From: hal@finney.org
Date: Mon Dec 03 2001 - 16:35:58 MST
Anders writes:
> The equlibrium condition is that G M_sun / R^2 - G M_jup / (R-R_jup)^2 -
> omega^2 R = 0
Right, and there is a simplification that omega^2 R_jup = G M_sun/R_jup^2
that is, centrifugal force equals gravitational attraction at Jupiter,
so this gives us omega^2 = G M_sun / R_jup^3 and we can remove omega^2
from the equation. We can also remove G since it is now in every term
and get M_sun/R^2 - M_jup/(R-R_jup)^2 - M_sun R/R_jup^3 = 0.
This could be further simplified by defining m = M_jup/M_sun and
r = R / R_jup to get rid of units, giving us:
1/r^2 - m / (1-r)^2 - r = 0
This produces a 5th degree equation in r. I solved it numerically in
Mathematica and it turns out that for Jupiter's mass and orbit that the L1
point is close to 1/15 of the way in towards the sun. Spike's calculation
neglecting orbital motion got 1/32.
Specifically for Jupiter's orbital radius of 7.8E11 m, the L1 point is
at 7.28E11 m from the Sun or 5.2E10 m from Jupiter. That's about 1/3 AU
inward.
Hal
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