Thanks, Grant=2E Your value of eccentricity works just fine=2E
I was using e =3D sqrt(1  (b*b)/(a*a)), where a and b are the semimajor a=
nd
semiminor axis of the ellipse, respectively=2E I used the moon's volumetri=
c
mean radius to come up with these values:
a =3D 186 + 1737=2E1 =3D 1917=2E1
b =3D 183 + 1737=2E1 =3D 1914=2E1
then e =3D 0=2E05592
Obviously, this is not at all correct=2E But why isn't it? Where's the fau=
lt
in my math? Have I been dillusional for half of my life?
Thanks=2E
Original Message:

From: Grant Hutchison granthutchison@...=2Eco=2Euk
Date: Sun, 28 Sep 2003 22:22:18 +0100
To: cweisbrod@...=2Eca, celestiadevelopers@...=2Esourceforge=2Enet
Subject: [Celestiadevelopers] Re: Apollo 11 Orbit
Clint:
The problem is with your calculated eccentricity=2E The eccentricity is eq=
ual
to (1  perihelion/semimajor axis): for the orbit you describe, and using
the moon's equatorial radius in Celestia, that's (1  1920=2E5/1922) =3D
0=2E00078=2E
So the following definition produces good apo and pericynthions:
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D
"Apollo 11" "Sol/Earth/Moon"
{
Class "spacecraft"
Mesh "apollo=2E3ds"
Radius 0=2E010
EllipticalOrbit
{
Period 0=2E0873
SemiMajorAxis 1922
Eccentricity 0=2E00078
Inclination 5
}
Albedo 0=2E10
}
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D
(The apparent shift in orbital centre you found using your eccentricity is=
just a property of the geometry of ellipses  for a given semimajor axis,
pericentre and apocentre will alter by the *same amount* when you alter th=
e
eccentricity=2E)
Grant

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