Download Kepler Orbits for Binary Systems

Lecture 7: Kepler’s Law for Binary Systems
1
REVIEW: Newton’s Universal Law of Gravitation
Analytic Results
You are all familiar with Newton’s Universal Law of Gravity as applied to the solar system. The
magnitude of the gravity force between the Sun and the Earth is given by
FG = G
MS ME
r2
where r is the distance between the center of the Sun and the center of the Earth. The MS
is the mass of the Sun and the ME is the mass of the Earth. The constant G is the universal
gravity constant which in the MKS system is 6.67 × 10−11 N-m2 /kg2 .
In classical mechanics courses the solar system orbits are analyzed in polar coordinates (r, θ),
especially since the polar coordinate formulation makes very obvious the fact that the angular
momentum L = µr2 θ̇ is being conserved for the reduced mass µ = MS ME /(MS + ME ). The
analytic solutions are obtained for the orbital motion equation r(θ) in the center-of-mass frame.
These orbital motion equations are the equations for conic sections (circles, ellipses, parabolas,
and hyperbolas). The planetary orbits are exactly ellipses with the Sun at one focus. The
eccentricities of the ellipses (see Table 4.1 on page 98) are 0.10 or less, except for Mercury
(0.206) and the former planet Pluto (0.248). A circular orbit would have an eccentricity exactly
zero. And except for Pluto, the remaining eight official planet orbits are all in the same plane.
Cartesian Coordinate Formulation
For the moment we assume that the Sun’s mass is infinitely heavy and we want to compute
the Earth’s orbit in Cartesian coordinates. Since the Sun is more than five orders of magnitude
more massive than the Sun, and even three orders more massive than Jupiter, this is a good
first approximation. In such Cartesian coordinates we can write the four first order differential
equation for the motion as
dvx
dy
= vx and
= vy
dt
dt
dvx
GMS x
dvy
GMS y
=−
and
=−
3
dt
r
dt
r3
The negative signs indicate that the forces are attractive. The x/r3 = (1/r2 )(x/r) represents the
original inverse square force multiplied by the cosine of the angle of the force direction in order
to get the Fx component (Figure 4.1). We can solve these equations numerically in the same
manner that we solved the projectile motion equations. The only difference is the formulation
of the functions representing the derivatives of the velocity components. Here these derivatives
are seen to depend upon the position coordinates and not the velocity coordinates.
Lecture 7: Kepler’s Law for Binary Systems
2
REVIEW: Numerical Solution for the Earth’s Orbit
Astronomical Units of Length and Time
In solar system examples it is very convenient to not use the MKS units for Mass, Length and
Time, but instead use the astronomical units. So one uses the Solar mass, the AU for length,
and the year for time. The AU is the mean distance of the Earth’s orbit, ≈ 1.5 × 1011 m. An
immediate benefit of these units is the expression for GMS can be obtained from the Earth’s
mean (circular) orbit numbers:
ME v 2
MS ME
= FG = G
=⇒ GMS = v 2 r = 4π 2 AU3 /Year2
2
r
r
This last equation has the Earth’s orbit as a circle of one AU taking one year (≈ π ×107 seconds).
Numerical Algorithms
One cannot use the simplest Euler method to solve the four differential equations. It is just not
accurate enough for reasonable step sizes. Instead there is the modified Euler-Cromer method:
4π 2 xi
∆t
ri3
xx,i+1 = xx,i + vx,i+1 ∆t
4π 2 yi
vy,i+1 = vy,i − 3 ∆t
ri
yy,i+1 = yy,i + vy,i+1 ∆t
vx,i+1 = vx,i −
In the Euler-Cromer method, one must first evaluate the new velocity component, and then
use that new velocity component to evaluate the new position component. On page 97 of the
textbook there is a nice pseudocode design for an Euler-Cromer program.
There is no problem using the RK4 algorithm as before. In fact, in my own C++ program
solutions, I have the identical calculateRK4 functions as for the corresponding projectile motion
programs. The essential thing that changes is what is evaluated in the derivative functions that
are being called by, and are external to, the calculateRK4 function.
Checking the results
As with solving all mechanics problems, one must provide initial conditions in order to start
the solution to the differential equations. This is not completely trivial as one will not want
to specify zero velocity and position at the origin (Sun) as the initial conditions for the orbit.
Instead one can specify for the Earth the initial x position as 1 AU, the initial y position as 0,
the initial vx as 0, and the initial vy as 2π AU/year. These initial conditions will result in a
circular orbit as shown in Figure 4.2 on page 99.
In viewing the planetary motion
q solutions, one does not look at the separate x(t) and y(t) results.
Instead, one looks at r(t) = (x2 + y 2 ) for the orbits. Even better, it is nice to animate the
motion plotting r(t) in successive time steps. It is possible to do this animation without too
much effort using Mathematica to read the ASCII file which is produced by a program.
Lecture 7: Kepler’s Law for Binary Systems
3
Conserved Quantities
Energy
Newton’s Universal Law of Gravity is a conservative force. It can be derived from a potential
energy function of distance
m1 m2
V (r) = −G
r12
where m1 and m2 are the two masses and r12 is the distance between them.
So in a solar system where one mass is the mass of the Sun MS and the second mass is the mass
of a planet MP , and we assume that MS >> MP such that the Sun is essentially stationary,
then the total energy of the system is
1
MS MP
Etotal = MP vP2 − G
2
rSP
Even when we relax the assumption M1 >> M2 , or we include three bodies, energy is still
conserved as long as we account for the motion of all the masses.
Escape speed
If a system as negative total energy, then it is a bound system. All planets and other objects in
the Solar system which have closed, repeating orbits are bound with negative total energy. The
escape speed vescape is the speed necessary to leave the Solar system. It can be easily computed
by setting the total energy to 0 which yields
1
MS M P
2
MP vescape
=G
=⇒ vescape =
2
rSP
s
2MS G
rSP
Notice that the escape speed is independent of the mass of the object, and depends only on the
mass of the Sun and the distance between the object and the Sun.
What is the escape speed of the Earth from the Sun when the Earth is at its mean orbital
distance? Try giving values close to but less than the escape speed in the onePlanet program
and see what happens. Assume that the Earth starts always at 1 AU distance.
Class Lab Work (use the onePlanet program, or your own program if you have one)
Next move on to the other planets in the Table 4.1 on page 98, starting with Venus and then
Mars, Mercury, and Jupiter. First assume that the planets start at the radius value given in the
Table, and see if you can reproduce the orbital times and the eccentricities listed. Integration
step sizes of 0.002 AU can be used, but you can explore how stable the results are as compared
with 0.02 AU or 0.0005 AU. You should open an EXCEL window to keep track of your results.
At the end of the class you will deposit that EXCEL window (with your name on it) in the
Drop Box. For the final exercise with the onePlanet program, do exercise 4.4 to work out
the Halley’s comet orbit which has a period of 76 years and a closest distance to the Sun
of 0.59 AU. In other words, find the speed that Halley’s comet has when it is at 0.59 AU,
and that speed results in a 76 year orbit. The mass does not matter, but you can look at
http://www.seds.org/nineplanets/nineplanets/halley.html to figure out an estimate.
Lecture 7: Kepler’s Law for Binary Systems
4
Binary Systems
Differential Equations of Motion
For a binary system for which one object is not extremely more massive than the other object,
it is necessary to allow both objects to move. However, one knows that the center-of-mass of an
isolated system does not accelerate. So one can look at the coordinates of the motion in terms
of the center-of-mass. In that system, there is a one-body equation of motion in terms of the
m2
reduced mass µ = mm11+m
, and the system can still be solved analytically.
2
In our computational approach, we can move to the two body system without too much difficulty.
We consider that the coordinates of the two bodies are with respect to the center-of-mass. We
further assume that the linear momentum of the system is zero.
dv1x
= v1x
dt
dv1x
GMS x1
=−
2
dt
r1 r12
dv2x
= v2x
dt
dv2x
GMS x2
=−
2
dt
r2 r12
and
and
and
and
d1y
= v1y
dt
dv1y
GMS y1
=−
2
dt
r1 r12
d2y
= v2y
dt
dv2y
GMS y2
=−
2
dt
r2 r12
The vector coordinates are r~1 = x1 ı̂ + y1 ĵ and r~2 = x2 ı̂ + y2 ĵ and r12 = |~
r1 − r~2 |.
The easiest system to study is the equal mass system, for which I developed a movie animation
that we will view in class.
Second class exercise and homework assignment
Develop a two-body program as in exercise 4.7 to study the motion of a binary system. Allow
for the two bodies to be of unequal masses, and assume that the second body has equal and
opposite linear momentum compared to the first mass. The center-of-mass of the system is taken
to be the origin of the coordinate system.
Next week we will expand this program to the three body system.