Lecture 5 - Orbital Mechanics I (9/24/98)

Prev Newton --- | --- Barycenter Next


Chapter 1-4, 1-5, P1-7 (ZG4)

Isaac Newton
? Key Question: How do Kepler's Laws follow from Newton's Laws?
! Key Principle: Newton's Law of Universal Gravitation
# Key Problem: Find the mass of Jupiter from the orbits of its moons.
@ Key Quote: "... and I'm floating in a tin can, far above the world." - Major Tom


  1. Steps Toward Universal Graviatation
  2. Acceleration and Weight
  3. Kepler's Laws Revisited I
  4. Orbits Explained

Infinitesimal Steps and Numerical Simulations

In the previous lecture, we used the notion of "infinitesimals" in the calculus of derivatives to derive the purely geometric centripetal acceleration when applied to a circular orbit. This is of more than academic interest, in that in systems more complicated than two mutually gravitating bodies the only method for solution is by numerically simulating the motions of the bodies with finite steps - essentially what we have done in our simple derivation!

To recap, we found that if we start at time t=t_0 at the position and velocity given by:

t = t0
x = r
y = 0
vx = 0
vy = v

After a time-step dt (technically delta-t) we have

t = t0 dt
x = r dx = - a dt2 / 2
y = 0 dy = v dt
vx = 0 dvx = a dt
vy = v dvy = 0

And thus after our single time-step dt we have the updated position

t = t0 dt t = t0 + dt
x = r dx = - a dt2 / 2 x = r - a dt2 / 2
y = 0 dy = v dt y = v dt
vx = 0 dvx = a dt vx = a dt
vy = v dvy = 0 vy = v

Note that this is only valid in the limit that dt -> 0, in that we have assumed that dx uses the average value of v_x over the step, and that dv_y = 0. These are both equivalent to the linear approximation in that we assume that the velocities grow linear with time under the approximation of a constant force over the timestep in the -x direction. There are higher order corrections to this which would include the changing aspect of the force in the movement.

What errors does this introduce? We noted that to remain on the circle x^2 + y^2 = r^2, the position after the time step must be given by

( r - a dt2 / 2 )2 + ( v dt )2 = r2

or after expanding the squares on the left hand side

r2 - r a dt2 + a2 dt4 / 4 + v2 dt2 = r2.

Cancelling the r^2 on each side, dividing through by -r dt^2, and isolating a on the left side, we get the useful expression

a = v2 / r + a2 dt2 / 4r.

If we take the ananlytic limit dt->0 then we get the usual expression for the centripetal acceleration

a = v2 / r

required to keep the body in a circular orbit. We can therefore think of the extra term

da = a2 dt2 / 4r

as an error term which scales as the square of the timestep. If we were to use this simple scheme to update (x,y) as above, we would make errors that would slowly compond, as the act like a force error. (Q: How does this error in acceleration translate to errors in the velocity (v_x,v_y) and eventually position (x,y)?)

This simple finite step scheme, with some upgrading, is essentially what is used in astrophysics when we want to numerically model systems with more than 2 bodies! This is called the N-Body problem, where the N refers to the arbitrariness of the number of bodies or particles in the system. Currently, we routinely run simulations with N=10^6 particles, and are pushing hard on simulations with N=10^8! Whew.

For an example of N-body simulations in cosmology, see the NCSA "Cosmos in a Computer" Expo.

Prev Prev Lecture --- Next Next Lecture --- Index Astr11 Index --- Home Astr11 Home

smyers@nrao.edu Steven T. Myers