[amsat-bb] Re: Path to HEO
Patrick Strasser
patrick at wirklich.priv.at
Wed May 1 23:20:44 PDT 2013
schrieb Ken Ernandes on 2013-04-30 09:49:
> I note the disclaimer at the bottom, so I'll help with the incorrect assumptions.
>
> 1. g = 9.81 m/sec only applies to one Earth radius (i.e., the Earth's surface). Gravitational acceleration drops of as an inverse square of the radius.
> 2. GEO altitude is close to 36000 km, but GEO radius is approximately 42164 km (you must add the Earth's radius to altitude to get orbital radius)
Gotcha!
> Gravity can be simplified by using a constant MU = 398600.4418 km^3/sec^2
>
> For any radius you can compute g by:
>
> g = MU / r^2
>
> But make sure you use radius and not altitude. Mean Earth radius at mid-latitudes is approximately 6371 km and is 6378 at the equator.
>
> The speed (v) for an elliptical orbit can be computed from the current radius (r) and semi-major axis (a):
>
> v = sqrt(MU*(2/r - 1/a))
>
> This can be simplified for a circular orbit (r = a):
>
> v = sqrt(MU/a)
These are the formulas they did not tell us at school, thank you!
> The more important thing is what Dan Schultz pointed out. At 300 km altitude, atmospheric drag is a significant factor in a continuous drain of orbital energy. This is less at 500 km and almost insignificant starting around 800 km. The drop off is because drag is:
>
> 1. proportional to atmospheric density which drops off quickly with increased altitude.
> 2. also proportional to the square of the speed relative to the atmosphere.
>
> If you can get above 800 km without using up a lot of your fuel, you have a chance to make something workable.
>From the last few cubesat rides I reckon that going there instead of
only to 300 km is not that impossible.
Regards
Patrick
--
Engineers motto: cheap, good, fast: choose any two
Patrick Strasser <patrick at wirklich priv at>
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