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Re: Satelite Motion



Thus c(new) = c(old)/2 but r_max = c(new)/(1-e) => e = 1/2 in new
orbit since c(old) = R. (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.

I just realized a more elementary way of getting the above results.
You *might* make this fly in an introductory class:

U(new) = -G*2M*m/R
K(new) = K(old) = G*M*m/2R (obtained using F=ma for the old circular orbit)

thus E(new) = U(new)+K(new) = -3GMm/2R

Now for the *only* non-elementary step you need:

E = -D/2a for any bound orbit where D = numerator of force law

For an attempt to derive this at an introductory level, see
<http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Gravity/GravityME-Dervation.html>.
But it's still a bit daunting. (Anyone else have a suggestion of how
to tackle this step in an intro class?)

In the new orbit, D = G*2M*m.

Equating the two expressions for E above, you immediately get a =
2R/3 and thus the new period. But r_max = R. Hence r_min = 2a - r_max
= R/3. You can go on to get eccentricity, semi-minor axis, speeds at
aphelion and perihelion, or whatever else you like.

--
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/