David.Smith@cmu-cs-ius.arpa (04/02/84)
My skepticism about using the space station for satellite repair was admittedly from a chemical rocket mindset. How old-fashioned of me not to think of ion rockets. (Previous discussions concluded that a solar sail is not useful below about 1000 miles.) Any low-thrust solution will take quite a while, since to be effective it has to be applied at or near the orbit crossovers. How about the following, for low orbit plane changes. This is based on an article I read in (I think) Astronautics and Aeronautics not too many months ago. Move the perigee to the crossing node, and brake off enough speed to drop it into the sensible (but still very thin) atmosphere. Use wing lift normal to the flight path to effect the plane change. This may take several passes. Once the plane is sufficiently changed, fire at apogee to raise the perigee. Of course, the drag at perigee will drop the apogee, unless thrusters fire during the maneuver. As I recall, they expected lift/drag to be about 4. So you get a major plane change for 1/4 the propellant that would have been required for a purely propulsive maneuver. A boon whether you use chemical or electric propulsion. David Smith P.S. Up with the winged space tug!
David.Smith@cmu-cs-ius.arpa (04/23/84)
In my original message about using aerodynamic lift for orbital plane changes, I neglected an important fact. If L/D is 4, then you get four times the effective thrust for an equivalent rate of fuel expenditure. But this does not mean that the fuel requirement is one-quarter what it would have been. Rather, it means that for equivalent delta-V, the required mass ratio is the fourth root of what it would have been. This follows from the fact that deltaV = Vexhaust * ln(mass ratio). Let's take the problem of transferring between the orbits of two space stations, each with orbital inclination of 28.5 degrees (latitude of Cape Canaveral), but launched 12 hours apart, so that the tug has to make a 57 degree plane change. As I recall, that came out to about a 16,400 mph delta-V. And let's be old fashioned and use chemical rockets. The specific impulse of hydrogen-oxygen rockets is often listed as 425 "seconds." This is 425 seconds of kilogram-force per kilogram of propellant. Make that 425 * 9.8 = 4165 newton-seconds per kilogram. Use the formula of f = ma = m dv/dt, and consider that instead of accelerating a finite block of exhaust over some time, we kick out an infinitesimal of mass at instantaneous exhaust velocity. This changes the force formula to f = Ve dm/dt. Combine this with 4165 n-s/kg, and we get Ve = 4165 meters per second, or 9319 mph. If we want to get that 16400 mph with pure propulsion, then the mass ratio must be exp(16400/9319) = 5.81. Using the L/D=4 maneuver, the mass ratio must be 5.81**(1/4), or 1.55. The aerodynamic plane change cuts the required propellant by a factor of 8.7. This analysis ignores other factors, such as: mass of wings reduced mass of tanks fuel to lower and raise perigee fuel and structure that would have been required to put 8.7 times the propellant into orbit (either from the earth or the moon) -- David Smith