perm filename ASTRO7.ART[ESS,JMC] blob
sn#179972 filedate 1975-10-07 generic text, type T, neo UTF8
We shall derive formulas for the time required to accomplish
an interstellar journey of distance %s% on the basis of the following
technology: energy from a nuclear reactor (fission or fusion) is used
to expel mass at a velocity that is varied during the mission in an
optimal way. We shall assume that the performance of the system
(reactor + rocket) is characacterized by a number %p% equal to the
power the system can handle per unit mass of apparatus. The
plausible values of %p% are between one watt/kilogram and 1000
watts/kilogram. Since the time turns out proportional to %p-1/3%,
this will only mean a factor of ten in the time required to
accomplish a given journey.
We introduce symbols as follows:
s = length of journey
T = time of journey
t is a time variable
M = initial mass of the system
m0 = final mass of the system
m is a mass variable
α = M/m0 = the mass ratio
w is an exhaust velocity variable
a(t) is the acceleration
p = power available for unit mass of system
P is a power variable.
Conservation of momoentum and conservation of energy give the
following equations:
.
1) -mw = ma
. 2
2) P = -1/2 m w .
.
Solving for %w% in 1), substituting in 2) and solving for %m%
gives
3)
We now distinguish two cases: in a single stage rocket, we
have
4)
expressing the fact that the power available is proportional to the
final mass of the system.
In a continuously staged rocket, we have
4') P = pm,
expressing the fact that the power available is proportional to the
current mass. (We suppose that every so often a nuclear reactor or
a rocket is taken out of service, vaporized and expelled as working
fluid).
In the two cases, we get the equations
5)
and
5')
Taking into account the initial and final masses we get the
following results by integration:
6)
and
6')
Using %α%=M/m % and setting for the two cases
7) q = p(1 - 1/α)
and
7') q = p log α,
we get the following equation valid in both cases:
8)
Assuming that the journey begins and ends at rest we have
9)
The final distance is given by
10) s =
from which
11) s =
follows by integration by parts.
Our goal is now to determine the acceleration profile %a(t)%
satisfying equations 8),9) and 11) so that %T% is minimized for a
given %s. Before doing this, however, we shall treat the simple case
in which we use a constant magnitude acceleration reversed in sign at
the midpoint of the journey. This assumptions gives from 8) and 11)
12)
and
13) s =
Solving for %T% and %a% gives
14)
and
15) a =
We shall now labor mightily to optimize %a(t), but the eager
reader is warned that this only changes the co-efficient 2 in
equation 14) to 1.817 which might not be considered worth either the
mathematics or the engineering. Well, onward!
Instead of holding %s% fixed and minimizing %T, we take the
equivalent but simpler problem of maximizing %s% holding %T%
constant. Using Lagrange multipliers and taking variations gives
16)
and since this must hold for arbitrary variations %%a(t), we have
17)
Combining 17) with 8), 9), and 11) gives (if we have finally
gotten the algebra right)
18)
and
19) T =
Thus if we optimize acceleration and use continuous staging,
we get
20) T =
As a numerical example, we let %s%=%10 ~ 100 light years,
p%=%1000, and α%=%e%%~%3000. We then get
T = .9 10 sec = 3000 years.
Remarks: The zero in acceleration at the midpoint means that the
calculation is invalid at that time because with constant power, that
corresponds to infinite exhaust velocity. The actual optimum profile
involves emitting only the waste products of the nuclear reactor near
the midpoint.
Since the time is proportional to the 2/3 power of the
distance, clearly the formula is incorrect for long distances. It
becomes incorrect when it recommends exhaust velocities greater that
corresponding to emitting only the waste products of the nuclear
reaction.