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.