Asri-unix.734
net.space
utcsrgv!utzoo!decvax!ucbvax!ARPAVAX:C70:sri-unix!Ciccarelli@PARC-MAXC
Tue Feb 9 10:42:18 1982
Re: Horseshoe (and other) orbits
About "horseshoe" orbits, actually about orbits in general... The higher orbit is
not faster (higher *velocity*) but of higher *energy*. As the trailing moon
(lower, faster, less energy) catches the leading moon (higher, slower, more
energy) it takes some of the leading moon's energy, swapping orbits. It's
counterintuitive -- higher energy does NOT imply higher orbital velocity. I dug
up the formulas for your reference...
---------------
The formula for orbital velocity (circular orbit approximation) is:
V = V0 * SQRT( Earth radius / Orbit radius), where
V0 = 7.86 KM/SEC ("Circular velocity at Earth's surface),
Earth radius = 6400 KM (approx.), and
SQRT is the Square-root operation (of course)
You can check the formula for the three familiar orbit radii; calculate the
velocity using the formula, then see that this velocity gives the right orbital
period [distance = 2 * PI * Orbit radius; divide by velocity to get the period].
1) near-earth: Orbit radius = Earth radius ==> V = V0 [gives 90 min. period]
2) geosynchronous: Orbit radius = 40,000 KM (approx.) [gives 24-hr period]
3) lunar: Orbit radius = 400,000 KM (approx.) [gives 28-day period]
This also gives the expected result of V=0 at very great distances.
---------------
The formula for orbit energy is:
Total Energy = - (G * M1 * M2 / 2 * Orbit Radius), where
G = Newton's gravitational constant, and
M1, M2 = masses (i.e. Earth and satellite)
Note that:
1) the total energy is negative (the physical interpretation is that the orbit is
"bound", i.e. the satellite has less energy than that required to escape).
2) Orbit radius appears in the denominator again, thus the total energy will
become greater (still negative, but closer to zero) as orbit radius increases. At
very great distance, the energy goes to zero (as expected).
[ Source: "Introductory Astronomy and Astrophysics", by Smith and Jacobs ]
/John
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