Suppose you have an object in orbit that requires a force of 0.0001356 mkg/sec2 for a full year. How much reaction mass do you need at 40,000 m/sec reaction mass velocity?
Note: the application is keeping a small space settlement (Kalpana One) in LEO without the orbit degrading.Params:Area or cross section 162.5000 m2 (250 m x 315 m cylinder in the worst attitude, the area of the thermal arrays is ignored for now)mass 1,805,500 tons (2 tons shielding per m2)altitude 700 km (circular orbit)
Note: I used http://spacience.blogspot.com/2012/03/howto-calculate-drag-in-leo-using.html for the methodology (but not the code), but using a different approach would be helpful

What happens when you move a conductor thru a magnetic field? It converts the energy of motion into electrical flow thru the conductor,,,now reverse that...see? No need for reaction mass...
Gary 7

While the editorial comments can be interesting, what I really need is confirmation that I havent screwed up the analysis and calculations somehow. I know its real work to figure out how to do the calcs and do them, but I would really appreciate it if someone would.
BTW: the mass doesnt matter. Dont know what I was thinking. Only the cross-sectional area, altitude, and reaction mass exhaust velocity are needed for calculating mass of the reaction mass.
On Jul 3, 2014, at 11:56 AM, john strickland jkstrickl@... wrote: I would no sooner put a space habitat in LEO than I would put an ice cream cone in an oven.
This is an incompatible location for several reasons.
If abandoned, the strong structure required by a colony would partially survive re-entry and cause damage on the Earths surface.
Until the space debris problem in LEO is cleaned up, impacts on the large surface area of the colony would greatly add to the debris problem.
John

You dont need the rocket equation. The mass of the propellant is a tiny fraction of the mass of the settlement, so you can just ignore it.
All that is needed is the mass of the propellant per second needed to produce 0.0001358 mkg/sec2 of force then multiply by the number of seconds in a year.
Ive done the calcs. Ive got an answer, but I may have screwed something up. If someone else does the calcs, ideally in a different way, and gets more-or-less the same answer then Im probably right. If they get a radically different result then there is a problem somewhere.
On Jul 3, 2014, at 12:16 PM, john strickland jkstrickl@... wrote: We would need to figure out how to convert the rocket equation, which is based on ISp or exhaust velocity and mass values, into an equation based on thrust and time and mass.
There may be equations which bridge this gap.
It took me most of a day in 2010 to decipher the rocket equation, which is a relatively simple algebra equation, identifying all of the input values, the input and output units, and then match one of Gopal's values to 4 digits, which proved I did it right!. I can now use it, and have it in Excel cells, but I still do not understand WHY it works.
I would ask Gordon Woodcock or Henry Spencer what equation type to use. Go to someone who is an actual physics and math major and knows instinctively what to do.

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