Scenario Create new scenario around Moon

ghrasko

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I am returning back to Orbiter for a very long break and should re-learn a lot of things. Now I try to re-create the distant retrograde orbit (DRO) of Artemis-1 Orbiter when they had filmed Earth eclipsed by Moon as you can see it here Moon eclipses Earth. I use Orbiter 2016 on Windows 10.
I try to define the Orbital elements in Scenario editor (rought estimations at the moment):

Orbit reference: Moon
Frame: ref. equator
MJD: 59911 (2022 nov 28, 00:00)
SMa: 60 000 km
Ecc: 0
Inc: 7 deg
LAN: 0 deg
LPe: 0 deg
eps: 256 deg

My main problem is that it seems it is saving this as a Sun orbit. The current position is OK, but if I close and re-open the Orbital elements dialog, it reverts back to Sun orbit reference. If I run the scenario with time acceleration, it is obvious thatit is not a Moon orbit.

I have the feeling that I totally misunderstand how to use the Scenario editor.
 

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Ripley

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Why don't you start from a copy of the default "Landed at Brighton Beach" scenario file?
Just copy the scn file an go edit it.
 

C3PO

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Lunar DRO IS a 3-body orbit with the Sun as the major contribitor. IIRC Scott Manley made a video explaining it fairly well.

PS: If you set up an orbit around the Earth at the same altitude as the Moon, but 180 deg. from it, you will also find that the Sun is the major contribitor.
You can check the percentage on the OrbitMFD by swithing the reference manually. (Jupiter will be around 4%) 🤯
 

ghrasko

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Lunar DRO IS a 3-body orbit with the Sun as the major contribitor. IIRC Scott Manley made a video explaining it fairly well.

PS: If you set up an orbit around the Earth at the same altitude as the Moon, but 180 deg. from it, you will also find that the Sun is the major contribitor.
You can check the percentage on the OrbitMFD by swithing the reference manually. (Jupiter will be around 4%) 🤯
Wow, sure, I am stupid! This is why Orbiter is a brilliant tool to learn. Thanks.
 

ghrasko

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Well, so far I had failed to figure out, how could I define the proper speed vector once I had positioned the spaceship beypnd the L2 point (70 000 km from Moon). The space vector should lay on the Moon - Earth orbital plane and that is not a reference frame option. I think, I should calculate a bit.

This is Manley's related video (starting at 5 min): presentation of DRO. He is defining a magical speed vector (0.679 km/s), though I see 0.481 km/s from the article referenced by him.
 
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