

that is, the vector cross pro duct of twovectors is p erp endicular to b oth of

them, r and r are b oth p erp endicular to L, i.e. all the motion o ccurs in a

plane perpendicular to L (rememb er that r and r are prop ortional to r ,

1 2

hence parallel to r ).

Adopting co ordinates in the plane of the orbit, we can write in Cartesian

co ordinates:

L = (0; 0;L)

r = (x; y ; 0)

r = (_x; y;_ 0) :

The momentum equation b ecomes

M L

xy_ y x_ = L =

M M

1 2 69

with an energy equation

1 E GM M

2 2

E = (_x +_y ) = :

2 2 1=2

2 (x + y ) M M

1 2

In principle the ab ovetwo equations are the coupled di erential equations

for the two unknowns x(t) and y (t). It is simpler, however, to exploit the

planar geometry in polar co ordinates r and .

r(t+ dt) radial direction ^ r δθ r δδr r azimuthal angle direction r(t) ^ θ δθ

particle

0 trajectory

If r (t) ! r (t + dt) in time dt, then, to rst order,

(comp onentof r along original radial direction) = r

(comp onentofr along direction of azimuth) = r :

Therefore if, at any given time, in cylindrical co ordinates (r;;z),

r = (r; 0; 0) r e

r r (t + t) r(t)

r = = ;r ; 0 =_re +re

t t t

rr = 0;0;r

2 2 2

_ _

r r = r_ + r 70 . . r θ r eθ .

r er

\Centrifugal form of Pythagoras"

We nd that the angular momentum equation is

M L

r = L =

M M

1 2

while the energy equation is

GM E 1 M

2 2 2

r_ + r = E = :

2 r M M

1 2

The angular momentum equation can b e interpreted geometrically

t + dt

r(t + dt) dA dθ(t) t

r(t)

In interval t to t + dt, the vector b etween the ob jects sweeps at area dA,

dA = r rd :

Therefore, the rate of change of area is

dA 1

= r

dt 2

whichby the rst equation b ecomes

1 M L dA

= = constant. L =

dt 2 M M 2

1 2 71

Thus, it is a simple consequence of the Law of momentum conservation that

the rate of sweeping out of area is a constant.

This is Kepler's Second Law (1609). Kepler discovered it empirically |

he had no idea what physics was b ehind it. Kepler was Tycho's \graduate

student" (in mo dern terms), but Tycho would never let him lo ok at all the

data. Tycho died in 1601, and Kepler got the data then. Even to day many

observers would rather take their data to the grave than let theorists analyze

them. (Or so it seems to theorists!)

Pro ceeding, we can get one equation for r (t)by eliminating from these

equations. Then the equations can b e rewritten

dr M 2GM M L

= 2 E +

dt M M r M M r

1 2 1 2

d M L

= :

dt M M r

1 2

4.2.3 Shap es of the orbits

The ab ove equations could b e solved for r (t) and (t). The answer comes

out in \elliptic integral functions." But the shape of the orbit can b e derived

by eliminating time from the equations to get an equation for r ( ):

# "

1=2

2 2

dr r 2(GM ) (L=)

= 2(E=)+

d (L=) r r

where we are now using the \reduced mass"

M M =M:

1 2

This is a rst order di erential equation that can b e integrated

Z Z

(L=) dr

= d =

h i 0

1=2

2(GM ) (L=)

r 2(E=)+

r r 72

where = constantofintegration. De ne a constant

2 2

(L=) L

r = =

(GM ) GM

which sets a scale for the orbit, and a second constant,

2 2

2(E=)(L=) 2EL

=1+ =1+ :

2 2 3

(GM ) (GM )

Then the integral can b e rewritten

r dr

=( )

1=2

2 2 0

r 1

which, via the substitution

cos u =1

b ecomes (try it!)

1 1

= (1 cos( )) du =( ) )u=( ) )

0 0 0

r r

the plus-or-minus is thrown in gratuitously, b ecause it corresp onds to ,an

unknown constant, changing by ;sowe can write the result with the more

convenient + sign,

1 1

= (1 + cos ( )) :

r r

This is the equation of a conic section.

There are three sp ecial cases whichwenow consider.

Case of <1:

> 0 for all ; so the separation b etween the two masses If <1, then

remains nite: the ob jects remain bound with

r r

0 0

r :

1+ 1 73

The orbit in this case, is an el lipse.Ifwe put

x = r cos ( )

y = r sin( )

and eliminate and r , the equation of the orbit b ecomes

1 x 1

1+ =

2 2 1=2 2 2 1=2

(x + y ) r (x + y )

which simpli es to

x + a y

+ =1

a b

where

r r

0 0

a = ; b =

2 2 1=2

1 (1 )

are semi-ma jor and semi-minor axes.

= θθ0 + π/2 x = a(1 − ε ), y = 0 b r + ε θ = θ εa 0 r = r0 /(1 ), x a

x = − a(1 + ε), y = 0 = − ε θ = θ + π

r r0 /(1 ), 0

2 1=2

The axial ratio of the ellipse is b=a =(1 ) . The quantitiy r is called

the \semi-latus rectum." Note that <1 implies, from the de nition

2EL

2 3

(GM )

that E<0, i.e. the total energy of a bound system is negative. 74

We can also see that the orbit is periodic and closed. Let us nd the

p erio d:

1 1

(1 + cos ( )) =

r r

d (L=)

dt r

are the equations governing (t); we can write this as a single equation for

(t);

Z Z

r d

= dt :

(L=) (1 + cos ( ))

Now, in one rep eating orbit,

increases from ! +2

t increases from t ! t + ; ;= p erio d, so

r d

= :

(L=) (1 + cos )

1 1t 2dt

The substitution t = tan ; cos = ) renders this integral (so d =

2 2

2 1+t 1+t

tractable. Then

r 2

2 3=2

(L=) (1 )

1=2

Eliminating r and in terms of physical variables: (L=)=(GM r ) and

0 0

r = a(1 ), a the semi-ma jor axis,

3=2

=2 :

1=2

(GM )

Note, incidentally (you can check) that a dep ends only on the total energy

a = GM M =(2E ), while dep ends on b oth the energy and the angular

1 2

momentum.For b ound orbits wehavenow proved Kepler's laws:

1. The planets move in ellipses with the sun (the center of mass) at one

fo cus:

1 1

= (1 + cos ( ))

r r

0 75

with <1(= 0 is circle).

2. A line from the Sun to a planet sweeps out equal areas in equal times:

dA 1

= (L=) :

dt 2

(Note, this also works for unb ound orbits.)

3. The square of the p erio d of revolution is prop ortional to the cub e of

the semi-ma jor axis

a GM

: = or ! =

2 GM a

(Note this involves the total mass M = M + M .)

1 2

Case of >1:

Again from the de nition,

2EL

=1+

2 3

(GM )

we see that E>0, so the total energy of system is positive. This will imply

that the orbit is unb ound:

1 1

= (1 + cos( ))

r r

is a hyp erb ola: x = r cos ( ), y = r sin( ) gives, again

0 0

1 1 x

= 1+

2 2 1=2 2 2 1=2

(x + y ) r (x + y )

2 2

y (x a)

=1 )

2 2

a b

r r

0 0

a= b=

2 2 1=2

1 ( 1)

2 1=2

b = a( 1) 76 y

r0 M2 x M1

y = 0, x =(ε − 1) a

M0 is at one focus; Hyperbola; single-pass orbit.

Case of = 1:

From the de nition of , =1 ) E = 0, i.e. total energy of system is

zero. In this case the orbit is a parabola:

1 1

= (1 + cos( ))

r r

2 2

) y = r 2r x

0 0

r 0 M2 x M r 1 0/2

Parabola. Single pass orbit. 77

4.2.4 Orbital elements

From an observational p oint of view, a complete description of a Keplerian

orbit (e.g., of a planet or asteroid) is one that establishes its size, shap e, and

orientation in 3 dimensional space, and also gives enought information to

determine where along the orbit the ob ject is at any given time. One such

complete description consists of 6 \orbital elements." (There must b e 6,

b ecause the complete description is equivalent to giving a vector of p ositions

and a vector of velo cities at some ducial t = 0.)

In the following picture, the actual orbit is the inner ellipse, while the

\tilted circles" are just to show (separately) the orbital plane and ecliptic

plane (equator of the co ordinate system used). The symb ol is the \ rst

p oint of Aries," the origin of longitude in these co ordinates. Do not b e fo oled

by an optical illusion: The orbit is in the orbital plane and is an ellipse |

not a circle | whose ma jor axis is A SA. The p erihelion (closest approachof

the orbit to the sun) is A, while P is the p osition of the planet \rightnow." 78

The 6 orbital elements are de ned as [2 \shap e" parameters]:

semima jor axis a half the length of AA

2 2 1=2

eccentricity e (1 b =a ) if b is the semiminor axis

[ 2 parameters de ne the orbital plane: ]

inclination i angle ( QN B )between the orbital

plane and the ecliptic plane

longitude of the angle ( SN ) from rst p oint of Aries

ascending no de to the line of no des N N . \Ascending"

means pick the no de where the planet

crosses the ecliptic from south (b elow)

to north (ab ove.)

[ 1 parameter orients the orbit in its plane: ]

argument of p erihelion ! angle ( NSA ) from the ascending

no de, measured in the plane of the

orbit, to the p erihelion p oint A.

[ 1 parameter sp eci es orbital phase: ]

time of p erihelion passage T one of the precise times that the ob ject

passes through to the p oint A

Note that P , the p erio d is not an orbital element, since you can compute it

from a by

3=2

P (in years) = [a (in AU) ] :

Instead of ! , p eople sometimes quote the \longitude of p erihelion"

6 6

$ +! = SN + NSA :

(The symbol $ is actually a weird form of \pi," but most astronomers call it

\p omega"!) Note that the two summed angles are measured along two di er- 79

ent planes! This is not real ly the longitude of anything, but it approximates

it when the inclination i is small.

Also, instead of T , p eople sometimes give the \longitude" L of the planet

at a sp eci ed ep o ch (time). Here the longitude is again a kind of phoney:

6 6

L N + NP :

In case you want to compute lo cations of the planets, here are the values

of their orbital elements, called \ephimerides" (the plural of \ephemeris"):

Planets: Mean Elements

(For ep o ch J2000:0 = JD24515450 = 2000 January 1.5)

Planet Inclination (i) Eccentricity (e)

0 00

Mercury 7 00 17: 95051 0.205631752 49 14

0 00

Venus 3 23 40: 07828 0.006771881 91 42

Earth 0.0 0.016708617 15 40

0 00

Mars 1 50 59: 01532 0.093400619 94 74

0 00

Jupiter 1 18 11: 77079 0.048494851 21 99

0 00

Saturn 2 29 19: 96115 0.055508621 71 72

0 00

Uranus 0 46 23: 50621 0.046295898 51 25

0 00

Neptune 1 46 11: 82795 0.008988094 86 52

0 00

Pluto 17 08 31: 8 0.249050

Mean Longitude Mean Longitude Mean Longitude

of No de ( ) of Perihelion ($ ) of Ep o ch (L)

0 00 0 00 0 00

Mercury 48 19 51: 21495 77 27 22: 02855 252 15 03: 25985

0 00 0 00 0 00

Venus 76 40 47: 71268 131 33 49: 34607 181 58 47: 28304

0 00 0 00

Earth 0.0 102 56 14: 45310 100 27 59: 21464

0 00 0 00 0 00

Mars 49 33 29: 13554 336 03 36: 84233 355 25 59: 78866

0 00 0 00 0 00

Jupiter 100 27 51: 98631 14 19 52: 71326 34 21 05: 34211

0 00 0 00 0 00

Saturn 113 39 55: 88533 93 03 24: 43421 50 04 38: 89695

0 00 0 00 0 00

Uranus 74 00 21: 41002 173 00 18: 57320 314 03 18: 01840

0 00 0 00 0 00

Neptune 131 47 02: 60528 48 07 25: 28581 304 20 55: 19574

0 00 0 00 0 00

Pluto 110 17 49: 7 224 08 05: 5 238 44 38: 2 80

Mean Distance (AU) Mean Distance (10 m)

Mercury 0.387098309 8 0.579090830

Venus 0.723329820 0 1.08208601

Earth 1.000001017 8 1.49598023

Mars 1.523679341 9 2.27939186

Jupiter 5.202603191 3 7.78298361

Saturn 9.554909595 7 14.29394133

Uranus 19.218446061 8 28.75038615

Neptune 30.110386869 4 45.04449769

Pluto 39.544674 59.157990

4.2.5 The mass of the Sun and the masses of binary stars

Wehave seen that the sidereal p erio d of a planet, , is related to the semi-

ma jor axis of its orbit, a,by

2 3

= a ;

where

M = M + m :

planet

This formula has b een applied to calculate GM on the last column of the

following table.

Observed Inferred

Sidereal Semi-ma jor

p erio d axis of orbit GM

26 3 1

Planet (days) (AU) (10 cm s )

Mercury 87.969 0.387099 1.32714

Venus 224.701 0.723332 1.32713

Earth 365.256 1.000000 1.32713

Mars 686.980 1.523691 1.32712

Jupiter 4332.589 5.202803 1.32839

Saturn 10759.22 9.53884 1.32750

Uranus 30685.4 19.1819 1.32715

Neptune 60189 30.0578 1.32723

Pluto 90465 39.44 1.32727 81

Therefore, from the low-mass planets,

26 3 1

GM =1:32713 10 cm s :

This is known very accurately,to 1 part in 10 . Amazingly, G itself,

from lab oratory exp eriments, is only known to ab out 1 part in 10 as (6:670

8 3 1 1

0:004) 10 cm s g . So our knowledge of the mass of the Sun is limited

by lab oratory physics, not astronomical observations!

M =(1:989 0:001) 10 gm :

We can also estimate the mass of Jupiter, M :

26 3 1

G(M + M ) = 1:32839 10 cm s

26 3 1

G(M + M ) = 1:32713 10 cm s ;

where M is the mass of the Earth. Jupiter has ab out an 0.1% e ect on

GM .Thus M 0:1% of solar mass. Furthermore,

23 3 1

G(M M )=1:26 10 cm s

J E

M M

J E

=0:000949 ;

which gives us

M 0:000949M ' 1:89 10 g :

This is fairly inaccurate (1 part in 10 ), and it is a go o d illustration of an

imp ortant principle: normally,we measure planet masses by their e ects on

the satel lites or one another. Arti cial satellites (e.g., Voyager I I at Uranus)

are also go o d!

This works for our Sun, but what ab out other stars? Their planets are

to o faint: there are no low-mass test particles that we can see around these 82

stars. But 50% of all stars are in multiple star systems. So we can use the

relative motions of binary stars to get stellar masses.

The p erio ds of binary stars vary typically from a few hours (e.g., 6

for Mirzam, alias Ursa Ma joris) to hundreds of years (e.g., 700 yr for 61

Cygni). Some are even shorter, but for these it is likely that the stars are

highly distorted by tidal e ects and their masses will b e sub ject to systematic

errors (e.g., \semidetached" or \contact" binaries).

Avariety of data may be available for a binary star. Wemay see the

motion of one star around the other on the sky (which is rare), or wemay

see only variations in the velo cities of the stars (as seen from Doppler shifts

in their sp ectra | this is more common). Binary stars may b e single-lined

(see light from only one star) or double-lined.

Consider, for example, the second case, a \double-lined sp ectroscopic

binary star." Here, we see the two stars' sp ectra sup erimp osed, and we

can measure the radial velo cities for b oth stars, but the stars are to o close

together for us to see their orbits by measuring the changing p ositions of the

stars relative to one another. Then what we see are star velo cities like

v 1 system velocity not,

K2 in general, zero velocity,

0.511.5

time, t/τ 83

Immediately,we see that wehave three observables:

the orbital p erio d,

the p eak velo city of the less massive star 1, K (the more massive star

moves more slowly)

the p eak velo city of star 2, K

The shap e of the curves is here sinusoidal, which implies circular orbits,

" = 0. Other shap es are p ossible for O " 1.

What can we get from these data? To simplify things, supp ose the stars

are in a circular orbit, " = 0 (not necessary | we could t " from the velo city

data if wewanted to, but this makes things harder). Then our orbit equations

are

M r + M r =0 center of mass at rest

1 1 2 2

d L= 2

= = constant angular velo city ; circular orbit. = constant=

dt r

But the p erio d is related to the separation of the stars, r (the same as the

semima jor axis, for a circular orbit), by

2 3

4 r

= for circular orbits,

and the sp eeds of the stars in their orbits are v = r and v = r . If the

1 1 2 2

orbital plane is parallel to the line of sight, the p eak velo cities wewould see

for the stars are r and r . In general, however, this will not b e the case

1 2

(see gure on the next page).

At inclination angle i (i = angle b etween line of sight and normal to

orbital plane) 84

K = r sin i = r sin i

1 1 1

K = r sin i = r sin i:

2 2 2

If i =0,we see no velo city.Thus, the separation of the stars is

(K + K )

1 2

: r = r + r =

1 2

2 sin i

Using the previous formula for and eliminating r , the p erio d equation then

gives a total mass

(K + K )

1 2

M + M = :

1 2

2G sin i

The center of the mass provides the constraint M r = M r , so the mass

1 1 2 2

ratio is

M r K

2 1 1

= = :

M r K

1 2 2

Putting these together,

3 2

M sin i = (K + K ) K

1 1 2 2

2 3

(K + K ) K : M sin i =

1 2 1 2

2G 85

Here the right hand sides are completely known, so we infer the left-

hand sides. These data only give masses if we know the inclination of the

orbit, i. But, in general, we will not know this, so our data only measure a

combination of masses and i.

For one sp ecial group of stars, wedogetM and M directly | from

1 2

\eclipsing" binaries, where the light curve shows an eclipse (see gure).

1.2 τ 1 2 secondary +F ) 1 .8 eclipse

F/(F .6 .4 primary eclipse

light flux, .2

0 0 .5 1 1.5

time, t /τ

Here, the stars are eclipsing one another . This scenario is only p ossible

if i 90 (provided that the stellar radii separation, whichwe can check

in the ab ove equations; if this is not true, the stars are so distorted that we

cannot use this metho d anyway). Thus, sin i 1. Errors in i have little

e ect on the masses derived, b ecause sin i changes slowly at i 90 .

In fact, the b est data on masses of stars comes from eclipsing sp ectroscopic

double-lined binaries, even though these are hard to nd.

An alternative approach is to measure many noneclipsing binaries, assume

i to b e selected at random, and use statistics to derive the distribution of

masses, M and M .

1 2 86

4.2.6 Sup ernovae in binary systems

It is quite common for a sup ernova to b e in a binary system. In fact, some

sup ernovae are caused by a companion star's dumping mass on a star, until

the latter explo des (section 4.3.4, b elow). What happ ens to the binary system

when the explosion o ccurs? It turns out that the force of the explosion is not

an imp ortant e ect on the star, but its mass loss is.

Before the sup ernovawehave (say) two stars in circular orbit:

M v1 2 r1 r2

M CoM

1 v2

Putting the CoM at rest, wehave

M v +M v = 0

1 1 2 2

M r +M r = 0:

1 1 2 2

Let M b e the mass that explo des. Typically it is the more massive star that

explo des, so M >M . When it explo des, it quickly (and spherically,say)

1 2

blows o most of its mass, so that its new mass M is

= M M M

and wehave 87 expanding shell − ′ of mass M1 M M 1 v1 2 centered on M1

old new M ′ v

1 CoM CoM 2

A rst imp ortant e ect is that the remaining binary is not at rest. Let's

calculate its new CoM velo city v :

0 0

M v + M v =(M +M )v :

1 2 2 2 c

1 1

Or, using v = M v =M and M = M M ,

1 2 2 1 1

(M M ) v + M v =(M M +M )v

1 2 2 2 1 2 c

giving

v : v =

2 c

M (M + M M)

1 1 2

Notice that as M ! M (all-star explosion!), v ! v ,asitmust.

1 c 2

Typical values mightbe M =10M ,M =5M ,M =8:5M , leaving

1 2

a1:5M neutron star for M ,so

8:55

v = v =0:65v :

c 2 2

10 6:5 88

For close binaries, v can b e hundreds of kilometers p er second, so the system

really takes o !

In fact, we need to check whether the binary remains b ound at all! The

easiest way to do this is to see if its total energy in the new CoM is p ositive

or negative just after the explosion:

0 1 0 1

! !

total kinetic

total energy p otential

B C B C

E = = kinetic energy +

@ A @ A

in new CoM energy

energy of CoM

1 1 1 GM M

0 2 2 0 2

= M v + M v (M + M ) v :

2 2

1 1 2 1 c

2 2 2 r

Recall that (Section 4.2.2) the separation vector r and the relativevelo city

v v satisfy Kepler's Laws with the total mass. So for circular orbits,

2 1

G(M + M )

1 2

=(v v ) :

2 1

Substituting this, and v previously obtained, and also eliminating M in

favor of M M and v in favor of v M =M ,we get

1 1 2 2 1

1 1 1 M v

2 2

0 2

E = (M M ) + M v (M + M M )

1 2 1 2

2 M 2 2

" #

MM v (M M)M M

2 2 1 2 2

1+ v :

M (M + M M) M + M M

1 1 2 1 2 1

Now a fairly amazing algebraic simpli cation (I used the \Simplify [ ]"

function in Mathematica on the computer) gives

(M M )(M + M ) 1

1 1 2

0 2

E = M v (M + M 2M ) :

2 1 2

2 M (M + M M )

1 2

Notice that all the terms are p ositive (since M

We see that a condition for E to b e p ositive| unbinding the binary | is

M> (M +M )

1 2

2 89

that is, loss of more than exactly half the total mass of the system. In the

previous numerical example wehave

8:5> (10 + 5)

so the neutron star is unb ound and ies away at high velo city. In fact,

pulsars (which are neutron stars) are often seen to b e leaving the galactic

plane, where they are formed, at high velo city for just this reason.

You mightwonder if there isn't a simpler way to derive the ab ove \half of

total mass" result. There is if you are comfortable with expressing the energy

of a 2-b o dy system in terms of the reduced mass (as in ab ove section 4.2.2).

Then (as Mike Lecar p ointed out to me) you can write the total energy of a

moving or stationary binary system as

GM 1 1

2 2

Mv + v E =

CoM

2 2 r

where M is total mass, v is the center of mass velo city, and v is the

CoM

relativevelo city (that is, jr j). The rst term has no e ect on whether the

binary is b ound or unb ound. Rather, it is just a question of whether the

second term is p ositive or negative. For an initial circular orbit, wehave

GM 1 GM 1 GM

2 2

so that v = : v =

r 2 r 2 r

After the sup ernova, v is (instantaneously) the same, while and M change.

You can see that losing more than half of M causes v =2 GM =r to change

sign (b ecome unb ound).

4.3 Tides and Ro che e ects

When two b o dies are in orbit around each other, the otherwise spherical grav-

itational eld around each b o dy is distorted by the gravitational attraction 90

of the other b o dy.For the Earth-Mo on system (e.g.) this causes the surface

of the o ceans (and to some extent the solid Earth as well) to b e deformed

into the familiar tidal shap e:

least gravitational greatest gravitational Earth attraction attraction Moon

tidal deformation produced

Our goal is to understand the ab ove picture quantitatively.

The surface of the o cean is an equipotential, b ecause if it were not |

that is, if there were a p otential gradient along the surface | there would b e

a force causing the water to ow \downhill" until it lled up the p otential

\valley."

But is it a purely gravitational p otential wemust consider? No, b ecause

the \stationary frame" in which things can come to equilibrium is the rotat-

ing frame in which the Earth and Mo on are xed. (We are here assuming

a circular orbit. Things would b e more complicated if the orbit were sig-

ni cantly eccentric.) So there is also a centrifugal force and corresp onding

centrifugal p otential. If the orbital angular velo cityis !, with

G(M + M )

1 2

! = ;

then the centrifugal acceleration at a p osition r with resp ect to the CoM

is ! (! r ), so the centrifugal p otential (whose negative gradientisthe

1 1

2 2 2

acceleration) is j! r j or, in the orbital plane, ! r . The total p otential

2 2

in the rotating frame is thus

GM GM 1

1 2

2 2

= ! r :

jr r j jr r j 2

1 2 91

The gure shows an example of such a p otential. Notice that the downhill

gradient tries to make a test particle either fall into one of the p otential wells

of the masses or b e ung o to in nityby centrifugal force. We will come

back to this general case b elow in Section 4.3.4.

4.3.1 Weak tides

The Mo on (M ) is far from the Earth (M ) and can b e regarded as a p oint

2 1

mass. Since the size of the Earth is likewise small compared to the Earth-

Mo on distance R,we can write the Law of Cosines

2 2 2

D = R + r 2Rr cos

and then expand in the small ratio r=R. 92 D a θ 0 r2 r CoM R 1 Moon

Earth sub-lunar point

GM GM GM

2 2 2

= =

2 2 1=2

jr r j D (R + a 2Ra cos )

1=2

GM a a

= 1+ 2 cos :

R R R

We need to do the binomial expansion to second order to consistently pick

2 2

up all terms of order a =R . This gives

" #

GM a 1 a a

= 1+ cos + (3 cos 1) + O :

R R 2 R R

(If you know ab out multip oles, you will see the Legendre p olynomials in

cos lurking here.) Note that the term (a=R) cos actually varies linearly in

the z direction from M to M , since z = a cos , so the gradient of this part

1 2

of the p otential is a constant force.

a θ

Now, for the centrifugal p otential term, we again apply the law of cosines,

# "

1 1 M M

2 2

2 2 2 2

! r = ! R + a 2 R a cos :

2 2 M + M M + M

1 2 1 2 M1

a θ Earth Moon M 2

M CoM 2 R

M1+ M2

So, collecting the three terms in the overall p otential , and substituting for

! (formula in 4.3),

" #

GM GM a 1 a

1 2

= 1+ cos + 3 cos 1

a R R 2 R

" #

1 G(M + M ) M M

1 2 2 2

2 2

R + a 2 Ra cos :

2 R M + M M + M

1 2 1 2

Lo ok carefully and you will see that the term in (a=R) cos exactly cancels

out. This is not coincidence: it is b ecause the constant force term is exactly

canceled by the centrifugal force that keeps the b o dies in a circular orbit.

Also note that there are terms with no a or dep endence: Since these are

just constants, they pro duce no gradients and can b e ignored. So we get,

combining the remaining terms,

GM 1 Ga

(a; )= 3M cos + M + constant:

2 1

a 2 R

This is the lo cal tidal p otential near the Earth. To get a b etter feeling for

its shap e, let us expand a in terms of height h ab ove the radius of the Earth

(\mean sea level"):

! !

2h h

2 2 1 1

1+ a = R + h; a =R + ; a = R 1 + :

R R

2 1

Here wehave used the binomial expansion to get a and a to rst order in

h=R .Now, to this order,

! !

GM h 1 GR 2h

(h; )= 1 1+ (3M cos +M )+constant :

2 1

R R 2 R R

We can now absorb into the constant all terms that dep end on neither h nor

. Doing this, and with some rearranging of terms (putting all multipliers of

cos together, and all remaining multipliers of h together), we get

" ! !#

4 4

GM R M R 3 h

1 2

(h; ) = h 1 cos + + constant

4 3

R R M R 2 R

" #

3 M R

g h R cos + constant:

2 M R

Here the approximate equality means we are neglecting terms that are smaller

than the dominant terms by either factors of R =R or h=R (b oth b eing

assumed small). Note that g is the acceleration of gravity at the surface of

the Earth.

The surface of the o cean (in the gure at the b eginning of 4.2) is an

equip otential, so it must have(h; ) = constant, implying

3 M R

h = R cos + constant:

2 M R

Since cos varies b etween 1 (\front and back" of Earth) and zero (\sides" of

Earth) the other factors give the height of the tide (high minus low). Putting

in M =M 1=81, R 6400 km, R 380000 km , we get h =54cm.

2 1

The Sun also generates a signi cant tide that, to lowest order, just adds

with that of the mo on (but with a di erent origin for the direction , p ointing

to the Sun). For the Sun

3 6400 km

h = (332000) 6400 km = 25 cm:

2 1:5 10 km 95

Notice that we never assumed that M =M was either 1or1, so our

1 2

formulas apply b oth for the Sun and for the Mo on. It is a numerical coin-

cidence that the tides they raise are of the same order (though some have

sp eculated that the resulting more-variable tides pro duced could have b een

somehow necessary to mix the o ceans in a way that furthered the evolution

of life | or its emergence from the sea).

4.3.2 Tidal drag and the lengthening of the day

Real tides are not exactly the size wehave calculated, primarily b ecause the

Earth is rotating. Thus the continents are \dragged through" the o ceanic

tidal bulges, causing water to pile up on the continental shelves, owinto

and out of bays, etc. There can also b e resonance e ects in some cases like

the famous BayofFundy.

This kind of \friction" b etween the rotating Earth and the tidal bulges

pro duces two e ects.

1. The tidal bulge is, on average, dragged ahead of where it would b e if

it \p ointed at" the Mo on.

ϕ 13 /day

360 /day

2. The drag slows down the rotation of the Earth.

Since angular momentum must b e conserved, as the Earth loses angular mo-

mentum, it must b e gained by the Mo on's orbital motion! How do es this 96

come ab out? The non-symmetric bulge (angle ' in ab ove picture) causes

a gravitational torque back on the Mo on which transfers exactly the right

amount of angular momentum. This must work out exactly b ecause angu-

lar momentum is truly conserved! (Note that energy is not here conserved

b ecause friction is presentbetween the o cean and the rotating Earth.)

Increasing the Mo on's orbital angular momentum causes its orbit to grad-

ually move outward, and the length of the month gets longer. In fact we know

from growth scales in fossil corals (in which annual and monthly patterns can

b e discerned) that the day and month were b oth once substantially shorter

than now: Earth has slowed down and Mo on has moved farther away,aswe

predict.

Where will this pro cess end? As the Earth's rotation slows down, the

Earth will ultimately come to rotate onceper month (the then-length of the

month) and keep one face always towards the Mo on. In this state, called

\tidally lo cked," there is no drag on the tidal bulge: it will p oint exactly at

the Mo on as Earth and Mo on rotate. (In fact, this corotation the current

state the Mo on is in with resp ect to the Earth.)

What will the ultimate length of the day (and month) b e? Let ! =

2=(1 day), ! =2=1 month, ! =2=(ultimate day and month). Then

2 f

conservation of angular momentum gives (1 = Earth , 2 = Mo on ):

(angular momentum now) = (angular momentum then)

M M M M

1 2 1 2

2 2

I ! + I ! + ! R = I ! + I ! + ! R

1 1 2 2 2 1 f 2 f f

M M

while Kepler's law gives the future radius R of the Mo on's orbit,

2=3

R !

f 2

R !

f 97

to go o d approximation, since M 81:3M ,wehaveM M =M M . Also

1 2 1 2 2

we can neglect the spin angular momentum of the Mo on now (the Earth's

alone b eing so much larger), and we can neglect both spins in the nal state.

(If you doubt these approximations, you can substitute back the answer we

will get and check retroactively that they are justi ed.) Also we will use the

moment of inertia of a uniform sphere, even though this is not quite a true

assumption,

I = M R :

1 1

Then wehave

1=3

! 2

2 2 2 2

M R ! +M ! R =M ! R =M ! R

1 1 2 2 2 f 2 2

1 f

5 !

yielding

# " # "

2 2

M R ! + M ! R

! 2 M R !

1 1 2 2

2 1 1 1

= 1+ =

! M ! R 5 M R !

f 2 2 2 2

" #

2 6400

= 1+ 81:3 28 =1:99 :

5 380000

So the nal length of day and month will b e 28 1:99=54days, and the

2=3

nal mo on's orbital radius will b e its currentvalue times (1:99) or ab out

590000 km . The current rate of lengthening of the day is ab out 20% p er 10

years, so it will take 10 yr for the pro cess to go to completion. (By

then, the Sun will haveswollen to a giant and incinerated Earth and Mo on

anyway.)

Incidently, although our sp eci c example has b een the Earth-Mo on sys-

tem, exactly these e ects o ccur in close binary stars: Tidal drag causes them

to come into \corotation," tidally lo cked to each other. 98

4.3.3 Ro che stability limit for satellites

Let us now apply the previous result for the tidal p otential (slightchange in

notion),

GM Gr 1

+ constant (r;)= 3M cos + M

2 1

r 2 R

to the case where M is a small satellite orbiting close to a large parent (star

or planet) M ,so M M .

2 1 2

Satellite:

M2 BrA M1zaxis

It turns out that if R is to o small, the satellite is torn apart by tidal forces.

To see this, let's calculate the gravitational acceleration at p oints A and B

in the gure. By symmetry it must b e along the z axis, so

" #

d GM 1 Gz

g = r = (3M )

z z 2

dz jz j 2 R

3GM M 3M GM

2 1 2 1

z + z = Gz + : =

3 3 3 3

jz j R jz j R

This is a restoring force only if the co ecientof z is negative. Otherwise the

force is away from the center of the satellite and the satellite is torn apart,

starting at its surface. Putting jz j = r and considering a satellite of mean

density ,so

M = r

3 99

we get the condition for Roche stability,

9 M

> :

4 R

Equivalently, no satellite of density can b e stable inside the Ro che radius

1=3

R = :

crit

If the parent b o dy (e.g., a planet) has the same density as the satellite, then

M =4R =3, and we get

1=3

R =3 R =1:44R :

crit 2 2

This is approximately what is going on with the rings of Saturn (and lesser

rings of the other outer planets). Material inside the Ro che limit cannot form

into mo ons b ecause of tidal forces from the parent body. (\Approximately"

b ecause our assumption of a spherical rather than deformed, satellite gives

not quite the rightnumerical co ecient.)

4.3.4 Ro che Lob e over ow

Let us go back (Section 4.3) to the basic formula for the tidal p otential of

two masses in the rotating CoM frame,

GM 1 GM

2 1

2 2

! r (r )=

jr r j jr r j 2

1 2

with (Kepler)

G(M + M )

1 2

! = :

jr r j

1 2 100

A more carefully drawn version of the p otential surface plot in 4.3 is the

following contour plot:

The equip otentials = const. for the Newtonian-plus-centrifugal p otential in the orbital plane of a binary

star system with a circular orbit. For the case shown here the stars have a mass ratio M : M = 10:1.

1 2

The equip otentials are lab elled by their values of measured in units of G(M + M )=R where R is the

1 2

separation of the centers of mass of the two stars. The innermost equip otential shown is the \Ro che lob e"

of each star. Inside eachRoche lob e, but outside the stellar surface, the p otential is dominated by the

\Coulomb" (1=r ) eld of the star, so the equip otentials are nearly spheres. The p otential has lo cal

stationary p oints (r = 0), called \Lagrange p oints," at the lo cations marked L .

The picture is drawn for the particular case M =M =0:1, but would b e

2 1

qualitatively similar for other mass ratios.

Fo cus attention on the contour marked \Ro che lob e" that de nes the

sadd le point between the two p otential minima. (Lo ok back at the surface

plot in 4.3 if you need to visualize the \saddle p oint.") In a binary star

system, eachRoche lob e is the maximum size that its resp ective star can b e.

If (due to stellar evolutionary e ects) one star tries to swell up to larger than

the Ro che lob e, it simply dumps mass through the \lip of the pitcher" (across

the saddle p oint) onto the other star. This is called \Ro che lob e over ow" 101

and can profoundly a ect the evolution of the binary star system.

For example, consider a case where M is initially greater than M . Higher

1 2

mass stars evolve faster, so M may rst swell up and dump mass onto

M (p ossibly also losing mass to in nity as a stellar wind). Eventually M

2 1

b ecomes a white dwarf star and starts to co ol. Later, as M evolves (now

faster b ecause it has gained mass) it mayswell up and Ro che-over ow back

onto the white dwarf, a pro cess that can b e observed as an X-ray source

(the X-rays b eing pro duced as the matter crushes down on the white dwarf

surface). Finally the excess mass dropp ed onto the white dwarf maybetoo

much for it to supp ort, and it may collapse to a neutron star, in the pro cess

blowing o part of its mass as a sup ernova explosion!

Notice that the saddle p oint marked L is a very slightly higher contour

than L .Thus, a star that swells very slowly will dump preferentially onto

its companion, but more rapid swelling can over ow b oth L and L simul-

1 2

taneously. Matter that go es over the \lip" L is ung out to in nity in what

would lo ok like an expanding spiral trail (rememb er the co ordinate system

is rotating!). This is thought to b e happ ening in so-called \violent mass

transfer binaries" of which examples are W Serp entis and Lyrae.

4.3.5 E ect of mass transfer on binary orbits

Supp ose that M is lling its Ro che lob e and dumping mass onto M . This

1 2

pro cess conserves mass and angular momentum, but not energy (since the

mass crashing down onto M dissipates its energy into heat.

Since M is gaining mass, wehave

_ _

M =M 0:

2 1 102

Note that the total mass M is constant.

Now write the angular momentum for a circular orbit,

1=2

GM M M

1 2

2 2 1=2 1=2 1=2 1=2 1=2

L = R =R = R G =(M M R )(G M ) :

1 2

3 1=2

R M

Nowwe conserve angular momentum:

dL 1

1=2 1=2 1=2 1=2 1=2

_ _ _

0= = G M M M R + M M R + M M RR :

1 2 1 2 1 2

dt 2

_ _ _

Solving for R and eliminating M in favor of M we get

1 2

M M

2 1

_ _

M : R =2R

M M

1 2

So, if the lighter star is losing mass, M 0, we

1 2 2

get R>0 and the stars draw gradually apart. Often this turns o the mass

ow, since it puts M deep er into its Ro che lob e. Or, the mass ow pro ceeds

only on the stellar evolutionary timescale that it takes M to keep swelling

to ll the ever-larger Ro che lob e.

Things are di erent if the heavier star is losing mass, M >M ! Then

1 2

R is negative. The mass ow causes the stars to get closer, whichtypically

increases the mass ow, and so on in a catastrophic instability. This could

theoretically end when the stars reach equal mass. But in practice, the

mass ow is often so violent that hydro dynamic friction results in the stars'

merging.

4.3.6 Accretion disks

Gas at rest at the \lip" L of the Ro che lob e has a sp eci c angular momentum

(that is, angular momentum p er unit mass)

L = jr r j :

L1 CoM 103

Referring to the gure in 4.3.4, you can see that jr r j can b e signif-

L1 CoM

icantly less than r (the 2-b o dy separation). A consequence is that gas that

ows over L has only enough angular momentum to go into a circular orbit

well inside the other Ro che lob e | if it do es not impact the surface of the

other star rst. If the other star is a white dwarf or neutron star, so-called

\compact ob jects," it is small enough to allow the formation of an \accretion

disk" of gas (see gure). Once in the accretion disk, gas slowly spirals into

the compact ob ject by the action of viscous forces (including magnetic and

turbulent viscosity).

Ω

Disk "Normal DDsk star"

4.3.7 The Lagrange p oints

Referring again to the contour plot in 4.3.4, let us now take the masses to

b e safely tucked for inside their Ro che lob es, as for the Earth and Mo on,

or Sun and Jupiter. Wenow ask the question: Are there any equilibrium

p oints where a third orbiting test b o dy could b e placed, and where it would 104

maintain a constant p osition in the rotating frame (that is, co-orbit with the

same p erio d as the two main b o dies at constant p osition relative to them)?

The condition for such an equilibrium is that there b e no acceleration

at the chosen p osition, that is, the gradient of the p otential (including

centrifugal term) must vanish,

r=0:

Gradients vanish, in general, at extrema and at saddle p oints. In the contour

plot you can see that there are 3 saddlep oints co-linear with masses M and

M . It is easy to see that there must always b e these three (for any mass

ratio) as the following graph illustrates:

Gravitational force due to M1

Negative of centrifugal force L O L1 2 x L 3 A B Gravitational force

due to M2

You can see that gravitational force and (negative) centrifugal force must

always cross at exactly three p oints, where

F = F or F + F =0:

grav centrif grav centrif

These three p oints, L ;L ;L are the rst three \Lagrange p oints" where

1 2 3

a test mass can orbit. However, since they are saddle p oints (see contour 105

picture) all three are unstable: if the mass is slightly p erturb ed, it falls into

one of the p otential wells or is ung o to in nity.

More interesting are the two maxima of the p otential, lab elled L and L

4 5

in the contour plot. These are the so-called stable Lagrange points. Lo oking

at the contour plot, or the surface plot at the b eginning of 4.3, you can see

that these maxima o ccur along the ridge-line of a long, banana-shap ed ridge

between the p otential well of the combined masses and the centrifugal force

p otential that decreases toward in nity.

If a test mass is placed at L or L , it will stay there. If it is p erturb ed

4 5

slightly, it will execute (in the rotating frame) a stable, though not necessarily

closed, orbit that go es around L or L .

4 5

You mightwonder how such an orbit can b e stable if it is going around

a p otential maximum, not minimum. The answer is that, in the rotating

frame, there is a Coriolis force ! v , so that a test particle's equation of

motion is actually

= r ! v :

If ! in the contour picture is out of the page (M and M orbiting counter-

2 1

clo ckwise), then a particle pushed \outward" from L or L exp eriences a

4 5

clo ckwise force and go es into a small (stable) clo ckwise orbit around L or

L resp ectively.

The b est real-life example of ob jects orbiting a stable Lagrange p oint

are the Tro jan asteroids which orbit L and L of the Sun-Jupiter system.

4 5

Between 1906 and 1908, four such asteroids were found: the numb er has

now increased to several hundred (see gure). These asteroids are named

for the hero es from Homer's Iliad and are collectively called the Tro jans.

Those that precede Jupiter (at L ) are named for the Greek hero es (plus the

4 106

Tro jan spy, Hektor), and those that follow Jupiter (at L ) are named for the

Tro jan warriors along with the Greek spy,Patro clus. Some of the Tro jans

make complicated orbits taking as long as 140 years to meander around the

\banana" of the p otential (meanwhile, of course, orbiting the Sun every 11.86

years, just like Jupiter). [Ab ell 7 Ed., 18.3]

Wehave not yet determined the location of L and L in the gure. The

4 5

amazing fact is that, indep endent of the mass ratio M : M , L and L are

1 2 4 5

exactly at the vertices of equilateral triangles formed with M and M . This

1 2

is so remarkable that we should at least verify it for fun: 107 P mass m

F1 F2

ψ ψ 1 2

A 60 O 60 B R CoM M M1 2

pR (1−p)R

Wewant to see that F + F exactly balances the centrifugal term mr ! .

1 2

The magnitudes of F and F are

1 2

Gm(1 p)M GmM GmpM GmM

2 1

= F = = F =

2 1

2 2 2 2

R R R R

where p M =M and M = M + M . So the condition for force balance

2 1 1 2

p erp endicular to r is

F sin = F sin :

1 1 2 2

Apply the Law of Sines to the triangle AOP,

sin 60 3 =2 sin

= = :

r r pR 108

sin = 3pR=(2r )

and likewise using triangle BOP,

sin = 3(1 p)R=(2r ) :

Using the previous formulas for the magnitudes F and F , the force check

1 2

b ecomes

p p

3pR G(1 p)M 3(1 p)R GmpM

F sin = = = F sin :

1 1 2 2

2 2

2r R 2r R

Hey,itchecks! So nowwe need to check the comp onent paral lel to r (which

includes the centrifugal term):

F cos + F cos = mr !

1 1 2 2

Law of cosines on AOP:

2 2 2 2

r + R p R

cos =

2rR

Law of cosines on BOP:

2 2 2 2

r + R (1 p) R

cos =

2rR

Kepler's law:

! = :

" #" #

2 2 2 2

Gm(1 p)M r + R p R

R 2rR

" #

2 2 2 2

r + R (1 p) R GmpM GM mr

+ =

2 3

R 2rR R

2 2 2 2 2 2 2 2

(1 p)(r + R p R )+p[r +R (1 p) R ]

= r

2 2 2 2

r + R (1 p + p ) = 2r : 109

Hmm. Is this true for all values of p?Yes! It is the Law of Cosines applied

to the triangle BOP (using cos 60 =1=2):

2 2 2 2 2

r = R +(1 p) R 2(1 p)R cos 60

2 2

= R (1 p + p ) :

So the force balance is exact for b oth comp onents.

4.4 The virial theorem

Many-b o dy (that is, more than 2-b o dy) dynamics is something that comes

later in the course, but we need to derive one imp ortant result, the virial

theorem,now. Just for fun, here is a fairly fancy derivation, taughttomeby

George Rybicki. If this is ab oveyour level, just study it in a general way.

Supp ose wehaveN b o dies all interacting. Their kinetic energy is

T = m v

i=1

and their gravitational p otential energy is

X X

Gm m

i j

V = :

jr r j

i j

i6=j

Newton's force lawis

m v =(rV) :

i i

r=r

De ne a quantity (a bit like a moment of inertia, but around a point, not an

axis)

I m r :

i=1

Di erentiate with resp ect to time twice:

I = m v r

i i i

i 110

X X

I = m v r + m v v

i i i i i i

i i

= r (rV ) +2T:

i r =r

Now here is the tricky part: The virial theorem comes ab out b ecause the

p otential energy V has a scaling relation that describ es how its numerical

value would change if all p ositions r were stretched by some factor . In

particular, you can see rightaway that

V (r ;r ;:::;r )= V(r ;r ;:::;r ) :

1 2 N 1 2 N

Di erentiating with resp ect to , and then setting =1,we get

r (rV ) = V:

i r =r

Thus

I = V +2T:

This is called the time dep endent virial theorem. It says, roughly, that if

V +2T>0, the mean square size of the system must eventually b e increasing

while if V +2T<0 the size must eventually b e decreasing. (Rememb er that

V is negative, for this to make sense.)

It also follows, logically enough, that if a gravitating system is in equilib-

rium, neither increasing nor decreasing steadily in size, it must have

hV +2Ti =0 or 2hT i = hV i :

Here the angle brackets denote the long-time average.To see this formally,

average the time dep endent theorem over a long time T :

Z Z

T T

1 1 1

_ _

(V +2T)dt = Idt = [I(T)I(0)] : hV +2Ti =

T T T

0 0 111

Now if all particles remain b ounded with b ounded velo cities for all time

(the de nition of an equilibrium system) then I (t) remains b ounded (see its

formula ab ove) and the right hand side go es to zero as T !1,thus proving

the desired time-averaged virial theorem. 112