Transcript Two-Body Systems

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
  ext  ext
MR  F1  F2
F2int
m2
r = r1 – r2
F2ext
r2
  int  ext

m1r1  F1  F1
  int  ext

m2 r2  F2  F2
m1
R
F1int
r1
F1ext
• Subtract for relative motion
 int  int
r  r  F1  F2
1
2
m1
m2
Reduced Mass
• The internal forces are equal
and opposite.
• Express the equation in terms of
a reduced mass m.
– m less than either m1, m2
– m approximately equals the
smaller mass when the other is
large.
 int  int

r  r  F  F  ( 1  1 ) F int
1
2
m1
m2
m1 m2
 int

r  r  ( m1  m2 ) F int  F
1
2
m1m2
m
m
m1m2
 m2
m1  m2
for m1  m2
Central Force
Qm   Fi
i
xi
qm
xi : x, y, z
– Spherical coordinates
– Generalized force
qm : r ,  , 
Qr   Fi
i
• The internal force can be
expressed in other coordinates.
xi
qr
Q  Q  0
• A force between two bodies can
only depend on r.
– Central force
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.
T  12 m (r 2  r 2 2  r 2 sin 2  2 )
d T T

 Qr

dt r r
d T T

0

dt  
d T T

0

dt  
Coordinate Reduction
• T doesn’t depend on  directly.
d T T

0

dt  
• The angular momentum about
the polar axis is constant.
d T
0

dt 
– Planar motion
– Include the polar axis in the
plane
• This leaves two coordinates.
– r, 
T
2
2


m
r
sin



T  12 m (r 2  r 2 2 )
constant
Angular Momentum
• T also doesn’t depend on 
directly.
– Constant angular momentum
– Angular momentum J to avoid
confusion with the Lagrangian
d T T

0
dt  
d T
0

dt 
T
2 

m
r
J

constant
Central Motion
• 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 Potential
• The central force can derive
from a potential.
d T T
V

 Qr  
dt r r
r
• Rewrite as differential equation
with angular momentum.
J 2 V
mr  3 
0
mr
r
• Central forces have an
equivalent Lagrangian.
2
J
L  12 mr 
V
2
2mr
2
Time Independence
• Change the time derivative to
an angle derivative.
d d d
J d

 2
dt dt d mr d
• Combine with the equation of
motion.
d T T

 Qr

dt r r
• The resulting equation
describes a trajectory.
J d T T

 Qr
2
mr d r r
Orbit Equation
• The solution to the differential equation for the trajectory
gives the general orbit equation.
J d [ 12 m (r 2  r 2 2 )] [ 12 m (r 2  r 2 2 )]

 Qr
2
mr d
r
r
J dr
J 2 J d J dr
J2
 mr ( 2 )  2
( 2
)  3  Qr
2
r d
mr
r d mr d
mr
1 d 1 dr
1 mQr
(
)

 2
2
2
3
r d r d
r
J
Let u = 1/r
d 2u
mQr

u


d 2
J 2u 2
Inverse Square Force
• The inverse square force is
central.
k < 0 for attractive force
–
F2int
m2
m
m1m2
m1  m2
• Choose constant of integration
so V() = 0.
r = r1 – r2
r2
Qr 
m1
R
F1int
r1
V
k
r2
k
r

V
r
Kepler Lagrangian
• The inverse square Lagrangian
can be expressed in polar
coordinates.
T  12 m (r 2  r 2 2 )
k
V
L  T  V  12 m (r 2  r 2 2 ) 
r
k
• L is independent of time.
– The total energy is a constant
of the motion.
– Orbit is symmetrical about an
apse.
J2 k
E  T  V  mr 

2
mr
r
1
2
2
1
2
r
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.
d 2u
mQr
mk

u




d 2
J 2u 2
J2
u
1
u
r
mk
J
2
[1  e cos(   0 )]
J2
s
mke
1 1
 [1  e cos(   0 )]
r es
r  e( s  r cos(   0 ))
Conic Sections
• The orbit equation describes a
conic section.
r
s

focus
r  e( s  r cos  )
– 0 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
e  (1 
2 EJ 2
mk
2
)
1
2
1 1
 (1  e cos  )
r es
r1 
r1
es
1 e
r2 
r
s

es
1 e
r2
Angular Momentum
• The change in area between
orbit and focus is dA/dt
dr
r
– Related to angular velocity
• The change is constant due to
constant angular momentum.
• This is Kepler’s 2nd law
A  12 rr  12 r 2
J  mr 2
J
A 
2m
Period and Ellipse
r
r1
s

r2
• The area for the whole ellipse
relates to the period.
– semimajor axis: a=(r1+r2)/2.
A  a 2 1  e 2  a 2 
A  a
3
2

J2
mk
 2a
3
A
m
T    2a 2 
A
k
3
2
2 EJ
2
mk 2
J
m

2m
k
• This is Kepler’s 3rd law.
– Relation holds for all orbits
– Constant depends on m, k
Effective Potential
• The problem can be treated in
one dimension only.
J2 k
E  mr 
  Tr  Veff
2
mr
r
– Just radial r term.
1
2
• Minimum in potential implies
bounded orbits.
Veff
– For k > 0, no minimum
– For E > 0, unbounded
Veff
0
Veff
r
unbounded
0
r
possibly
bounded
2
1
2
J2
k


2mr 2 r
Star Systems
• Star systems can
involve both single
and multiple stars.
• Binary stars are a case
of a two-body central
force problem.
• 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
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
(M1  M 2 ) P  a
2
3
a  a /  
( M1  M 2 ) P 2  a /  
M1a1  M 2 a2
a1  a2  a
• 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.
Spectroscopic Binaries
r1  VP / 2
r2  vP / 2
a  r1  r2
M1  M 2  a3 / P 2
M1 / M 2  r2 / r1  v / V
• Binary systems that are too
close require spectroscopy.
– Doppler shifted lines
– Velocity measurements
Eclipsing Binaries
• An orbit inclination of
nearly 90° to the observer
produces an eclipsing
binary.
• Light levels are used to
measure period and radii.