## celestia-developers

 [Celestia-developers] Re: Apollo 11 Orbit From: Grant Hutchison - 2003-09-28 21:22:20 ```Clint: The problem is with your calculated eccentricity. The eccentricity is equal to (1 - perihelion/semimajor axis): for the orbit you describe, and using the moon's equatorial radius in Celestia, that's (1 - 1920.5/1922) = 0.00078. So the following definition produces good apo- and peri-cynthions: =================== "Apollo 11" "Sol/Earth/Moon" { Class "spacecraft" Mesh "apollo.3ds" Radius 0.010 EllipticalOrbit { Period 0.0873 SemiMajorAxis 1922 Eccentricity 0.00078 Inclination 5 } Albedo 0.10 } ======================== (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 the eccentricity.) Grant ```
 Re: [Celestia-developers] Apollo 11 Orbit From: - 2003-09-29 19:52:14 ```Ohhhh! Good grief, of course! I've been thinking the elliptical orbit center and the center of [mass of] the moon are in the same place, rather than the center of the moon corresponding to one of the focii of the ellipse=2E Well, I feel damn silly, but relieved=2E :-) Too much Teletubbies and Sesame Street lately=2E :-) Thanks for the explanation, Grant=2E Clint=2E Original Message: ----------------- From: Grant Hutchison granthutchison@...=2Eco=2Euk Date: Mon, 29 Sep 2003 17:19:35 +0100 To: cweisbrod@...=2Eca, celestia-developers@...=2Esourceforge=2Enet Subject: Re: [Celestia-developers] Apollo 11 Orbit > Obviously, this is not at all correct=2E But why isn't it? Where's the f= ault > in my math? Have I been dillusional for half of my life? Your formula is correct, but your difficulty is with the numbers you're using for semimajor axis and semiminor axis=2E The semimajor axis is half = the long axis of the ellipse, and the semiminor axis is half the short axis - = so these two measurements are at right angles to each other, each taken from the geometrical centre of the ellipse=2E But the two numbers you've used a= re actually the pericentre and apocentre radii - the distances from one focus= of the ellipse to the two ends of the long axis=2E If you average these tw= o you'll get the semimajor axis, but then you pretty much need to work out t= he eccentricity in order to calculate the semiminor axis, which isn't really any use for orbital calculations anyway=2E Grant ------------------------------------------------------- This sf=2Enet email is sponsored by:ThinkGeek Welcome to geek heaven=2E http://thinkgeek=2Ecom/sf _______________________________________________ Celestia-developers mailing list Celestia-developers@...=2Esourceforge=2Enet https://lists=2Esourceforge=2Enet/lists/listinfo/celestia-developers -------------------------------------------------------------------- mail2web - Check your email from the web at http://mail2web=2Ecom/ =2E ```