Lecture Notes for AST 608Selected Topics in Astronomy & Astrophysics

Prepared by

Dr. Joseph M. Hahn

Saint Mary’s University

Department of Astronomy & Physics

January 2, 2004

Loosely speaking, dynamics = the study of the motion of bodies due to the

forces (principally gravity) exerted by other bodies.

The Two–Body Problem

from Chapter 2 of Murray & Dermott’s Solar System Dynamics

Lets solve the 2–body problem:

Begin with 2 gravitating masses m1 and m2

that have position vectors r1 and r2.

These could be two stars that orbit one another,

or a planet in orbit about one star,

or two unbound bodies that encounter each other.

1

let r = r2 − r1 = the bodies’ relative coordinate (1.1)

so F1 = +Gm1m2

r3r = m1r1 = grav’ force on m1 due to m2(1.2)

and F2 = −Gm1m2

r3r = m2r2 = force on m2 due to m1 (1.3)

thus r = r2 − r1 = −G(m1 + m2)

r3r = relative acceleration (1.4)

or r = − µ

r3r (1.5)

where µ = G(m1 + m2) (1.6)

(1.7)

This is the equation for the motion of m2 (the secondary body)

relative to m1 (the primary body).

Note that writing our equation of motion (EOM) in terms of the

relative coordinate r effectively places our origin on m1.

Does this choice of a coordinate system mean

that our reference frame is inertial?

The gravitational constant G is

G = 6.67 × 10−11 m3/kg/sec = 6.67 × 10−8 cm3/gm/sec.

Although the text uses MKS units, most astronomers use cgs units...

2

Integrals of the Motion

The following will derive several integrals (ie, constants) of the motion

that will be quite handy:

note that r × Eqn. (1.5) = r × r = 0 =d

dt(r× r) (1.8)

so r × r = h = constant (1.9)

(1.10)

This is the system’s angular momentum integral h = r× r

which is a constant vector that is perpendicular to both r and r

⇒ m1 and m2 are restricted to moving in a plane perpendicular to h.

h has units of ang’ mom’ per unit mass,

and is sometimes call the specific angular momentum

Keep in mind that h is not the system’s total angular momentum per unit

mass, due to the fact that we used a coordinate system that is not inertial.

The total ang’ mom’ is calculated in Section 2.7 of the text using center–of–

mass (COM) aka barycentric coordinates.

3

Since the motion is restricted to a plane,

we will proceed using polar coordinates:

Your elementary Mechanics class showed that

r = rr (1.11)

r = rr + rθθ (1.12)

r = (r − rθ2)r +

[

1

r

d

dt

(

r2θ)

]

θ (1.13)

where r, θ, and z are the usual unit vectors in cylindrical coordinates.

Thus h = r × r = r2θr × θ = r2θz (1.14)

and h = |h| = r2θ (1.15)

Kepler’s 2nd

The above leads to Kepler’s 2nd law of planetary motion:

the planet’s radius vector r sweeps out equal areas in equal times.

In time ∆t, the radius vector r → r + ∆r and sweeps out area ∆A:

∆A =1

2|r||r + ∆r| sin ∆θ ' 1

2r2∆θ (1.16)

ThusdA

dt=

∆A

∆t

∣

∣

∣

∣

∆t→0

=1

2r2θ =

1

2h = a constant (1.17)

Thus r has a constant areal velocity, ie,

r sweeps out equal areas in equal times.

4

Solve for the orbit

Recall that r = − µ

r3r (see Eqn. 1.5) (1.18)

whose radial part is r − rθ2 = − µ

r2(Eqn. 1.13) (1.19)

An easy solution is obtained by first replacing r with the variable u = 1/r,

and assume that u = u(θ) while θ = θ(t):

since r = u−1 (1.20)

r = −u−2du

dθ

dθ

dt= −r2du

dθ

h

r2since θ = h/r2 (1.21)

so r = −hdu

dθ(1.22)

and r = −hd2u

dθ2θ = −h2u2d

2u

dθ2(1.23)

So the equation of motion (1.19) becomes

−h2u2d2u

dθ2− u−1h2u4 = −µu2 (1.24)

ord2u

dθ2= −u +

µ

h2(1.25)

This familiar EOM has the same form as that is subject to gravity as it

dangles from a spring:

mx = −kx + mg where x = displacement, k = spring const’(1.26)

or x = −ω2x + g (1.27)

with sol’n x(t) = A cos(ωt − δ) + x0 (1.28)

(1.29)

Thus Eqn. (1.25) has solution

u(θ) = A cos(θ − ω) + B (1.30)

where A, B, ω are integration constants.

5

Since

d2u

dθ2= −A cos(θ − ω) = −A cos(θ − ω) − B +

µ

h2(1.31)

⇒ B =µ

h2(1.32)

Also, set A = eB where e is now the remaining integration constant. Then

u(θ) =µ

h2[1 + e cos(θ − ω)] (1.33)

so r(θ) = u−1 =h2/µ

1 + e cos(θ − ω)(1.34)

≡ p

1 + e cos(θ − ω)(1.35)

This is the equation for a conic section where the constant p = h2/µ is

known as the semilatus rectum.

A conic section is the curve obtained by the intersection of a plane and a

cone:

The constants a, e, ω are known as orbit elements

6

The type of conic section depends upon the orbit’s eccentricity e:

circle e = 0 p = a E < 0

ellipse 0 < e < 0 p = a(1 − e2) E < 0

parabola e = 1 p = 2q E = 0

hyperbola e > 0 p = a(1 − e2) E > 0

(1.36)

where the constant a is the orbit’s semimajor axis.

Later we will show that orbits with eccentricities e < 1 are bound,

ie, they have energies E < 0,

while a parabolic orbit with e = 1 is marginally bound with E = 0,

and a hyperbolic orbit with e > 1 is unbound with E > 0.

The bound, elliptic orbit is most relevant to planetary problems:

r(f) =a(1 − e2)

1 + e cos f(1.37)

where f = θ − ω = m2’s true anomaly (1.38)

θ = its longitude (1.39)

ω = its longitude of periapse (1.40)

m2 is closest to m1 at periapse, when f = 0 and r(0) ≡ q = a(1 − e),

and furthest at apoapse when f = π and r(π) ≡ Q = a(1 + e).

Note that we have recovered Kepler’s 1st Law of planetary motion:

a planet moves along an ellipse with the Sun at one focus.

7

Since h =√

µp, this also means that the system’s angular momentum

(actually, its angular momentum integral) depends on e, too.

For an elliptical orbit, h =√

µa(1 − e2).

Kepler’s 3rd Law: T 2 ∝ a3

Recall that

dA

dt=

1

2h = r’s areal velocity, Eqn. (1.17) (1.41)

If T = the planet’s orbital period, then∫ T

0

dA

dtdt = A =

∫ T

0

1

2hdt =

1

2hT (1.42)

⇒ T =2A

h(1.43)

where A = πab = area of ellipse (1.44)

= πa2√

1 − e2 (1.45)

and T =2πa2

√1 − e2

√

µa(1 − e2)(1.46)

= 2π

√

a3

µ= 2π

√

a3

G(m1 + m2)(1.47)

which confirm’s Kepler’s 3rd.

8

Energy Integral E

Recall that r = − µ

r3r (Eqn. 1.5) (1.48)

so r · r +µ

r3r · r = 0 (1.49)

since r = rr and r = rr + rθθ (1.50)

r · r = rr (1.51)

and r · r +µr

r2= 0 (1.52)

now note that v2 = r · r (1.53)

sodv2

dt= 2r · r (1.54)

or r · r =1

2

dv2

dt(1.55)

Next note thatd

dt

(

1

r

)

= − r

r2(1.56)

so r · r +µr

r2=

d

dt

(

1

2v2 − µ

r

)

= 0 (1.57)

ie E =1

2v2 − µ

r= constant energy integral (1.58)

= kinetic + potential E per mass (1.59)

Again, this is not the system’s total energy since our coordinate system is

not inertial (ie, the relative velocity v2 would need to be replaced with COM

velocities V1 and V2).

9

Now write E in terms of orbit elements:

First introduce the mean motion n = 2π/T =√

µ/a3

= the ‘average’ angular velocity of m2 about m1.

Thus µ = n2a3 and h =√

µa(1 − e2) = na2√

1 − e2.

First note that v2 = r · r = r2 + r2θ2 (1.60)

= r2 + r2f 2 (1.61)

and since r =a(1 − e2)

1 + e cos f, (1.62)

r = −a(1 − e2)(−e sin f)f

(1 + e cos f)2=

r2e sin ff

a(1 − e2)(1.63)

and since h = r2θ = r2f = na2√

1 − e2, (1.64)

r =e sin fna2

√1 − e2

a(1 − e2)=

nae sin f√1 − e2

(1.65)

Also rf =na2

√1 − e2

r=

na2√

1 − e2(1 + e cos f)

a(1 − e2)(1.66)

=na√1 − e2

(1 + e cos f) (1.67)

Thus v2 = r2 + r2f 2 (1.68)

=n2a2

1 − e2

[

e2 sin2 f + 1 + 2e cos f + e2 cos2 f]

(1.69)

=n2a2

1 − e2(e2 + 1 + 2e cos f + 1 − 1) (1.70)

=n2a2

1 − e2

[

e2 − 1 + 2a

r(1 − e2)

]

(1.71)

= n2a2

(

2a

r− 1

)

(1.72)

= µ

(

2

r− 1

a

)

(1.73)

10

so E =1

2v2 − µ

r=

µ

r− µ

2a− µ

r(1.74)

or E = − µ

2a= −G(m1 + m2)

2a= energy integral (1.75)

which of course is a constant since energy is conserved.

The other important integral is the angular momentum integral h

h =√

µa(1 − e2) =√

G(m1 + m2)a(1 − e2) (1.76)

Section 2.7 of the text computes the system’s total energy

and angular momentum in barycentric (ie, COM) coordinates:

E? = −Gm1m2

2a= µ?E (1.77)

L? =m1m2

m1 + m2

h = µ?h (1.78)

where µ? =m1m2

m1 + m2

= the system’s reduced mass (1.79)

please don’t confuse µ? and µ...

11

The Orbit Plane and the Reference Plane

Given m2’s coordinates r, θ, you can

compute its Cartesian coordinates (xorb, yorb, zorb) in the orbital plane:

However if you are investigating the motion of, say, a comet in an eccentric

& inclined orbit, then another coordinate system might be more useful.

One usually uses ecliptic coordinates when doing Solar System work.

12

Let the xec, yec, zec axes refer to the ecliptic coordinate system,

where the xec–yec plane is Earth’s orbital plane,

and the xeq, yeq, zeq system is aligned with Earth’s equatorial plane.

In this system, xec points towards the vernal equinox = the ascending node

of Earth’s orbit (Earth travels upwards through its equatorial plane)

which points in the direction of Ares (γ)

13

The Orbit Plane and the Ecliptic Plane

The angles:

i = inclination

Ω = ascending node (m2’s motion carries it up thru the node)

ω = argument of periapse

ω = Ω + ω = longitude of periapse

(note that Ω and ω are not measured in a common plane when i > 0!)

θ = ω + f = m2’s longitude

The following shows how to obtain m2’s motion,

ie, its position and velocity,

from the 6 orbit elements a, e, i, Ω, ω, f .

14

Transform from Orbital→Ecliptic Coordinates

m2’s ecliptic coordinates (xec, yec, zec) are obtained from

its orbit–plane coordinates (xorb, yorb, zorb) via a series of rotations.

1. Given m2’s polar coordinates rorb(θ), get rorb = (xorb, yorb, zorb).

15

2. Rotate rorb about the z axis by angle ω such that r′ = Rz(ω)rorb

where Rz = matrix for rotations about z axis (see page 50).

3. Next rotate r′ about the x′ axis by angle i such that r′′ = Rx(i)r′

16

4. Finally rotate r′′ about z axis by angle Ω such that rec = Rz(Ω)r′′

Thus

rec = Rz(Ω)Rx(i)Rz(ω)rorb (1.80)

The matrix math is done in Section 2.8, which yields

rec = rorb

cos Ω cos(ω + f) − sin Ω sin(ω + f) cos i

sin Ω cos(ω + f) + cos Ω sin(ω + f) cos i

sin(ω + f) sin i

(1.81)

Likewise, m2’s velocity is obtained from

vec = Rz(Ω)Rx(i)Rz(ω)vorb (1.82)

17

Calculating the velocity vorb

Get vorb from the orbit elements:

Start with r =a(1 − e2)

1 + e cos f(1.83)

where vorb = xx + yy (1.84)

x = r cos f and y = r sin f (1.85)

and x = r cos f − rf sin f and y = r sin f + rf cos f (1.86)

but r =nae sin f√

1 − e2(1.87)

and rf =na(1 + e cos f)√

1 − e2(see Eqn.s 1.65–1.67) (1.88)

thus x =na√1 − e2

(e sin f cos f − sin f − e sin f cos f) (1.89)

= −na sin f√1 − e2

(1.90)

similarly, y =na(e + cos f)√

1 − e2(1.91)

Given m2’s orbit elements, a, e, i, Ω, ω, f ,

you can compute the m2’s position and velocity via

formulae 1.81, 1.82, 1.89, 1.91.

18

Assignment #1

1. Confirm that Eqn. (1.81) follows from (1.80).

2. Get orbit elements for comet Wild–2 from the JPL Horizon’s website for

the time when NASA’s Stardust spacecraft encountered the comet.

Convert these orbit elements to ecliptic coordinates and velocities rec and

vec. To confirm your result, convert your coordinates and velocities back

into orbit elements (see Section 2.8 of text).

3. Sketch Wild–2’s orbit, and also indicate the position vectors rec for the

comet and the Earth at the moment of encounter.

4. What was the comet’s right ascension and declination during the

encounter?

Due date: Thursday January 15 at the start of class.

19

The motion as a function of time

Our main result is the equation for an elliptical orbit,

r(f) =a(1 − e2)

1 + e cos f(1.92)

which we obtained by assuming that r is a function of the angle f

and that f is a function of time, and then solving the EOM.

Thus far we have solved the first half of the problem and obtained r(f).

We still need to solve the second part for f(t) and hence r(t),

especially if we wish to solve the assignment...

The following does so via yet another long and seemingly indirect derivation...

Begin with an ellipse of semimajor axis a and eccentricity e

that resides within a circle of radius a:

20

m2 has Cartesian coordinates (x, y) relative to focus where m1 resides,

and has coordinates (X, y) relative to the ellipse center.

From the geometry,(

X

a

)2

+(y

b

)2

= 1 equation for the ellipse, with b = a√

1 − e2 (1.93)

and X = a cos E (1.94)

so(y

b

)2

= 1 − cos2 E = sin2 E (1.95)

where E = eccentric anomaly (1.96)

so y = a√

1 − e2 sin E (1.97)

and x = X − ae = a(cos E − e) (1.98)

so r2 = x2 + y2 = a2[cos2 E − 2e cos E + e2 + (1 − e2) sin2 E] (1.99)

= a2[1 − 2e cos E + e2(1 − sin2 E)] (1.100)

= a2[1 − 2e cos E + e2 cos2 E] (1.101)

= a2(1 − e cos E)2 (1.102)

thus r(E) = a(1 − e cos E) (1.103)

And since

r(f) =a(1 − e2)

1 + e cos f, (1.104)

setting r(f) = r(E) plus some further manipulation

(see page 33 of text) yields

tanf

2=

√

1 + e

1 − etan

E

2. (1.105)

Note that this formula does not suffer any sign ambiguities when relating

f ↔ E, since E and f are always on the same half of the ellipse.

21

The next step is to relate E ↔ t which will ultimately complete our task:

to find the relation between r, f and t.

Start with formulae 1.61, 1.73, and 1.67:

v2 = r2 + (rf)2 = µ

(

2

r− 1

a

)

= n2a3

(

2

r− 1

a

)

(1.106)

where rf =na(1 + e cos f)√

1 − e2=

na2√

1 − e2

r(1.107)

so v2 = r2 +n2a4(1 − e2)

r2= n2a2

(

2a

r− 1

)

(1.108)

and r2 = n2a2

[

2a

r− 1 − a2

r2(1 − e2)

]

(1.109)

=n2a2

r2

[

2ar − r2 − a2 + a2e2]

(1.110)

=n2a2

r2

[

a2e2 − (r − a)2]

(1.111)

and since r = a(1 − e cos E), (1.112)

r = ae sin EE, (1.113)

so r =na

r

√

a2e2 − a2e2 cos2 E (1.114)

=na2e

rsin E = ae sin EE (1.115)

thusdE

dt=

na

r=

n

1 − e cos E(1.116)

(1.117)

To solve this differential equation, form

(1 − e cos E)dE = ndt (1.118)

and integrate: E − e sin E = n(t − τ ) (1.119)

where the integration constant τ = time of periapse passage.

This relationship is used to relate t ↔ r and f , and provides m2’s motion

as a function of time.

22

This equation is usually written as

M = E − e sin E (1.120)

where M = n(t − τ ) = mean anomaly (1.121)

and is known as Kepler’s equation.

If you want to calculate m2’s postion at some time t, ie. for some value of

M , you must solve this transcendental equation numerically.

Section 2.4 of the text describes a numerical method for solving Kepler’s eqn.

However typing this equation into your favorite math software

(MAPLE, MATLAB, MATHEMATICA, etc)

and asking for a numerical solution is quite easy.

23

Recipe for calculating r and v from orbit elements

You are given m2’s orbit parameters a, e, i, Ω, ω, τ and time t.

1. Calculate:

the mean motion n =√

µ/a3 =√

G(m1 + m2)/a3

the mean anomaly M = n(t − τ )

You now have the set or orbit elements a, e, i, Ω, ω, M

that completely specifies m2’s r and v via the following formulae:

2. Solve Kepler’s eqn. M = E − e sin E numerically

for the mean anomaly E and get r(t) and f(t) via

r = a(1 − e cos E) (1.122)

and tanf

2=

√

1 + e

1 − etan

E

2(1.123)

3. Cartesian coordinates for m2’s position & velocity coordinates are

x = r cos f (1.124)

y = r sin f (1.125)

x = −na sin f√1 − e2

(1.126)

y =na(e + cos f)√

1 − e2. (1.127)

which provides rorb and vorb relative to the orbit plane.

4. Use eqns. (1.81)–(1.81) to obtain m2’s ecliptic coordinates rec and vec.

Getting the orbit elements a, e, i, Ω, ω, τ from rec and vec

is straightforward, and is described on pages 52–52.

24

Elliptic expansions of the orbit

Kepler’s eqn. (KE) relates m2’s position↔time via

M = n(t − τ ) = E − e sin E (1.128)

r = a(1 − e cos E). (1.129)

However the relationship between t and r(f) is difficult to use in analytic

studies, principally because KE is a transcendental function of E.

However useful approximate formulae for m2’s motion is possible when the

orbit is nearly circular, ie, in the limit that e 1.

This approximation is often accurate enough when describing the motion

of most of the planets, satellites, and the orbits of dust grains, planetary

ring particles, as well as for binary stars whose orbits have been tidally

circularized.

The following approximation is often marginally true for asteroids that

typically have e ∼ 0.1.

However it is not useful for cometary orbits, stars in clusters, stars in

galaxies, which often have substantially larger e’s.

25

Consider an orbit with e 1, and derive expressions for the true anomaly

f and the radial coordinate r/a as power series in e.

The following derives those expressions to an accuracy of O(e2).

However Section 2.5 of the text derives these and several other useful

formulae to greater accuracy using more rigorous (and laborious) methods...

Start by inserting

E = M + e sin E (1.130)

intor

a= 1 − e cos E (1.131)

= 1 − e cos(M + e sin E) (1.132)

With the trig identities

cos(A + B) = cos A cos B − sin A sin B (1.133)

sin(A + B) = sin A cos B + cos A sin B (1.134)

sin2 A =1

2(1 − cos 2A) (1.135)

cos2 A =1

2(1 + cos 2A) (1.136)

we getr

a= 1 − e cos M cos(e sin E) + e sin M sin(e sin E) (1.137)

Then use the Taylor series expansions

cos x ' 1 − 1

2x2 + O(x4) (1.138)

sin x ' x − 1

6x3 + O(x5) (1.139)

to obtainr

a' 1 − e cos M + e2 sin2 E + O(e3) (1.140)

but E ' M + O(e) so (1.141)r

a' 1 − e cos M +

1

2e2(1 − cos 2M) + O(e3) (1.142)

where M = n(t − τ ) (1.143)

26

The series expansion for f is obtained from

h = r2θ = r2f = na2√

1 − e2 (1.144)

so df =na2

r2

√

1 − e2ndt (1.145)

=√

1 − e2

[

1 − e cos M +1

2e2(1 − cos 2M) + O(e3)

]−2

dM (1.146)

Next, Taylor expand the right hand side (RHS) to O(e2).

This is laborious to do by hand, but a cinch with MAPLE:

crux: /home/jhahn >maple

> eqn:=sqrt(1-e^2)/(1-e*cos(M)+(e^2/2)*(1-cos(2*M)))^2;

√1 − e2

(1 − e cos(M) + 1/2 e2 (1 − cos(2 M)))2(1.147)

> approx:=series(eqn,e,3);

1 + 2 cos(M)e +

(

−3

2+ cos(2 M) + 3 (cos(M))2

)

e2 + O(

e3)

(1.148)

> combine(approx,trig);

1 + 2 cos(M)e + 5/2 cos(2 M)e2 + O(

e3)

(1.149)

Thus

df '[

1 + 2 cos(M)e + 5/2 cos(2 M)e2]

dM (1.150)

can be integrated to yield

f ' M + 2e sin M +5

4e2 sin(2M) + O(e3) (1.151)

Higher–order expressions for r/a and f are given in Section 2.5.

We will use these formulae later when we examine the effects of a planet’s

perturbations upon, say, an asteroid in a low–e orbit,

or the motion of one satellite as it is perturbed by another.

27

The epicyclic approximation

Assume m2 is in a nearly circular orbit, so

r(t) ' a − ae cos nt ≡ a + x(t) where x(t) = −ae cos nt (1.152)

f(t) ' nt + 2e sin nt ≡ nt +y(t)

awhere y(t) = 2ae sin nt (1.153)

The text calls this the guiding center approximation,

but is also known as epicyclic motion.

The guiding center is the point G which travels about m1

on a circular orbit of radius a with a constant angular rate n.

Meanwhile, m2 is at the point x, y, which revolves around point G

in the opposite sense with a radial amplitude ae

(which is known as m2’s epicyclic amplitude),

and a tangential amplitude 2ae.

Note that these formulae are only appropriate for the nearly Keplerian prob-

lem, namely, for the motion of m2 as it orbits in a gravitational potential

that varies as r−1.

When we consider the motion of stars as they orbit in a galaxy’s non–

Keplerian potential, we will derive modified expressions for x and y.

28