Reaction drives and deltavees


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 initialtofinal 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 (rho^{2 u/c}  1)/(rho^{2 u/c} + 1)

equation 9


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