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