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C00002 00002	THE FEASIBILITY OF INTERSTELLAR TRAVEL
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THE FEASIBILITY OF INTERSTELLAR TRAVEL


	Optimists have proposed many schemes for interstellar travel,
usually aimed at reaching nearer stars within a human lifetime, but
these schemes usually involve extrapolations of present science.
Pessimists, finding flaws in these schemes have often concluded that
interstellar travel is forever infeasible.  The purpose of this article
is to show that interstellar travel is entirely feasible based on
present science and quite small extrapolations of present technology,
provided travel times of several hundred to several thousand years
are accepted.  Naturally, no-one is going to start journeys that
take a thousand years within the next few hundred years unless fleeing
a danger, because a faster method of travel may be discovered
that will permit an earlier arrival with a later start.
However, since interstellar travel in a few thousand years is
feasible with even our present technology and
our solar system will support life
for several billion years, it is hard to imagine that journeys of many
thousand years will not be undertaken over the course of this time
unless all of mankind comes under a dictatorship that forbids it.

	The technology proposed is to use a nuclear fission reactor
to generate electricity which is used to expel a working fluid
at an exhaust velocity optimized over time for the journey being
undertaken.  If the exhaust velocity is chosen too low, the mass
ratio of the rocket system is unacceptable, and if the exhaust
velocity is too high, very little thrust can be obtained given
plausible assumptions about the power that can be handled.  In
fact, the exhaust velocity used should vary during the journey.

	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 and maintaining the validity of 9).
Introducing Lagrange multipliers, we must hold stationary the integral

	16)

subject to the conditions 8) and 9) for arbitrary variations of a(t).
This gives

	17)

which must hold for arbitrary variations  a(t).  Therefore

18)

	Combining  18) with 8), 9), and 11) gives (if we have finally
gotten the algebra right)


19)


and


20)	T =


	Thus  if we optimize acceleration and use continuous staging,
we get


21)	T =


	As  a  numerical  example,  we let  s = 10   meters ~ 100 light years,
p = 1000 watts/kilogram, 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.
We can calculate this condition most readily for the case of constant
acceleration and continuous staging.  Solving equations 1) and 2) for
w  gives


22)	w =


which leads finally to


23)	w =


	As a numerical example, choose  α = e  and  p = 1000 again,
and solve for s getting


24)	s =


in MKS units.  The limiting case of a fission reactor using all
fissionable material and expelling only the fission products as
reaction mass and using 200,000,000 electron volts as the energy
of fission gives  s = 1700 light years, i.e. the equations of
this article are valid for journeys shorter than this.  Longer
journeys are energy limited, and the formulas customary for
chemical rockets apply.  If controlled fusion is feasible as
a source of energy and gives more energy per gram of fuel, then
the formulas apply out to correspondingly longer distances.

	The conclusion is that the galaxy could be occupied by
humanity in a time small compared to the time during which our
solar system will support life.