Post on 12-Apr-2016
description
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 mF
mFrr
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
FmFrr
intint
21
2121 )( FF
mmmmrr
221
21 mmm
mm
21 mm for
Central Force
,,:,,:
rqzyxx
qxFQ
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
qxFQ
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.
rQrT
rT
dtd
0
TT
dtd
0
TT
dtd
)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
dtd
0T
dtd
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
dtd
0T
dtd
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.
rVQ
rT
rT
dtd
r
Vr
JrL 2
22
21
2
03
2
rV
rJr
Time Independence
• Change the time derivative to an angle derivative.
• Combine with the equation of motion.
• The resulting equation describes a trajectory.
dd
rJ
dd
dtd
dtd
2
rQrT
rT
dtd
rQrT
rT
dd
rJ
2
Orbit Equation
Let u = 1/r
rQr
rrr
rrdd
rJ
)]([)]([ 222
21222
21
2
rQr
Jddr
rJ
dd
rJ
rJr
drd
rJ
3
2
222
22 )()(
2322
1)1(1JQ
rddr
rdd
rr
222
2
uJQu
dud 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
rV
rQr
2
21
21
mmmm
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 )( 222
21
rrJrVTE
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
JuJQu
dud r
))cos(( 0 rser
)]cos(1[ 02 e
Ju
eJs
2
r
u 1
)]cos(1[110 e
esr
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
eesr
12e
esr
11
21
2
2
)21(EJe
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
2JA
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 JaJaA
2
2222 21
EJaeaA
2
32 a
AAT
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
rrJVeff
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.