notes for celestial-mechanics, j. hedberg © 2018

Celestial Mechanics1. Basic Orbits

1. Why circles?2. Tycho Brahe3. Johannus Kepler4. Kepler's 3 laws of orbiting bodies

2. Newtonian Mechanics3. Newton's Laws

1. Law of Gravitation2. The shell theorem.3. Work and Energy

4. Potential Energy1. Potential Plot2. Escape Speed

5. Kepler's Laws1. Cartesian & Polar Coordinates2. 1st Law - Ellipses

6. The Virial Theorem1. The Vis Viva equation2. Hohmann Transfer Orbit

7. Lagrange Points

Basic Orbits

Why circles?Copernicus was right: the earth and the other planets do orbit around the sun. However,he was still operating under the assumption that all the celestial bodies had to move inperfect circles. This turned out to be a rather major flaw in the framework.

Other astronomers of the era were unable to match the predictions of the Copernicansystem with observations. More 'tweaks' were added to the Copernican systems ofcircular heliocentric orbits in order to 'save the appearances'. In order to make themodel match with observations, they still had to employ epicycles and so forth.However, the seed was planted. The sun should be at the center!

Tycho Brahe

Fig.1The CopernicanSystem consisted of circularorbits centered around thesun.

PHY 454 - celestial-mechanics - J. Hedberg - 2018

updated on 2018-09-06

Late 16th century. Hybrid system: geoheliocentric model had the planets going aroundthe sun, but the sun still went around the Earth. Close...

Johannus KeplerKepler was rad

Kepler played a major role in the 17th-century scientific revolution. He is best known forhis laws of planetary motion, based on his works Astronomia nova, Harmonices Mundi,and Epitome of Copernican Astronomy. These works also provided one of thefoundations for Newton's theory of universal gravitation.

Kepler's 3 laws of orbiting bodies1. A planet orbits the sun in an ellipse. The Sun is at one focus of that ellipse.2. A line connecting a planet to the Sun sweeps out equal areas in equal times3. The sqaure of a planet's orbital period is proportional to the cube of the average distance between the planetand the sun: .

Fig.2Danish astronomerTycho Brahe [1546-1601].Made many very goodmeasurements of the starsand planets.

Fig.3Germanmathematician, astronomer,and astrologer. [1571-1630]Kepler was Tycho'sassistant and was believedin the Copernican system.

∝P 2 a3



Ellipses

r’ r

a

b

ae FF’

An ellipse satisfies this equation:

is the semi-majoraxis and are the distances to the ellipse from the two focalpoints, and . represents the eccentricity of the ellipse ( )

The semi-majoraxis is one-half the length of the long (i.e. major) axis of the ellipse.

Conic SectionsThe circle is one of several conic sections. (There's really nothing all that special about circles.)

See Appolonius - On Conics.

Elliptical OrbitsThe first law for Kepler holds that the orbiting body (i.e. planet) will trace out anelliptical path as it orbits the central body (i.e. the sun). The central body islocated at the principlefocus. When the distance between the planet and thesun is the shortest, the planet is said to be at perihelion, or the point of closesapproach. When that distance is greatest, the planet is located the aphelion.

Polar Coordinates

r+ = 2ar′ (1)

a

r r′ F F ′

e 0 ≤ e ≤ 1

principlefocus

aphelion perihelion

orbitingbody

centralbody

Fig.4

r =a(1 − )e2

1 + e cos θ(2)



The polar coordinate equation can be obtain by starting with the basic ellipseequation Eq. Take the point at the top of the semi-minor axis, where

. Since we know that . Using the Pythagoreantheorem, we can say that:

Putting in for yields:

For polar coordinates, we'll use and , being the distance from theprinciple focus and the angle measured counterclockwise from major axis ofthe ellipse. Again, from Pythagorus:

Expanding the 2nd RHS term:

and using the fact that , we can obtain the polar equation forthe ellipse shown above in eq .

Deviation from a Circles

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

e = 0

e = 0.0934

Mars' orbit [ ] looks a lot like a circle.

Equal Areas in Equal TimesKepler's second law describes the areas swept out by an orbiting planet in agiven time and says that for any given interval of time, the areas swept out willalways be equal.

r’ r

a

b

ae

F

F’

Fig.5Ellipse - the point where r = r′

(1)r = r′ r+ = 2ar′ r = a

= +r2 b2 a2e2

a r

= (1 − )b2 a2 e2

rr’

F

F’ θ

ae

Ellipse - and polar coordinates and .r θ

r θ r

θ

= θ+ (2ae+ r cos θr′2 r2 sin2 )2

r′2 = θ+ 4 + 2aer cos θ+r2 sin2 a2e2 r2 cos= + 4ae (ae+ r cos θ)r2

=r′2 (2a− r)2

(2)

e = 0.0934

focus

t1

t2

areaorbiting

body

focus

t1

t2area

Fig.6The blue areas in these figureswill be the same if is the same.−t2 t1



The other conicsParabolas:

Hyperbolas:

Newtonian Mechanics

Newton's LawsIn words:

1. No change in velocity if there is no net force acting.2. The net force acting on a body will cause a proportional acceleration inversely proportional to the mass of thatbody.3. Forces come in equal and opposite pairs

In symbols:1. if 2. 3.

Law of GravitationThe scalar form:

And in vector form:

This is Newton's LawofUniversalGravitation . It considers two masses: and and the distance between them . Also included is , the UniversalGravitationalConstant. ( m kg s From nist

http://physics.nist.gov/cgi-bin/cuu/Value?bg). The direction of the force is always attractive and is along a lineconnecting the centers of the two masses.

is the force applied on object 2 by object 1. is the distance between the two objects ( )

And is the unit vector pointing from object 1 to object to 2.

e = 1

r =2p

1 + cos θ(5)

e > 1

r =a( − 1)e2

1 + e cos θ(6)

Δv = 0 = 0Fnet

= = maFnet ∑ni=1 Fi

= −FAB FBA

1 2

Fig.7Two masses and a gravitationalinteraction

F = GMm

r2

= −GF21Mm

r2r

M m

r G G = 6.674 08 × 10−11 3 -1 -2

F21

r |r12|2

r12

=r12−r2 r1

| − |r2 r1



http://physics.nist.gov/cgi-bin/cuu/Value?bg

The shell theorem.

Show that a spherical mass (of density ) acts as a point mass for objectsoutside.

Consider a mass a distance away from the center of a spherical mass . The radius of the spherical mass is . First, the force from the ring which

a mass is:

We can re-express the mass of the ring in terms of its density . (The density is at most only a function of theradius, i.e. spherically symmetric)

The volume of a ring is just its thickness: , times its width: , times its circumference: whichequals:

The can be expressed as:

and can be written as the following (from Pythagorus)

Let's put all these back into the force from the ring:

Then, we'll have to integrate over both variables. will range from 0 to and from 0 to to get all the ringsthat make up the entire sphere.

Integrating over yields:

Example Problem #1:

ρ(R)

m r

M R0

dM

d = G cosϕFringm dMring

s2

ρ(R)

d = ρ(R) dMring Vring

dR Rdθ 2πR sin θ

d = ρ(R) 2π sin θ dR dθMring R2

cosϕ

cosϕ =r− R cos θ

s

s

s = =+ θ(r− R cos θ)2 R2 sin2− −−−−−−−−−−−−−−−−−−√ − 2rR cos θ+r2 R2− −−−−−−−−−−−−−−√

dFring = G cosϕm dMring

s2

= Gm [ρ(R) 2π sin θ dR dθ] × [ ] × [ ]R2 1s2

r− R cos θs

dR R0 dθ π

F = 2πGm[ dθ dR]∫ R0

0∫ π

0

ρ(R)(r sin θ− sin θ cos θ)R2 R3

( + − 2rR cos θ)r2 R2 3/2(9)

θ

F = 4π ρ(R) dRR0

2



However, the mass of a shell of thickness is just:

which is just quanity inside the integrand. Thus the force from on shell will be,

This force acts as if it is located at the center of the shell.

We can then integrate over all the mass shells and the above becomes:

which is just the equation for the force between two point masses.

little g

Let's consider the gravitational force a distance above the surface of the earth. Since , we can see that theacceleration will be equal to:

We generally call this value 'little g'.

Work and Energy

The change in potential energy is given by the negative of the work done bygravity. Expressing it in the integral above allows for changing force as afunction of position (in contrast to the elementary definition:

Potential Energy

F = 4π ρ(R) dRGm

r2∫ R0

O

R2 (10)

dR

d = 4π ρ(R)dRMshell R2

d =FshellGm dMshell

r2

F = GmM

r2(11)

F = GmM⊕

( + hR⊕ )2(12)

h F = ma

a

a = G = gM⊕

R2⊕

(13)

z

x

yM

r i

r f

drm

F

Fig.8A point mass moves from to in a gravitational field.

m rirf

− = ΔU = − F ⋅ drUf Ui ∫ rf

ri

W = F ⋅ d)



Find the work done on an object by gravity as it moves from point to infinity,where the potential energy is defined as 0.

The force is always against the displacement, so

We can pull the constants out of the integral, and we are left with somethingvery simple to integrate:

Evaluating at the limits of and , we see that the work done is:

From the original definition of work and potential energy:

thus, since we can say: And now we have a potential energy function for masses at a distance from the center of the Earth:

Potential Plot

Escape SpeedHow fast to get something to reach infinity, (and stop there)?

Fig.9What is the potential energy atpoint ?P

P

W = F(r) ⋅ dr∫ ∞

R

F(r) ⋅ dr = − drG mME

r2

W = −G m dr =ME ∫ ∞

R

1r2

[ ]G mME

r

∞

R

∞ R

W = = 0 − = −[ ]G mME

r

∞

R

G mME

R

G mME

R

ΔU = − = −WUf Ui

= 0U∞ U =W r

U = −G mMe

r

r

U

The potential energy as a function ofposition for a body in a gravitationalinteraction.



This is the total energy of a particle: kinetic plus potential. Since the particle will be at and will have stoppedthen the total energy must be zero. However, due to conservation of energy, the total mechanical energy will be thesame. Thus, at all times. Or:

which leads to

For Earth, = 11182.4 m/s. Apollo 11 traveled at 10423 m/s [nasa ref]. It got a little help from the moons gravitytoo.

Kepler's Laws

Cartesian & Polar CoordinatesIn polar coordinates, we have different unit vectors: and .

and

We can also express some derivates:

And using the chain rule, we can also express time derivatives:

These will be useful later.

1st Law - EllipsesThe angular momentum per unit mass will be given by:

E = m −G12

v2Mm

r(15)

r = ∞

m −G = 012 v2 Mm

r

m = G12

v2Mm

r

=vescape2GMr

− −−−−√ (16)

vescape

y

x

z

y x

m

θ

r

θ

cartesian polar

Fig.10

θ r

= cos θ+ sin θr i j

= − sin θ+ cos θθ i j

= − sin θ+ cos θ =drdθ

i j θ (19)

= − cos θ− sin θ = −dθ

dθi j r (20)

= =drdt

drdθ

dθ

dtθdθ

dt(21)

= = −dθ

dt

dθ

dθ

dθ

dtrdθ

dt(22)

=L 2 dθ



https://history.nasa.gov/SP-4029/Apollo_18-24_Translunar_Injection.htm

and the centrally directed gravitational force is given by:

The acceleration of the orbiting body is therefore:

but

We can put eq into the acceleration equation

Combining this with

Integrating both sides of this equation yields:

The is a constant of integration that depends on the initial conditions, which we may choose.

We'll say that at , the orbiter is at perihelion and we can orient the axesso that its motion is entirely in the +y direction, in other words: where is a constant. Now:

We can take the dot product with of both sides:

Simplifying, using the definitions above (i.e. ):

=L

mr2dθ

dt(23)

= − = mFGMm

r2r

dvdt

(24)

a = = −dvdt

GM

r2r (25)

= −r ( )dθ

dt

−1dθ

dt(26)

(26) (25)

=dvdt

GM

r2( )dθ

dt

−1dθ

dt(27)

(23)

=L

GMm

dvdt

dθ

dt(28)

= +L

GMmv θ e (29)

e

y

xsun

Fig.11The initial conditions of anorbiting body.

t= 0= ee j e

= + eL

GMmv θ j

θ

⋅ = ⋅ + e ⋅L

GMmv θ θ θ j θ

(18)

= 1 + e cos θL

GMmvt (32)



( ) is the tangentialvelocity of the orbiter. From the definition of angular momentum:

which when put back into , yields:

or solving for :

This is just the polar coordinates expression for a conic section.

The Virial TheoremThe Virialtheorem shows that for a gravitationally bound system in equilibrium, the total energy is one-half the timeaveraged potential energy.

Consider the quantity .

where is the linear momentum and the position of some particle . The product rule gives the following if wetake the time derivative of the right hand side.

We could also say:

where is the moment of inertia of the total number of particles:

Now,

Thus:

vt L = r × p

mr = Lvt (33)

(32)

= 1 + e cos θL2

GM rm2(34)

r

r =L2

GM (1 + e cos θ)m2(35)

Q

Q ≡ ⋅∑i

pi ri (36)

pi ri i

= ( ⋅ + ⋅ )dQ

dt∑i

dpidt

ri pidridt

(37)

= ⋅ = ( ) =dQ

dt

d

dt∑i

mi

dridt

rid

dt∑i

12d

dtmir

2i

12

Id2

dt2(38)

I

I = ∑i

mir2i (39)

− ⋅ = ⋅12

Id2

dt2∑i

pidridt

∑i

dpidt

ri (40)

− ⋅ = − ⋅ = −2 = −2K∑i

pidridt

∑i

mivi vi ∑i

12miv

2i (41)



where we have used the 2nd law of Newton ( )

Let represent the force of interaction between two particles in the system. The force on due to . If you have aparticle , then we need to add up the force contributions from every other particle where .

We can re-write the position of the th particle as:

which leads to

Since the 3rd law of Newton says: , the first term on the R.H.S. in the above equation will be zero. Thus:

Assuming only gravitational interactions:

we can write

The potential energy between the and particles is just the term in the sums:

− 2K = ⋅12

Id2

dt2∑i

Fi ri (42)

F = dpdt

Fij i j

i j j ≠ i

⋅ = ⋅∑i

Fi ri ∑i

⎛⎝⎜⎜∑

jj≠i

Fij

⎞⎠⎟⎟ ri (43)

i

= ( + ) + ( − )ri12

ri rj12

ri rj (44)

⋅ = ⋅ ( + ) + ⋅ ( − )∑i

Fi ri12

∑i

⎛⎝⎜⎜∑

jj≠i

Fij

⎞⎠⎟⎟ ri rj

12

∑i

⎛⎝⎜⎜∑

jj≠i

Fij

⎞⎠⎟⎟ ri rj (45)

= −Fij Fji

⋅ = ⋅ ( − )∑i

Fi ri12

∑i

∑jj≠i

Fij ri rj (46)

= GFijmimj

r2ijr ij (47)

⋅∑i

Fi ri = − G12

∑i

∑jj≠i

mimj

r3ij( − )rj ri

2

− G12

∑i

∑jj≠i

mimj

rij

(48)

(49)

jUi i j

= −GUijmimj

jri(50)



Thus,

Now we can fill out eq:

where we have replaced all the quantities with the time averaged values. The average of is going to be:

but for periodic motions like orbits:

So

which leads finally to

or in terms of the total energy

The Virial Theorem

or in terms of the total energy

⋅ = = U∑i

Fi ri12

∑i

∑jj≠i

Uij (51)

(42)

⟨ ⟩ − 2 ⟨K⟩ = ⟨U⟩12

Id2

dt2(52)

Id2

dt2

⟨ ⟩Id2dt2

= dt1τ

∫ τ

0

Id2

dt2

= ( − )1τ

dI

dt

∣∣∣τ

dI

dt

∣∣∣0

(53)

(54)

=dI

dt

∣∣∣τ

dI

dt

∣∣∣0

(55)

⟨ ⟩ = 0Id2

dt2(56)

−2⟨K⟩ = ⟨U⟩

E

⟨E⟩ = ⟨U⟩12

−2⟨K⟩ = ⟨U⟩

E



The Virial Theorem: UsesIt allows for statistical results for many body systems.

For example, with galaxies: Since and , we can obtain:

which relates two observable values, and , with a non observable, but very interesting value, the mass of thegalaxy.

The Vis Viva equation

Hohmann Transfer Orbit

Lagrange PointsThe basic free body diagram showing the forces on a small body near twolarger ones.

There are five Lagrange points. Some are interesting for technologyapplications.

⟨E⟩ = ⟨U⟩12

K = M12 v2 U = GM 2

R

M =Rv2

G

v R

= GM( − )v22r

1a

Fsun Fearth

objectsun earth

R

r

Fig.12The basic FBD and L1 point

sun L1L1 L2L2

L4L4

L3L3

L5L5

Fig.13All 5 Lagrange points



notes for celestial-mechanics, j. hedberg © 2018

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