Re: Satelite Motion

["Carl E. Mungan" <mungan@USNA.EDU>]

David Abineri wrote:

> If the earth's gravity doubled, how would the moon's orbit be affected?
>
> I am getting tangled up in having too many variables and it is not at
> all obvious what might happen. Would the results be different if it
> happens instantaneously rather than incrementally?

Orbital motion can be described by two constants. These constants can
be geometric (eg. semi-major & semi-minor axes) or physical (eg.
maximum speed & period) or a combination thereof, but the point is
that 2 constants are enough.

From a pedagogical point of view, the preferred choice is the energy
E and angular momentum L, because of the conservation laws.

In particular, you double Earth's mass M, but *instantaneously* do
not change Moon's position and velocity. Consequently, L does not
change. But the potential energy and hence E decreases (recall that
PE is negative and you made it more negative). The Moon therefore
transfers from essentially a circular orbit into an elliptical orbit
where apogee is the initial position and perigee is substantially
closer than that.

So the new orbit is specified by: r_max = R (apogee distance =
original Earth-Moon distance), L unchanged, GMm (constant numerator
of the force law) doubled.

The numbers are given below, but are probably beyond an intro course.
So I would end here in that context.

Elliptical orbits are specified in polar form by r=c/[1+e*cos(phi)]
where e = eccentricity and c = L^2/(GMm*m) with unchanged Moon mass
m<<M.

Thus c(new) = c(old)/2 but r_max = c(new)/(1-e) => e = 1/2 in new
orbit. (Very eccentric!)
Or if you prefer to know the new perigee distance, it is r_min =
c(new)/(1+e) = R/3.

Note that a = (r_min + r_max)/2 = 2R/3. (Another nice way to get this
is to use the vis-viva equation: the speed squared at apogee is GM/R
in the original orbit and 2G*2M/R-G*2M/a in the new orbit. Equate
these and solve for a.) Thus ratio of new to old period of Moon is
(a/R)^1.5 * (M_old/M_new)^0.5 = 2/3^1.5 = 10.6 days. Carl

ps: If you subtract the mass discontinuously, ie. scooping off a
little bit of Earth's mass each time the Moon is back to apogee, then
the above results will still hold. However, if you spread the
subtraction uniformly around many orbits, you keep changing where you
poke the orbit inward. I *think* what will happen then is that the
orbit will stay close to circular. In that case we still have L = mvr
= constant, but since GMm/r^2 = mv^2/r = L^2/mr^3 we conclude that Mr
= const. and consequently that r halves when we double M. The results
*are* different. Now the new period is instead quartered to 6.9 days.)
--
Carl E. Mungan, Asst. Prof. of Physics  410-293-6680 (O) -3729 (F)
      U.S. Naval Academy, Stop 9C, Annapolis, MD 21402-5040
mailto:mungan@usna.edu       http://usna.edu/Users/physics/mungan/