# [amsat-bb] Re: Path to HEO

Stefan Wagener wageners at gmail.com
Mon Apr 29 18:26:25 PDT 2013

```Thanks,

I like simple math. A 3U Cubesat can have deployable solar cells which will
give you between 45 and 70 watts of power. That should drive an ion engine.
Will need an active attitude control system and should start at the right
orbit. Getting +-3kg to GTO is a \$100.000? Don't know what the satellite
would cost but we are now in a \$ range that could be funded.

Thoughts?

Stefan

On Mon, Apr 29, 2013 at 2:50 PM, Patrick Strasser
<patrick at wirklich.priv.at>wrote:

> schrieb Stefan Wagener on 2013-04-29 01:29:
>
> > LEO (low earth orbit) from 400 to 2000km or something like that
> > HEO (high earth orbit) >20000km
> >
> > Going from 310 to 700km in a year is not doing us anything. We need a
> > highly elliptical orbit (Apogee ~60000km, Perigee 900km) similar to AO-40
> > to allow for cross continental communication.
>
> Let's check with some maths (*):
>
> Energy at 300 km:
> -----------------
>
> Total Energy = kinetic energy + potential energy
> E_kin     + E_pot =
> (m*v^2)/2 + m*g*h
>
> For simplicity, we choose
>  mass as 1 kg,
>  h = 300km,
>  v = 1st cosmic velocity =~ 7100m/s
>
> 1*7200² + 1*9.81*300x10^3 = 2.8148x10^7 [Joule]
>
> Potential Energy is some 10% of the total energy.
> As this is for one kg of mass, and m goes linear in the above equations,
> you can scale with the mass of your satellite.
>
> Energy at 36000 km:
> -------------------
>
> Speed from radius and time for one orbit (1 day=84600 sec)
>  v = 2*r*pi/t = 2*36x10^6*3.14159265/84600 = 2673.7 m/s
>
> E_kin + E_pot =
> (m*v^2)/2 + m*g*h =
> 3.57x10^6 + 3.53^8 =~ 3.56×10^8 [Joule]
>
> Now kinetic energy is only about 1% of the total energy!
>
> A LEO has about 8% the energy of an GEO. The satellite needs 3.29x10^8
> J/Kg Energy to get from LEO to GEO.
>
> Lets say it's 10kg and has 50 W of power for thrust.
>  3.3x10^8 * 10 = 3.3x10^9 J thrust
>
> 1 Joule is 1 Watt / 1 second, 1 Watt second = 1 Joule
> 1 Watt day = 84600 Joule = 8.46x10^4 Joule
>
> Our 50 Watt ion drive can increase the energy by 4.23x10^6 Joule a day.
> How much days will LEO to GEO take:
>
> We have some 10^8 divided by some 10^6, it's a matter of months, the
> calculation says 77.8. But we should be satisfied to get an order of
> magnitude after the rough assumtions and estimations made before.
> If I made a mistake above, maybe this is off by an factor of 10, then
> it's 2 year. Still fine!
>
> Of course you have to count in the gas you want to ionize, which reduces
> the weight over time (but I was really bad at differential equations and
> would not get that right), and maybe the weight and power estimations
> are not very realistic, and using steady thrust instead of short
> impulses decreases efficiency in orbit changes, and changing from polar
> to equatorial orbit takes extra energy, and maybe an elliptic orbit
> takes less energy, and maybe some inaccuracies more. But this does not
> matter:
>
> In the end, it seems that changing from LEO to GEO or HEO is possible in
> sensible time.
>
> Regards
>
> Patrick
>
> (*) Disclaimer: This is High School maths, please double check and
> correct my calculations
> --
> Engineers motto: cheap, good, fast: choose any two
> Patrick Strasser <patrick at wirklich priv at>
>
```