celestial mechanics - tufts university

Celestial Mechanics

Kepler’s Laws:I. A planet orbits the Sun in an ellipse, with the Sun at one focus of

the ellipseII. A line connecting a planet to the Sun sweeps out equal areas in

equal time intervals

Kepler’s Laws:I. A planet orbits the Sun in an ellipse, with the Sun at one focus of

the ellipseII. A line connecting a planet to the Sun sweeps out equal areas in

equal time intervalsIII. P[yr]2=a[AU]3, with P orbital period of the planet, a average

distance of the planet from the Sun

log(a) = (2/3) log(P)

BELLIPSE:defined by a set of points satisfying the equationr+r’=2a

Eccentricity:e = FF’/2a0<e<1

If F=F’, then r=r’, then r=aeq. of a CIRCLE (e=0)

A = ⇡ab

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b2 = a2 (1-e2)Relation between semi-major, semi-minor, and ellipticity A = ⇡ab





b2 = a2 (1-e2)Relation between semi-major, semi-minor, and ellipticity

r =a(1� e2)

1 + e cos ✓, 0 e < 1

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If a, e, and theta are known, the position from the focus can be derived

A = ⇡ab





b2 = a2 (1-e2)Relation between semi-major, semi-minor, and ellipticity

r =a(1� e2)

1 + e cos ✓, 0 e < 1

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If a, e, and theta are known, the position from the focus can be derived

PerihelionAphelion

Area of ellipse: A = ⇡ab


Conic curves

E<0E<0

E total energyE<0—> bound systemE>0 —> unbound system

E=0 E>0

0 e < 1

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r =2p

1 + cos ✓

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r =a(e2 � 1)

1 + e cos ✓

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e > 1

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Each type of conic section is related to a specific form of celestial motion

Newton’s Laws:I. An object at rest will remain at rest and an object in motion will

remain in motion in a straight line at a constant speed unless acted upon by an external force (law of inertia)

= the momentum of an object remains constant unless it experiences an external force

p = mv

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II. The net force (the sum of all forces) acting on an object is proportional to the object’s mass and its resultant acceleration

Fnet =nX

i=1

Fi = ma

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m = constant

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a =dv

dt

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Fnet =dp

dt

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The net force on an object is equal to the time rate of change of its momentum





III. For every action there is an equal and opposite reaction

F12 = �F21

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Action and reaction are forces acting on DIFFERENT objects

m1

M2





III. For every action there is an equal and opposite reaction

Newton’s Law of Universal Gravitation:

F = GMm

r2, G = 6.673⇥ 10�11 N m2

kg2

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Work and EnergyThe amount of energy (the work) needed to raise an object of mass m to a height h against a gravitational force is equal to the change in the potential energy of the system:

Work Integral:

For the gravitational force on m being due to a mass M located at the orgin:

Uf � Ui = �U ⌘ �Z rf

ri

F · dr

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F · dr = �Fdr

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�U =

Z rf

ri

GmM

r2dr

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U = �GMm

r(where Uf = 0 at rf = 1)

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Work and EnergyWork must be performed on a massive object if its speed is to be changed:

Kinetic energy of an object

The work done on the particle results in an equivalent change in the particle’s kinetic energy (conservation of energy)

W ⌘ ��U = . . . =1

2mv2f �

1

2mv2i

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K ⌘ 1

2mv2

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W = �K =1

2m(v2f � v2i )

<latexit sha1_base64="ZK8ogzfLRU62PVIv/IqSku4lnSM=">AAACJXicdZDNSgMxFIUz/lv/qi7dXCyCLiwz42AVFERdCG4UrBXaWjJppg1NZoYkI5RhXsaNr+LGhSKCK1/FTFtBRQ8Evpx7L8k9fsyZ0rb9bo2NT0xOTc/MFubmFxaXissr1ypKJKFVEvFI3vhYUc5CWtVMc3oTS4qFz2nN753k9dodlYpF4ZXux7QpcCdkASNYG6tVPKjBITROKdcYznMMJCapk6VuBgI2724NtNKGFBBksA3f7izbahVLdnl/b9f1dsEu23bFcZ0c3Iq344FjnFwlNNJFq/jSaEckETTUhGOl6o4d62aKpWaE06zQSBSNMenhDq0bDLGgqpkOtsxgwzhtCCJpTqhh4H6fSLFQqi980ymw7qrftdz8q1ZPdLDXTFkYJ5qGZPhQkHDQEeSRQZtJSjTvG8BEMvNXIF1sctIm2IIJ4WtT+B+u3bLjlfcvvdLR8SiOGbSG1tEmclAFHaEzdIGqiKB79Iie0Yv1YD1Zr9bbsHXMGs2soh+yPj4B/zCjOA==</latexit>

E = K + U =1

2mv2 �G

Mm

r

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Energy of a particle of mass m with velocity v at a distance r from the center of a larger mass M

At r = 1, v = 0

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E = 0

<latexit sha1_base64="CT6cClPX2pe8f54ocXZuft1KuNk=">AAAB6nicbVDLSgNBEOyNrxhfUY9eBoPgKexKQD0IQRE8RjQPSJYwO5kkQ2Znl5leISz5BC8eFPHqF3nzb5wke9DEgoaiqpvuriCWwqDrfju5ldW19Y38ZmFre2d3r7h/0DBRohmvs0hGuhVQw6VQvI4CJW/FmtMwkLwZjG6mfvOJayMi9YjjmPshHSjRF4yilR5ur9xuseSW3RnIMvEyUoIMtW7xq9OLWBJyhUxSY9qeG6OfUo2CST4pdBLDY8pGdMDblioacuOns1Mn5MQqPdKPtC2FZKb+nkhpaMw4DGxnSHFoFr2p+J/XTrB/4adCxQlyxeaL+okkGJHp36QnNGcox5ZQpoW9lbAh1ZShTadgQ/AWX14mjbOyVylf3ldK1essjjwcwTGcggfnUIU7qEEdGAzgGV7hzZHOi/PufMxbc042cwh/4Hz+AIyhjVY=</latexit>

1

2mv2 = G

Mm

r

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vesc =p2GM/r

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The escape velocity is independent on m

Center of mass

position vector R as the weighted average of the position vectors of the individual massesR ⌘ m1r1 +m2r2 + . . .

m1 +m2 + . . .

<latexit sha1_base64="fMSt6o8AXStkQ+4cw7noGE2qKCw=">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</latexit>

MR =nX

i=1

miri

<latexit sha1_base64="+a5n9KPdoewtV8CoRJuHucpB+6U=">AAACEnicbVDLSsNAFJ34rPUVdelmsAi6KYkU1EWh6MaNUMU+oIlhMp20Q2cmYWYilJBvcOOvuHGhiFtX7vwbp4+Fth64cDjnXu69J0wYVdpxvq2FxaXlldXCWnF9Y3Nr297Zbao4lZg0cMxi2Q6RIowK0tBUM9JOJEE8ZKQVDi5HfuuBSEVjcaeHCfE56gkaUYy0kQL7+BpmXhjB2xxWoadSHmS06ub3mcgzHtCJKQOa54FdcsrOGHCeuFNSAlPUA/vL68Y45URozJBSHddJtJ8hqSlmJC96qSIJwgPUIx1DBeJE+dn4pRweGqULo1iaEhqO1d8TGeJKDXloOjnSfTXrjcT/vE6qozM/oyJJNRF4sihKGdQxHOUDu1QSrNnQEIQlNbdC3EcSYW1SLJoQ3NmX50nzpOxWyuc3lVLtYhpHAeyDA3AEXHAKauAK1EEDYPAInsEreLOerBfr3fqYtC5Y05k98AfW5w8cqp0v</latexit>

r = r02 � r01

<latexit sha1_base64="rc5i/ethk5ardnrxjxKSA+EL9Ho=">AAACHnicbZDLSgMxFIYz9VbrbdSlm2AR3FhmSkVdCEU3LivYC3TGkkkzbWiSGZKMUIZ5Eje+ihsXigiu9G1M21lo6w+Bj/+cw8n5g5hRpR3n2yosLa+srhXXSxubW9s79u5eS0WJxKSJIxbJToAUYVSQpqaakU4sCeIBI+1gdD2ptx+IVDQSd3ocE5+jgaAhxUgbq2efpl4QQpnBSzij+9SLJeUk66XVLIMni7abZT277FScqeAiuDmUQa5Gz/70+hFOOBEaM6RU13Vi7adIaooZyUpeokiM8AgNSNegQJwoP52el8Ej4/RhGEnzhIZT9/dEirhSYx6YTo70UM3XJuZ/tW6iw3M/pSJONBF4tihMGNQRnGQF+1QSrNnYAMKSmr9CPEQSYW0SLZkQ3PmTF6FVrbi1ysVtrVy/yuMoggNwCI6BC85AHdyABmgCDB7BM3gFb9aT9WK9Wx+z1oKVz+yDP7K+fgDtiqJs</latexit>

displacement vector

MV =nX

i=1

mivi

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R is the position of the center of mass of the system, and V is the velocity of the center of mass. P=MV is the momentum of the center of mass.

Center of massr = r02 � r01


displacement vector

If all forces acting on individual particles in the system are due to other particles contained within the system (Newton’s 3rd law):

F =dP

dt= 0

<latexit sha1_base64="ACqJ6asCsSkyNABAkTXBkPk4DVA=">AAACCnicbZDLSsNAFIYn9VbrLerSzWgRXJVECupCKArisoK9QBPKZDJph04mYWYilJC1G1/FjQtF3PoE7nwbJ2kW2vrDwM93zuHM+b2YUaks69uoLC2vrK5V12sbm1vbO+buXldGicCkgyMWib6HJGGUk46iipF+LAgKPUZ63uQ6r/ceiJA04vdqGhM3RCNOA4qR0mhoHqaOF8CbDF5CJxAIp34B2lmW+iqn1tCsWw2rEFw0dmnqoFR7aH45foSTkHCFGZJyYFuxclMkFMWMZDUnkSRGeIJGZKAtRyGRblqcksFjTXwYREI/rmBBf0+kKJRyGnq6M0RqLOdrOfyvNkhUcO6mlMeJIhzPFgUJgyqCeS7Qp4JgxabaICyo/ivEY6QTUTq9mg7Bnj950XRPG3azcXHXrLeuyjiq4AAcgRNggzPQAregDToAg0fwDF7Bm/FkvBjvxsestWKUM/vgj4zPHz+UmV0=</latexit>

i.e., the center of mass does not accelerate if no external force exists, i.e., the reference frame associated to the center of mass is inertial



displacement vector

It is therefore convenient to place the coordinate center at the center of mass, choosing R=0. Substituting r2=r1+r, and defining the reduced mass (for a binary system) as: µ ⌘ m1m2

m1 +m2

<latexit sha1_base64="MeYeCdQMw6EK5pIRkzmTPySUZLQ=">AAACDXicbZDLSgMxFIYz9VbrbdSlm2AVBKHMlIK6K7pxWcFeoFOGTJppQ5PMmGQKZegLuPFV3LhQxK17d76N6XQW2vpD4OM/53By/iBmVGnH+bYKK6tr6xvFzdLW9s7unr1/0FJRIjFp4ohFshMgRRgVpKmpZqQTS4J4wEg7GN3M6u0xkYpG4l5PYtLjaCBoSDHSxvLtE48n0CMPCR1DL5QIp9x3Ifer0wzOM/TtslNxMsFlcHMog1wN3/7y+hFOOBEaM6RU13Vi3UuR1BQzMi15iSIxwiM0IF2DAnGieml2zRSeGqcPw0iaJzTM3N8TKeJKTXhgOjnSQ7VYm5n/1bqJDi97KRVxoonA80VhwqCO4Cwa2KeSYM0mBhCW1PwV4iEymWgTYMmE4C6evAytasWtVa7uauX6dR5HERyBY3AGXHAB6uAWNEATYPAInsEreLOerBfr3fqYtxasfOYQ/JH1+QN47JqM</latexit>

r1 = � µ

m1r

<latexit sha1_base64="Mz0X+FvTPfV/UeEmxHrgtNfdbj8=">AAACDHicbVDLSgMxFL3js9ZX1aWbYBHcWGakoC6EohuXFewDOqVk0kwbmmSGJCOUYT7Ajb/ixoUibv0Ad/6NaTsLbT0QOJxzLjf3BDFn2rjut7O0vLK6tl7YKG5ube/slvb2mzpKFKENEvFItQOsKWeSNgwznLZjRbEIOG0Fo5uJ33qgSrNI3ptxTLsCDyQLGcHGSr1SOfWDEKmel6ErdOqHCpPUF0mWCivNvMym3Io7BVokXk7KkKPeK335/YgkgkpDONa647mx6aZYGUY4zYp+ommMyQgPaMdSiQXV3XR6TIaOrdJHYaTskwZN1d8TKRZaj0VgkwKboZ73JuJ/Xicx4UU3ZTJODJVktihMODIRmjSD+kxRYvjYEkwUs39FZIhtIcb2V7QlePMnL5LmWcWrVi7vquXadV5HAQ7hCE7Ag3OowS3UoQEEHuEZXuHNeXJenHfnYxZdcvKZA/gD5/MHslGazg==</latexit>

r2 =µ

m2r

<latexit sha1_base64="F7FDZH6b2DQtvTf/CbuOo7I7uZE=">AAACC3icbVDLSsNAFJ3UV62vqEs3Q4vgqiSloC6EohuXFewDmhAm00k7dGYSZiZCCdm78VfcuFDErT/gzr9x2mahrQcuHM65l3vvCRNGlXacb6u0tr6xuVXeruzs7u0f2IdHXRWnEpMOjlks+yFShFFBOppqRvqJJIiHjPTCyc3M7z0QqWgs7vU0IT5HI0EjipE2UmBXMy+MoAwaObyCXiQRzjye5hk3ysLKA7vm1J054CpxC1IDBdqB/eUNY5xyIjRmSKmB6yTaz5DUFDOSV7xUkQThCRqRgaECcaL8bP5LDk+NMoRRLE0JDefq74kMcaWmPDSdHOmxWvZm4n/eINXRhZ9RkaSaCLxYFKUM6hjOgoFDKgnWbGoIwpKaWyEeIxOINvFVTAju8surpNuou8365V2z1rou4iiDE1AFZ8AF56AFbkEbdAAGj+AZvII368l6sd6tj0VrySpmjsEfWJ8/RGmamQ==</latexit>

E =1

2µv2 �G

Mµ

r

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i.e., the total energy of the system is equal to the kinetic energy of the reduced mass plus the potential energy of the reduced mass moving about a mass M located and fixed at the origin.



displacement vector

Total angular momentum:

L = m1r1 ⇥ v1 +m2r2 ⇥ v2 = µr⇥ v = r⇥ p

<latexit sha1_base64="aRHCMkp/UcNG4M0pCTuQ9OTg6OY=">AAACaHicbVHLSsQwFE3ra5zxUV+IuAkOgiAMbRHUhSC6ceFCwXGEaSlpJnWCSVuSdGAoxX905we48StMOxW0eiFwcs65uclJmDIqlW2/G+bc/MLiUmu53VlZXVu3NjYfZZIJTPo4YYl4CpEkjMakr6hi5CkVBPGQkUH4cl3qgwkRkibxg5qmxOfoOaYRxUhpKrBecy+M4G0BLyAPHFjtROAU0FOUEzkjJiVxrA3ut8FtGtzyBI9ntaEhl2LuCd4U0iKwunbPrgr+BU4NuqCuu8B680YJzjiJFWZIyqFjp8rPkVAUM1K0vUySFOEX9EyGGsZID/PzKqgCHmpmBKNE6BUrWLE/O3LEpZzyUDs5UmPZ1EryP22YqejMz2mcZorEeDYoyhhUCSxThyMqCFZsqgHCguq7QjxGAmGl/6atQ3CaT/4LHt2ec9I7vz/pXl7VcbTAPjgAR8ABp+AS3IA70AcYfBgdY9vYMT5Ny9w192ZW06h7tsCvMg++AGkvtYI=</latexit>

i.e., the total angular momentum equals the angular momentum of the reduced mass onlyThe two-body problem can be treated as an equivalent one-body problem with the reduced mass moving about a fixed mass M at a distance r.µ

<latexit sha1_base64="78EuS6LOXAG8aEjxmwocUWGHntw=">AAAB6nicdVDJSgNBEK2JW4xb1KOXxiB4CjMiqLegF48RzQLJEHo6PUmT7p6hu0YIIZ/gxYMiXv0ib/6NnUWI24OCx3tVVNWLUiks+v6Hl1taXlldy68XNja3tneKu3t1m2SG8RpLZGKaEbVcCs1rKFDyZmo4VZHkjWhwNfEb99xYkeg7HKY8VLSnRSwYRSfdtlXWKZaCsj8F8X+RL6sEc1Q7xfd2N2GZ4hqZpNa2Aj/FcEQNCib5uNDOLE8pG9AebzmqqeI2HE1PHZMjp3RJnBhXGslUXZwYUWXtUEWuU1Hs25/eRPzLa2UYn4cjodMMuWazRXEmCSZk8jfpCsMZyqEjlBnhbiWsTw1l6NIpLIbwP6mflIPT8sXNaalyOY8jDwdwCMcQwBlU4BqqUAMGPXiAJ3j2pPfovXivs9acN5/Zh2/w3j4BYr6N4w==</latexit>

dL

dt=

d

dt(r⇥ p) =

dr

dt⇥ p+ r⇥ dp

dt= v ⇥ p+ r⇥ F = 0

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L of a system is a constant for a central law force

Revisited Kepler’s 1st Law

r =L2/µ2

GM(1 + e cos ✓)General equation of a conic section

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i.e., the path of the reduced mass about the center of mass under the influence of gravity is a conic section.

Elliptical orbits result from an attractive r-2 central-force law (i.e., gravity) when the total energy of the system is less than 0 (E<0, bound system); parabolic trajectories when E=0; and hyperbolic trajectories when E>0 (unbound systems).

NOTE: both objects in a binary system move about the center of mass in ellipses, with the center of mass occupying one focus of each ellipse.

For closed planetary orbits, comparing the above with the previous equation of an ellipse, I obtain the total angular momentum of the system (max for e=0, i.e., circular motions):

L = µp

GMa(1� e2)

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Derivation of Kepler’s 2nd Law

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Derivation of Kepler’s 2nd Law

Kepler’s revised1st law

Derivation of Kepler’s 3rd Law

celestial mechanics - tufts university

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