Scenario Synchronous orbit scenario request

Kendo

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Is it possible for one of you kind gentlemen to do me a synchronous orbit scenario.
I have a large station model and it is supposed to be in a synchronous orbit around the Earth. I,m not sure how high, as nobody seems to know. It would be good a little higher than the ISS.

I,m presuming it will stay over one part of the Earths surface. If so, just put it over a nice part. The station is vertical.
You will be credited. Many many thanks. Ken.
 

garyw

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GeoSync is 35,786km - a lot more than a little higher than the ISS.
 

Kendo

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Thanks GaryW. In that case that height will do.
This is a fictitious station though.
 
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Kendo

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This is the only info re the station I have.

To an observer standing on the equator of the rotating body, an object so orbiting would generally appear to oscillate north and south. The one synchronous orbit which did not deviate from the body's equatorial plane, known as stationary because the orbiting object would appear to that observer to be suspended motionless overhead, was yet often understood to be the type intended by the general term.

Please be aware this is a fictitious station.
 
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Capt_hensley

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GWS needs either one as long as the scenario works. I prefer Geostationary as it has a perfect orbit with the equator. Enjoy, I did all the research for my "orbital slot" to find the perfect place that has at least 3 ground com sites within range of the station. I project a total of 5 ground stations will be available by the time my station is in orbit.

Having lots of ground stations is important for weather considerations when using laser communications as your primary data channel.
 

Lars789852

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Orbiter's default scenarios already include a geostationary orbit. The scenario is located in Navigation > Special Orbits > Geostationary orbit. Launch the scenario and there will be a DeltaGlider in a geostationary Orbit over Africa.

You can move your station to this orbit by intercepting and maybe docking with the DeltaGlider.

Watch the map, there won't be a green trajectory line, since it "hovers" always over the same spot.
 

Kendo

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Thanks Lars, I was forgetting that one. Perfect :cheers:
 

Capt_hensley

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1. But what if you want to spawn the station directly to your default location, in orbit, and not the DG4 location?

2. I havn't used this scenario, is it an equitorial orbit or inclined orbit at ### degrees?
 

Loru

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1. But what if you want to spawn the station directly to your default location, in orbit, and not the DG4 location?

If rest of elements are correct for geostationary orbit, you can adjust Mean longitude at epoch [eps] to place your station above point you want.

eps.jpg



2. I havn't used this scenario, is it an equitorial orbit or inclined orbit at ### degrees?

It's geostationary = equatorial at 42164.2km
 

Kendo

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Please excuse my ignorance, but does it have to be that altitude. ?
 

dgatsoulis

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Please excuse my ignorance, but does it have to be that altitude. ?

Yes. The period of an orbit is defined by the semi-major axis and the mass of the planet.

[math] T = 2 \cdot \pi \cdot \sqrt{\frac{a^3}{G\cdot M}} [/math]
In this case you want the period to be exactly one day (86164.10132 seconds).

So, solving for semi-major axis:

[math] a^3 = \frac{ T^2 \cdot G\cdot M}{4 \cdot\pi^2} \rightarrow \\ a = \left( \frac{ T^2 \cdot G\cdot M}{4 \cdot\pi^2}\right)^\frac{1}{3}[/math]
Plugin the numbers for Earth:

[math] a = \left( \frac{ 86164.10132^2 \cdot 6.67259\cdot 10^{-11}\cdot 5.973698968\cdot 10^{24}}{4 \cdot\pi^2}\right)^{\frac{1}{3}} \;= 42,164,173.089346 \; meters \simeq 42,164.17 \;km [/math]
That's the distance from the center of the Earth. If you want altitude you simply subtract Earth's radius. In Orbiter, Earth is a perfect sphere with a radius of 6371.01 km

[math] 42164.17 - 6371.01 = 35793.16 \; km[/math]
 
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