Two-Body Systems
Two-Body Force
• A two-body system can be defined with internal and external forces.– Center of mass R
– Equal external force
• Add to get the CM motion
• Subtract for relative motion
ext1
int111 FFrm
m2
r1
F2int
r2R m1
F1int
F1ext
F2extr = r1 – r2
ext2
int222 FFrm
ext2
ext1 FFRM
2
int2
1
int1
21 m
F
m
Frr
Reduced Mass
• The internal forces are equal and opposite.
• Express the equation in terms of a reduced mass .– less than either m1, m2
– approximately equals the smaller mass when the other is large.
int
212
int
1
int
21 )11
( Fmmm
F
m
Frr
intint
21
2121 )(
FF
mm
mmrr
221
21 mmm
mm
21 mm for
Central Force
,,:
,,:
rq
zyxx
q
xFQ
m
i
i m
iim
• The internal force can be expressed in other coordinates.– Spherical coordinates
– Generalized force
• A force between two bodies can only depend on r.– Central force
0
q
xFQ
i r
iir
Kinetic Energy
• The kinetic energy can be expressed in spherical coordinates.– Use reduced mass
• Lagrange’s equations can be written for a central force.– Central force need not be from a
potential.
rQr
T
r
T
dt
d
0
TT
dt
d
0
TT
dt
d
)sin( 22222221 rrrT
Coordinate Reduction
• T doesn’t depend on directly.
• The angular momentum about the polar axis is constant.– Planar motion
– Include the polar axis in the plane
• This leaves two coordinates.– r,
0
TT
dt
d
0T
dt
d
22 sinrT
constant
)( 22221 rrT
Angular Momentum
• T also doesn’t depend on directly.– Constant angular momentum
– Angular momentum J to avoid confusion with the Lagrangian
0
TT
dt
d
0T
dt
d
JrT
2 constant
• Central motion takes place in a plane.
– Force, velocity, and radius are coplanar
• Orbital angular momentum is constant.
• If the central force is time-independent, the orbit is symmetrical about an apse.
– Apse is where velocity is perpendicular to radius
Central Motion
Central Potential
• The central force can derive from a potential.
• Rewrite as differential equation with angular momentum.
• Central forces have an equivalent Lagrangian.
r
VQ
r
T
r
T
dt
dr
Vr
JrL
2
22
21
2
03
2
r
V
r
Jr
Time Independence
• Change the time derivative to an angle derivative.
• Combine with the equation of motion.
• The resulting equation describes a trajectory.
d
d
r
J
d
d
dt
d
dt
d2
rQr
T
r
T
dt
d
rQr
T
r
T
d
d
r
J
2
Orbit Equation
Let u = 1/r
rQr
rr
r
rr
d
d
r
J
)]([)]([ 222
21222
21
2
rQr
J
d
dr
r
J
d
d
r
J
r
Jr
d
rd
r
J
3
2
222
22)()(
2322
1)
1(
1
J
Q
rd
dr
rd
d
rr
222
2
uJ
Qu
d
ud r
• The solution to the differential equation for the trajectory gives the general orbit equation.
Inverse Square Force
• The inverse square force is central.– < 0 for attractive force
• Choose constant of integration so V() = 0.
rV
m2
r1
F2int
r2R m1
F1int
r = r1 – r2
r
V
rQr
2
21
21
mm
mm
Kepler Lagrangian
• The inverse square Lagrangian can be expressed in polar coordinates.
• L is independent of time.– The total energy is a constant
of the motion.
– Orbit is symmetrical about an apse.
rrrVTL
)( 22221
rr
JrVTE
2
2
212
21
)( 22221 rrT
rV
Kepler Orbits
• The right side of the orbit equation is constant.– Equation is integrable
– Integration constants: e, 0
– e related to initial energy
– Phase angle corresponds to orientation.
• The substitution can be reversed to get polar or Cartesian coordinates.
2222
2
JuJ
Qu
d
ud r
))cos(( 0 rser
)]cos(1[ 02
eJ
u
e
Js
2
r
u1
)]cos(1[11
0 eesr
Conic Sections
focus
r
s
)cos( rser
• The orbit equation describes a conic section.– init orientation (set to 0)
– s is the directrix.
• The constant e is the eccentricity.– sets the shape
– e < 1 ellipse
– e =1 parabola
– e >1 hyperbola
Apsidal Position
• Elliptical orbits have stable apses.– Kepler’s first law
– Minimum and maximum values of r
– Other orbits only have a minimum
• The energy is related to e:– Set r = r2, no velocity
)cos1(11 eesr
r
sr1 r2
e
esr
12e
esr
11
21
2
2
)2
1(EJ
e
Angular Momentum
• The change in area between orbit and focus is dA/dt– Related to angular velocity
• The change is constant due to constant angular momentum.
• This is Kepler’s 2nd law
2
JA
r
dr
221
21 rrrA
2rJ
Period and Ellipse
• The area for the whole ellipse relates to the period.– semimajor axis: a=(r1+r2)/2.
• This is Kepler’s 3rd law.– Relation holds for all orbits
– Constant depends on
22 2
322
3 Ja
JaA
2
2222 2
1
EJaeaA
2
32 a
A
AT
r
sr1 r2
Effective Potential
• The problem can be treated in one dimension only.– Just radial r term.
• Minimum in potential implies bounded orbits.– For > 0, no minimum
– For E > 0, unbounded
rr
JVeff
2
2
2
effr VTrr
JrE
2
2
212
21
Veff
0 r
Veff
0 r
unboundedpossibly bounded
Star Systems
• Star systems within 10 Pc have been cataloged by RECONS (Jan 2012).– Total systems 259
– Singles 185
– Doubles 55
– Triples 15
– Quadruples 3
– Quintuples 1
• Star systems can involve both single and multiple stars.
• Binary stars are a case of a two-body central force problem.
Visual Binaries
• Visual binaries occur when the centers are separated by more than 1”.– Atmospheric effects
• Apparent binaries occur when two stars are near the same coordinates but not close in space.
Binary Mass
• Kepler’s third law can be made unitless compared to the sun.
– Mass in solar masses
– Period in years
– Semimajor axis in AU
• Semimajor axis depends on knowing the distance and tilt.
• Separate masses come from observing the center.
3221 )( aPMM
/aa
2211 aMaM
/)( 221 aPMM
aaa 21
Spectroscopic Binaries
• Binary systems that are too close require spectroscopy.
– Doppler shifted lines
– Velocity measurements
2/1 VPr
21 rra
2/2 vPr
2321 / PaMM
VvrrMM /// 1221
Eclipsing Binaries
• An orbit inclination of nearly 90° to the observer produces an eclipsing binary.
• Light levels are used to measure period and radii.
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