Reaction drives and deltavees
 30Es10 Essays
A brief look at the mathematics reaction drives.

Reaction drives and deltavees.
In a reaction drive, thrust is derived from ejecting reaction mass out of the drive at a given speed. The momentum gained by consuming an amount of mass khi is
p = u khi equation 1
where u is a constant of proportionality, which in fact is merely the exhaust velocity of the drive (the speed at which the reaction mass is expelled). The exhaust velocity is an efficiency rating of the drive, and merely differs from the more commonly seen (though strangely defined) specific impulse by a constant.

Differentiating with respect to time t, we see that

dp/dt = u dkhi/dt equation 2
or
F = u kappa; equation 3
that is, the thrust F gained is the exhaust velocity multipled by the rate of fuel consumption kappa (measured in units of mass divided by of time, such as kg/s).

Since kappa is the rate of fuel consumption, it is also the rate of mass loss of the ship, -dm/dt. We can then rewrite the above equation as the following differential equation:

dp = -u dm. equation 4
In the Newtonian approximation, dp = m dv; the differential equation then becomes
m dv = -u dm equation 5
or
(1/m) dm = -(1/u) dv. equation 6
Integrating m from (mu + phi) (mass of payload plus mass of fuel) to mu (mass of payload without fuel), and v from zero to v (final deltavee), we find
v = u ln rho equation 7
or
rho = exp (v/u) equation 8
where rho == (mu + phi)/mu, and represents the initial-to-final mass ratio; (rho - 1) represents the amount of fuel required for the trip, expressed in units of the mass of the payload; thus a rho of 3 indicates a fuel requirement of twice the mass of the payload.

Taking special relativity into account, this equation becomes (left as an exercise for the reader)

v = c (rho2 u/c - 1)/(rho2 u/c + 1) equation 9
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