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

QQ

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.