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6.0 Space travel

       "Space, the final frontier. These are the voyages of the
   starship 'Enterprise'. Its continuing mission, to explore
   strange new worlds, to seek out new life and new civilizations,
   to boldly go where no one has gone before..."  This is of
   course the prologue of the television series "Star Trek: The
   Next Generation".  In the world of make believe its easy to travel
   at warp speed or faster than the speed of light, but we can't.
   One of the major problems of space travel is that warp drives
   are not possible.  Even worse, we can't at the present time
   even approach the speed of light.  Recall that the speed of
   light is about 300,000,000 meters per second or 300,000 km
   per second.  The highest projected velocities published so
   far are in the neighborhood of 300 kilometers per second
   [53, p.34] which is a mere 0.1 percent of the speed of light.
   The distances to be traveled in space are so large that travel
   times will be correspondingly long. Travel times to the outer
   planets are measured in years and travel times to the nearest
   stars are measured in generations.  Most estimates of the travel
   time to nearby destinations such as Mars are in the range of 8
   to 10 months.
       Traveling is really a very boring activity.  You are merely
   moving your physical body from one place to another.  Too bad
   that teleportation doesn't work.  Of course that doesn't impede
   science fiction writers or movie makers.  Space travel has
   the potential of being exceptionally boring.  Who would look forward
   to a nine month journey in microgravity which would be followed by
   another similar trip to come home?  Very few.
       We will show how the duration of space voyages can be
   reduced significantly.  The method to be shown includes artificial
   gravity for the crew members.   This is a major improvement over
   previous space trips because life will be much more like here
   on earth.  And most important of all, the bodies of the crew
   will be spared the debilitating effects of decalcification of
   their bones.  Decalcification is similar to osteoporosis only
   worse.  In microgravity, the calcium is flushed out of the bones
   at an alarming 1-2 percent per month [101, p.107].  Under those
   conditions a person would lose more than 10 percent of their
   bone calcium in a 9 month trip to Mars - and 10 percent more
   on the return trip (see section 6.7).
       There is little doubt that there are millions of people all
   over the world who are interested in space exploration.  But
   the reality is that 99 percent of those people (over 18)
   have to earn a living and at the present time there are no
   jobs in space.  There are in fact only a few tens of thousands of
   jobs in the satellite business and in the launch services
   industry, but those people are not really in "space".
       Many (but not all) famous explorers have had major
   sponsors who paid the bills and thereby allowed the explorers to
   seek their glory - such as Christopher Columbus or Lewis & Clark.
   We are now seeking sponsors for our real-life trips to the moon and
   Mars and beyond.   We face the same problem that Christopher Columbus
   did which was to convince his sponsors that he could bring back
   great wealth from the new world.   We shall do the same.
       We will convince our sponsors that we too can bring back great
   wealth from the moon and other space places.
   6.1  Rockets
       So far all of the satellites that have been put in orbit and
   all of the manned space excursions have been made with rockets.
   Although we have heard of dozens of exotic spaceship propulsion
   systems, not a single one (even the one we shall tout below) has
   gotten off the drawing boards and into space.  Sadly, the dumb
   rocket is perhaps the least efficient of the spaceship propulsion
   systems that have been proposed.  So why haven't the others been
   built?   Some require technology that we don't have.  Some require
   exotic fuels that we don't have.  Some are not practical for
   manned voyages.  Some won't work.  Some are too dangerous, etc.
       Rockets simply burn fuel, either solid or liquid, which
   converts said fuel into a gas which is expelled from the rocket
   at as high a velocity as possible.  The escaping gas then pushes
   the rocket in the opposite direction according to Newton's third
   law of motion.  The equation which describes the relationship of
   the final mass of the rocket (m), as compared to the rocket's
   initial mass (M) is :
*      M = m * exp( dv/g*Isp )         6.1-1
       where "dv" is the total velocity change of the mission and
   "Isp" is the specific impulse of the propulsion system and
   "g" is the acceleration of gravity.
       The total velocity change refers to the sum of all of the
   small velocity changes which constitute the mission. For example,
   an indirect trip to the moon would involve the following velocity
*  Table 6.1-1  Earth to Moon delta velocities
   Earth to low earth orbit (LEO)        9300 m/s
   Trans-Lunar injection (TLI)           3200 m/s
   Mid course correction                   80 m/s
   Lunar orbit insertion (LOI)            800 m/s
   Lunar landing                         2100 m/s
         total dv                   =   15480 m/s
        The most important point is that the higher the specific
   impulse the better.  A higher specific impulse means that less
   propellant is required which in turn translates to more payload.
   Usually (but not always) it also means a greater spaceship velocity
   which means at shorter trip time.   The space shuttle uses
   liquid oxygen and liquid hydrogen as fuel - specifically about 555
   MT (metric tons) of liquid oxygen [71, p.80] and 92.5 MT of liquid
   hydrogen [LB1, p.60].  These two liquids burn to produce water
   - a non-polluting compound.  The major problem with liquid oxygen
   and liquid hydrogen is that they are difficult to handle.  The
   temperature of liquid oxygen is -182.962C and the temperature of
   liquid hydrogen is -252.87C [70, p.532-3].  The specific
   impulse of this fuel mixture is about 475 seconds - which is
   very high when compared to most chemical reactions and exceeded
   by only a few - such as the following:
*  Table 6.1-2  High specific impulse chemical propellants
   Hydrogen and Florine            528 seconds
   Ozone and Hydrogen              607
   Florine and Lithium hydride     703
   Oxygen and Beryllium hydride    705  [66, p.43]
       Another very useful equation is the one which allows us
   to calculate the number of tons of fuel required for each ton
   of payload.  This equation will help us determine the costs of
   various proposed missions.  When any rocket propelled mission
   takes off its weight consists of the following components: (1)
   the payload, (2) the fuel, (3) the fuel tanks, and (4) the rocket
   engines.  When the mission is complete (single stage rocket only)
   the fuel is gone but everything else remains.  These relationships
   can be expressed in two equations as follows.
*      initial mass = fuel + tanks + engines + payload;        or
         mi         =  f       +  t + e + p;           6.1-2
       final mass   =        tanks + engines + payload;        or
         mf         =       t + e + p;                 6.1-3
       We want to solve for "fuel" in terms of "payload".  Equation
   6.1-1 allows us to eliminate the initial and final mass terms
   because the ratio of those two masses is the mass ratio.  Thus we
   have the equation:
*                mi  =  mass ratio  *  mf             or
       f + t + e + p = mass ratio * ( t + e + p );    6.1-4
   We will now use "mr" for mass ratio and "x*f" for "t+e" giving:
*          f + x*f + p = mr * ( x*f + p );            or
       f = p * ( mr - 1 ) / ( 1 + x  - mr * x );      6.1-5
       Note that "x*f" expresses the weight of the tanks and engines
   as a fraction of the weight of the fuel - i.e. 0 < x < 1.0.
       Now we will show a short example to clarify how equations 6.1-1
   and 6.1-5 are used.  Suppose we want to go from LEO to the surface
   of the moon.  What is the required velocity change?  It is given in
   table 6.1-1.  It is the sum of the last four numbers or 3200 + 80
   + 800 + 2100 = 6180.  We will use a liquid oxygen and liquid hydrogen
   fueled rocket with a specific impulse of 475 seconds.  The
   acceleration of gravity (g) is 9.8 m/s/s.  Now we calculate the mass
   ratio using equation 6.1-1.
*      M = m * exp( dv/g*Isp )         6.1-1
       M = m * exp( 6180/9.8*475 )      or
       M = m * exp( 1.32760 )           or
       M = m * 3.772
       The mass ratio is 3.772.  Now we want to know how many tons
   of fuel are required for each ton of payload.  All we need to know
   is what the weight of the engines and tanks will be as a percentage
   of the fuel weight.  Let's say that the tanks and engines weigh
   8% of the fuel.  Then we have:
*      f = p * ( mr - 1 ) / ( 1 + x  - mr * x );            6.1-5
       f = p * ( 3.772 - 1 ) / ( 1.08 - 3.772 * 0.08 );       or
       f = p * ( 2.772 ) / ( 0.77824 );           or
       f = p * ( 3.562 );
       The answer is that we need 3.562 tons of fuel for each ton
   of payload.  In this case that will be 6/7 liquid oxygen or 3.053
   tons of oxygen and the rest or 0.509 tons of liquid hydrogen.  One
   interesting observation that can be made from equation 6.1-5 is
   that certain missions are impossible with single stage rockets. If
*      mr * x  >=  1 + x     (mission impossible)          6.1-6
       where "x" is the weight fraction of tanks and engines and
   "mr" is the mass ratio -  then the mission is impossible.
       We would like to introduce one more equation which tells us
   the "payload fraction" or the fraction of the total weight that is
   payload.  That relationship is:
*      payload fraction = (1 + x * ( 1 - mr)) / mr         6.1-7
       where "x" is the weight fraction of tanks and engines and
   "mr" is the mass ratio.
       This equation can be derived in a manner similar to the previous
   one. We have omitted the derivation to avoid boring too many readers.
   6.2  Nuclear rockets
       "Nuclear rockets" refers to rockets that use nuclear power
   as an integral part of the propulsion system rather than merely
   to supply electricity for systems on board a research spacecraft
   such as Voyager 1 or 2.   There are at least five or six
   different types of nuclear propulsion schemes: nuclear thermal
   propulsion, nuclear electric propulsion, the Orion project, the
   fusion rocket, Project Daedalus, and my proposal detailed in a
   subsequent section.  Most have been known for many years.
   William R. Corliss gives detailed accounts of some of them in his
   excellent book, "Propulsion Systems for Space Flight" [Ref 50]
   written in 1960.  Another excellent book is "The Starflight Handbook",
   by Mallove and Matloff, published by John Wiley in 1989 [Ref 66].
   Their book lists large numbers of exotic space propulsion systems
   and discusses each one in fascinating detail [66, p.43-145].
       In this section I will briefly review three of the more
   serious proposals, leaving the others to be covered under the
   "exotic" schemes section.
       The first is nuclear thermal propulsion [50, p.153-161].  This
   concept is quite trivial.  It simply uses a fission type nuclear
   reactor to heat some propellant to a very high temperature before
   exhausting it from the engine.  Thus it is quite similar to the
   chemical rockets - except that it can attain higher specific
   impulses because of the lower molecular weight of the exhaust gases
   [AW 33, p.18].  The same article (Aviation Week of 4/18/91) projected
   a specific impulse of 1000 seconds [AW 33, p.19] for this type
   of propulsion system.  This is more than twice the specific impulse
   of LOX/LH2 - which is about 475 seconds.  The cost to develop a
   propulsion system of this type has been estimated at $7-$8 billion
   [AW 33, p.18] and SDI has already invested more than $40 million
   in its investigation of the concept.  Mallove & Matloff describe
   three variations of nuclear fission propulsion systems - namely:
   solid core, liquid core and gas core [66, p.44-7].  The specific
   impulse of these systems are 500-1100 seconds for the solid core,
   1300-1600 seconds for the liquid core, and 3000-7000 seconds for the
   gas core type [66, p.44-7].
       The second is nuclear electric propulsion [AW 32, p.24-5]. In
   this method, the propellant is heated so hot that it becomes a
   plasma which is then accelerated by electrostatic or electromagnetic
   fields to increase the exhaust velocity.  Since specific impulse
   is a function of exhaust velocity, this means a higher specific
   impulse.  Such a nuclear electric plasma rocket has already been
   built and its specific impulse is in the range of 800 to 30000
   seconds [AW 32, p.24].  Specialists claim the nuclear propulsion
   systems can reduce the travel time to Mars from nine months to
   five and perhaps as low as three months [AW 32, p.25].  Also see
   [50, p.173-182] on plasma jets.
       The third is basically the same as the second except that
   nuclear fusion is used instead of nuclear fission.  J.F. Santarius
   discusses spaceship propulsion by nuclear fusion using Helium-3
   as the fuel in [7, p.4-6].  A more recent article by Edward Teller
   et al (including J.F. Santarius) describes nuclear propulsion by
   fusion in a magnetic dipole [53, p.1-34]. That proposal also
   requires Helium-3.  Corliss also briefly mentions nuclear fusion
   [50, p.169-170], but not using Helium-3.  Again, as in the other
   schemes, the idea is to increase the specific impulse.  In both
   of the above mentioned fusion propulsion systems the specific
   impulse is very high - as much as 1,000,000 seconds.  Of course
   the thrust is extremely low but both of these systems are
   tunable and can theoretically reach Mars in about three months
   [53, p.34].  J.F. Santarius [7, p.5] gives a thrust to weight ratio
   of about 5e-5 for fusion propulsion, which is in agreement with
   figures of 1e-4 to 1e-5 as given in [66, p.48].
   6.3  Other exotic methods of space travel
       A list of 52 propulsion concepts courtesy of Dani Eder
   is given in "Mirror Matter" by R.L.Forward and J.Davis on pages
   222-223 [Ref 68].  This indicates that there is a wide variety
   of propulsion systems available.  I will only describe a few of
   these systems.  I have selected these few because they seem to be
   popular in the literature and not because they are relatively
   better or worse.  These schemes are: the interstellar ramjet,
   the solar sail, the matter-antimatter rocket, the ion rocket
   and the Orion project.
       The ion rocket is quite similar to the nuclear electric
   propulsion system in that it uses some device to ionize a neutral
   propellant to produce a plasma which is then electrostatically
   or electromagnetically accelerated to the desired velocity and
   then is finally neutralized again just before it is expelled from
   the exhaust nozzle.  See [50, p.195-215], or [68, p.221,225-6]
   or [66, p.43-4].
   Typical ion rockets have specific impulses between 5000 and
   25,000 seconds and thrust to weight ratios of 1e-4 to 1e-5
   [50, p.213].  If the ionizing device is a nuclear reactor, then
   the nuclear electric propulsion system could be called an ion
   rocket.  Similarly, if the electric power source is an array of
   solar cells, then this type of rocket is sometimes called a
   solar electric propulsion system or a photovoltaic ion rocket
   or electric propulsion.
       The Orion project was a propulsion system based on atomic
   bombs.  One simply tossed bombs out the back and then exploded
   them.  The drawbacks of this scheme are so obvious that it is
   difficult for me to understand how anyone could actually approve
   the expenditure of money on any part of it.  Not only does it
   produce a constant stream of highly radioactive and dangerous
   products, but it also generates wave after wave of intense and
   dangerous radiation.  And on top of that, only a small percentage
   of the power of the explosions can be converted into thrust to
   propell the spaceship.  See [66, p.61-6], or [50, p.167-9] or
   [68, p.226-228].  Project Daedalus was very similar except that
   nuclear fusion "bombs", actually tiny fuel pellets, were to be
   used instead of fission bombs [66, p.67-9].
       The solar sail scheme has received a lot of attention lately
   primarily due to the support of this scheme by Carl Sagan.  Solar
   sails are one of only four known "fuelless" propulsion systems.
   The basic scheme is to construct a very very large but also very
   thin (and therefore fragile) sail which reflects the light of the
   sun thus propelling the sail in the opposite direction.  Instead
   of using wind as a ship sailing on the ocean does, this sail
   harnesses the light of the sun to sail in space.  The only positive
   attribute of this system is its avoidance of the fuel problem.
   It is difficult to overestimate this factor, but one must keep in
   mind that the solar radiation decreases with the square of the
   distance from the sun.  This means that the light pressure that
   is available at two astronomical units is only one fourth of that
   available at one astronomical unit (earth's orbit).  Solar sails
   are discussed extensively in both [50, p.247-54], and [66, p.89-105].
   The thrust to weight ratio is about 0.0002 according to Corliss
   [50, p.251].  Again this leads to long slow trips.  A more optimistic
   appraisal of solar sails can be found in [SM 17] by K.E.Drexler.
   He has developed a technique for manufacturing ulta-thin foils
   which he claims allows a thrust to weight ratio of between 0.01
   and 0.1.  His article suggests a cost of 12 to 26 cents per
   square meter of sail area, but his sails will be measured in square
   kilometers not square meters.  Although this system may be suitable
   for freight missions it is too slow for human use.  Drexler
   estimates that a sail would take 50 days to move a payload from
   LEO to GEO [SM 17, p.435].  He estimates an acceleration of 0.05
   m/s/s for a foil of 50nm in thickness or an acceleration of 0.03
   m/s/s for a foil of 90nm in thickness [SM 17, p.433].  The earth's
   gravitational acceleration is 0.03 m/s/s at a distance of about
   115,000 km or about 30% of the way to the moon and is about
   0.226 m/s/s at GEO.  There are other problems with this scheme which
   need some attention, such as: how much will the solar wind damage the
   sail, or how much will dust particles or micrometeoroids damage
   the sail.  With such a large sail area, the probability of
   collisions with meteroids will surely be significant. The
   earth sweeps up 10,000 tons of micrometeoroids each year [34, p.9].
   In addition, one meteroid larger than 100 grams falls on each
   21,100 square kilometers of the earth each year [34, p.9].  That
   means that a sail whose area is 21,100 square kilometers would take
   about one hit per year from a meteoroid of 100 grams or larger.
       Other versions of this scheme call for sails to be driven
   by lasers or microwaves.  Still others use lasers here on earth
   to push spaceships in space.  All of these schemes are simply
   too slow to be useful for manned voyages in space.
       The interstellar ramjet is the other "fuelless" propulsion
   system.  This idea is usually attributed to Robert Bussard who
   first described it in "Galactic Matter and
   Interstellar flight" published in 1960 [Ref 100].
   His idea was to use a giant funnel or scoop to catch the
   hydrogen atoms which exist in very very low concentrations
   throughout interstellar space.  The funnel would guide the gas
   to a nuclear fusion reactor at the center which would fuse the
   hydrogen into helium and use the liberated energy to heat and
   expell the gas.  See [50, p.65-6], or [66, p.108-113].  Bussard
   assumed a density of 1000 hydrogen atoms per cubic centimeter.
   Unfortunately it now appears that estimate was orders of magnitude
   off [68, p.230].  Consequently a much larger scoop is required.
   The two attributes of this scheme which are most attractive are
   the fact that it doesn't need to carry any fuel and that it can
   theoretically achieve speeds close to that of light.  Of course
   its drawbacks are obvious.  By definition it is not for use
   within the solar system and even if you got outside of the solar
   system, it would take years to reach the closest star.
       The matter-antimatter rocket is perhaps the most exotic of
   all.  R.L. Forward and J. Davis have devoted an entire volume to
   the discussion of this idea [Ref 68, "Mirror Matter"].  For those
   people who have had little training in physics, I will briefly
   outline the situation.  All the things we see are composed of
   matter, but physicists have discovered that other kinds of particles
   exist which are called antimatter.  The first such particle was
   discovered at Cal Tech by Carl Anderson in 1932.  It was the
   anti-electron although it was named the positron.  The anti-proton
   was found at Lawrence Berkeley Laboratory in October of 1955.
   These anti-particles are almost identical to their normal matter
   twins and hence their names.  They differ in their electrical charge
   and their handedness [68, p.9].  Both the positron and the anti-proton
   are electrically charged particles, but they could be combined to
   produce anti-hydrogen which should be a stable gas with the same
   chemical characteristics as normal hydrogen.
       The most exciting property of antimatter
   is that when it comes in contact with normal matter, both
   particles are annihilated and other energetic particles are
   produced.  This forms the basis of a propulsion system.  The
   particles produced during annihilation contain all the energy
   of the starting particles according to Einstein's famous equation
   of mass and energy equivalence.
*      E = m * c * c;                               6.3-1
       where "c" is the speed of light, "m" is mass,
   and "E" is energy.
       These reaction products are moving at 94% of the speed
   of light and can be used to heat a working fluid (propellant)
   such as (normal) hydrogen or they could be used as reaction
   mass themselves [68, p.135].  Forward and Davis claim that the
   matter-antimatter rocket has a mass ratio (as described in
   equation 6.1-1) which "will never be worse than 4.9, ever!" [68,
   p.197].  This is clearly false.  One need only select a mission
   whose characteristic velocity is twice the speed of light or greater
   to see that.  Such a mission would be to visit a black hole and
       Whether or not this is possible is really irrelevant
   because the real problem is that we don't have sufficient
   antimatter to make it work.  Forward and Davis state that if
   antimatter can be manufactured for less than $3 million per
   milligram, that it will be financially competitive with nuclear
   propulsion [68, p.199].  It takes Fermilab about a week to produce
   1e-12 grams of antimatter [68, p.150].  At that rate it would take
   them 20,000,000 years to make a milligram.
   6.4  Spaceship propulsion by momentum transfer
       For years space scientists the world over have sought a
   propulsion system which would significantly reduce interplanetary
   travel times.  My colleague, Albert Wu, and I have discovered
   such a propulsion system.  This propulsion system works by
   transferring momentum from a series of "smart" projectiles
   launched from an electromagnetic projectile launcher located at
   the north pole of the moon, to the spaceship by catching the
   projectiles with another electromagnetic projectile launcher
   on board the spaceship.  A spaceship with a crew of 1000 can now
   travel to Mars in under two months and to Jupiter in under a year.
       The prime mover is nuclear power which converts mass to energy
   according to Einstein's equation (6.3-1) and drives the EMPLs
   which launch the projectiles.  As these fast moving projectiles are
   caught by the spaceship's EMPL, their momentum will be transferred
   to the spaceship according to Newton's third law of motion.  This
   reaction will accelerate the spaceship in the desired direction
   by an amount proportional to the ratio of the mass of the projectile
   to the mass of the spaceship.  (Technically speaking, all propulsion
   systems work by transferring momentum from the fuel to the spaceship
   but in this system the projectiles, the primary "fuel", are NOT
   carried on board the spaceship.)  As each projectile is caught, the
   combined mass will be increased by the mass of one projectile. This
   means that some of the momentum of subsequent projectiles will be
   used to accelerate projectiles that have already been caught.  That
   energy is wasted.  For that reason, the projectiles will be launched
   toward the spaceship in groups - each projectile separated by the
   time and distance required by the EMPL on board the spaceship to
   charge up its capacitors to catch the next projectile.  Once the
   spaceship has caught all of the projectiles in one group, it will
   proceed to launch them again in the direction opposite to the desired
   trajectory of the spaceship.  In this manner each projectile will
   contribute two small acceleration pulses to the spaceship.
       Let me explain some of the trade-offs that are involved here.
   The spaceship would consist of three major components: the EMPL,
   the nuclear power system, and the crew's quarters.  Each of these
   components has its own size, mass, and cost.  Clearly the size of
   crew's quarters will vary directly with the size of the crew.  That
   is, twice the crew will require roughly twice the size, mass, and
   cost of crew's quarters.  The same is not however true of the other
   two components because they are interdependent.  We have calculated
   that the spaceship will require an EMPL which will be about 6
   kilometers long and a nuclear power system capable of continously
   generating about 500Mw of electricity.  Both of these requirements
   are quite large and both correspond to large size, mass, and cost.
   Suppose we wanted to reduce the length of the EMPL by half to 3
   kilometers.  What would the consequences be?
*     Positive consequences
   1. roughly half the size, mass, and cost of the EMPL
      Negative consequences
   1. projectile acceleration would be twice as high
      which would mean -
   2. higher development and production costs for all
      projectiles to survive the higher acceleration.
   3. twice the acceleration of the spaceship when
      catching or launching each projectile which would mean -
   4. more stress on the crew during launching/catching and
   5. more stress to the spaceship which would mean more
      structural support which means more size, mass, and
   6. twice the power would be necessary which means roughly
      twice the size, mass, and cost for a 1000Mw nuclear
      power system - OR -
   7. twice the capacitor charging time between each projectile
      which would lengthen each flight by some number of days.
       It is clear that some additional study is warrented to try to
   determine the optimal parameters - preferably by some people who
   are more familiar with EMPL and nuclear power technology than we
   are.  Nevertheless we offer the following paragraphs to give you the
   flavor of how the system would work.
       The capacitor driven EMPL on the moon will be about 10 kilometers
   long and will be capable of launching one metric ton projectiles at
   velocities of up to 20 kilometers per second.  This will require
   something on the order of 200,000 MJ of energy for each launch -
   which will be supplied by a 2500 Mw nuclear power system.  The reactor
   will require about two minutes to charge the capacitors between the
   launching of each projectile.  The launcher itself will be mounted
   on a series of circular tracks, so that it can be rotated between
   each launch to compensate for the orbital motions of the earth and
   the moon.  The earth orbits the sun in a prograde direction at about
   29.8 kilometers per second and the moon orbits the earth also in a
   prograde direction at about 1.02 kilometers per second.  Thus for
   launches two minutes apart, the launcher would be rotated in the
   retrograde direction by 5 to 10 meters depending on the final velocity
   of the projectile.  Since the moon is locked in its orbit about the
   earth, the same part of the moon always faces earth.  This means that
   as the moon rotates there are only two points on its surface from
   which an observer should be able to constantly observe the sun (except
   for eclipses) - namely, the north and south poles.  Since it appears
   that there is significant interest in the south pole for astronomical
   observations, the north pole has been selected for the site of the
   lunar EMPL.
       In order to accelerate the spaceship to cruising velocity it will
   be necessary to throw several thousand projectiles from the moon to
   the spaceship (assuming none is reused).  Since each individual
   projectile will be separated by about two minutes and each group by
   several hours, this process will take several days.  Thus the moon
   will rotate significantly during the launching of the projectiles -
   perhaps as much as 45 degrees.  The tilt of the orbital axis of the
   moon relative to a normal to the ecliptic plane may prevent launching
   of projectiles during part of the moon's orbit, but this will be only
   a small percentage of the time and will not interfere with our plans.
       The entire spaceship will be rotated axially
   about the long axis of the EMPL to provide gyroscopic stabilization
   and artificial gravity for the crew.  In order to make the artificial
   gravity relatively uniform, the crew's quarters will be contructed
   in the shape of a ring.  The ship's EMPL will be about 6 kilometers
   long and the on board nuclear power system will be rated at about
   500 Mw.  These capabilities should allow the spaceship to launch or
   catch one metric ton projectiles at up to 10 kilometers per second
   at roughly two minute intervals.  Of course it could handle higher
   velocity projectiles at the cost of longer capacitor charging times.
       The anticipated mass of the spaceship is about 3000 metric tons,
   but could be increased to say 4000 MT if it were necessary.
   This includes 100 metric tons for the nuclear power system
   (0.2 tons/Mw), 1900 metric tons for the EMPL (317 kg/m), and 1.0 MT
   per crew member for each of the 1000 crew members.  The allocation
   per crew member includes the following: the crew member, his food
   production facilities, his equipment, his living quarters, etc.
       Let us take a look at those numbers.  R.S. Lewis states that Glen
   Seaborg has designed a flyable reactor not much larger than a desk
   which would be capable of producing 1500Mw of electricity [15, p.153].
   If this is not a joke (suppose he was referring to an atomic bomb)
   and if its lifetime is not too brief, it would surely weigh less
   than 100 MT and would serve our purpose.   In 1969 the AEC
   completed tests of a nuclear engine called NERVA.  That engine,
   which was no bigger than a medium sized truck, produced 1500Mw
   of thermal power [15, p.153].  That same engine was also tested
   successfully at a burst rating of 5000Mw [72, p.303].  Brookhaven
   National Laboratory has built a gas core particle bed reactor
   that can produce 200Mw from a 300kg 1.0 by 0.56 meter package
   [72, p.302].
       The story on the EMPL is less optimistic, but surely it too
   can be done.  From the above estimate of 200,000 MJ, you can
   calculate that this is about 34 MJ per meter of EMPL.  In 1974
   a 5 MJ capacitor occupied 63 cubic meters and weighed 144 tons
   [AW 23, p.89].  In 1985 a 5 MJ capacitor occupied 8.5 cubic
   meters [73, p.16].  And in 1990 that same capacitor occupied
   only 0.8 cubic meters [73, p.16].  But it still weighed 1.8 tons
   [AW 23, p.89].  Maxwell Industries Inc believes they can get
   that down to 200kg and 0.1 cubic meter [AW 23, p.89].  At that
   rate, 35MJ would be 1.4 MT and 0.7 cubic meters.  That would put
   us within about a factor of ten of the desired specification.
       Food will be grown on board in a hydroponic production facility.
   Due to the artificial gravity, the crew will be able to move
   about relatively normally - to eat, to sleep, and to accomplish
   simple bodily functions normally.  In addition, human muscles should
   not atrophy and extensive daily exercise programs will not be
   necessary.  Perhaps most importantly, the artificial gravity should
   be sufficient to inhibit the bones of the crew from decalcifing - a
   well known and very dangerous consequence of weightlessness.
       There are numerous reasons for carrying a large crew. Among
   them are the following:
*  1. It will transform the first Mars mission into the greatest
      international expedition of all time.
   2. It will cause an exponential increase in public interest
      and political support for the project.
   3. On board seats can be sold to the public worldwide to raise
      funds to support the project.
   4. The cost per person will be greatly reduced from that which
      would occur if a crew of fewer than 10 were sent on a
      similar mission.
   5. The knowledge and experience to be gained from a large
      crew is clearly much greater than could be accomplished
      with a small crew.
   6. The crew will not be a small elite group selected in some
      obscure suspicious way by unknown and untouchable
      bureaucrats or governments.
   7. Crew members need not be special in any way (except perhaps
      in not being seriously ill); however, since weight will be
      important, women may have a preference.
   6.4.1 EMPL acceleration
         The amount of acceleration required to produce a given final
   velocity  is  a function of two primary factors: first, the length
   of the EMPL and second, the strength of the electromagnetic fields
   which  is  in  turn a function of the power available. Table 6.4-1
   shows  the  acceleration  required in thousands of Earth gravities
   (Kg)  to achieve the given velocity, 'V', in kilometers per second
   for electromagnetic projectile launchers of various lengths from 2
   kilometers  to  10 kilometers. The columns labeled by 'T' give the
   time  in  seconds  necessary to achieve the given velocity. Recall
   from  basic  physics  that  velocity,  'V',  equals the product of
   acceleration,  'A',  times time, 'T', for a constant acceleration.
   Also for a constant rate of acceleration, distance, 'D', travelled
   equals  one  half  of  the  product of that acceleration times the
   square of the time. Expressed in equations we have:
*        V = A * T                   6.4-1
         D = 0.5 * A * T * T         6.4-2
   By  squaring  the  first equation and substituting into the second
   equation we get:
*        D = V * V / ( 2 * A )       6.4-3   or
         A = V * V / ( 2 * D )       6.4-4
   Equation  6.4-4  was  used to calculate the results of table 6.4-1
   which  were converted to Kgs by dividing the result by 9800 meters
   per  second  per second. By substituting for 'A' in equation 6.4-1
   from equation 6.4-4 we get:
*        T = 2 * D / V         6.4-5
   Table 6.4-1     Required Projectile Acceleration
                   Length of EMPL in Kilometers
             2            4            6            8           10
    V     Kg   T       Kg   T       Kg   T       Kg   T       Kg   T
    5    0.64 0.80    0.32 1.60    0.21 2.40    0.16 3.20    0.13 4.00
   10    2.55 0.40    1.28 0.80    0.85 1.20    0.64 1.60    0.51 2.00
   15    5.74 0.27    2.87 0.53    1.91 0.80    1.43 1.07    1.15 1.33
   20   10.20 0.20    5.10 0.40    3.40 0.60    2.55 0.80    2.04 1.00
   25   15.94 0.16    7.97 0.32    5.31 0.48    3.99 0.64    3.19 0.80
   30   22.96 0.13   11.48 0.27    7.65 0.40    5.74 0.53    4.59 0.67
   35   31.25 0.11   15.62 0.23   10.42 0.34    7.81 0.46    6.25 0.57
   40   40.82 0.10   20.41 0.20   13.61 0.30   10.20 0.40    8.16 0.50
   45   51.66 0.09   25.83 0.18   17.22 0.27   12.91 0.36   10.33 0.44
   50   63.78 0.08   31.89 0.16   21.26 0.24   15.94 0.32   12.76 0.40
   V - velocity in kilometers per second
   T - time in seconds
   Kg - thousands of earth gravity equivalents
       Obviously the time and effort required to construct an EMPL of
   two  kilometers  will be much less than that required to construct
   an  EMPL  of  ten  kilometers - but not one fifth of the time. The
   reason  is that the shorter EMPL is more complex and also requires
   much  more  power  to  drive  it  (unless  we drastically increase
   capacitor  charging  times).  More power means either larger solar
   arrays  or  a  larger nuclear power plant. In either case it means
   more  cost  and  extra  development  time.  There  are  additional
   problems  with  shorter  EMPLs.  First, due to the higher acceler-
   ation,  all  projectiles  will  need  to  be  more  robust   which
   increases  their  complexity, cost, and development time. The EMPL
   itself  must  be more robust. Very likely the capacitors will need
   to be larger which means they will be more costly and require more
   development time.
        For  the  above several reasons we recommend the 10 kilometer
   EMPL.  One  can  see  from  the  above  table  that we can reach a
   velocity  of  20  kilometers  per  second  in  one  second with an
   acceleration of about 2040 earth gravities.
   6.4.2  How many projectiles?
         It  is important to know how many of these smart projectiles
   are required to propel the spaceship. Table 6.4-2 shows the number
   of  projectiles  required,  'N', to increase the velocity, 'v', of
   the  spaceship  by  approximately  one  kilometer  per second. The
   column  labelled  'V'  is  the velocity of the projectiles as they
   leave  the first electromagnetic projectile launcher in kilometers
   per  second.  Of critical importance here is the ratio of the mass
   of the spaceship to the mass of the projectiles. Let that ratio be
   denoted by 'R', then we have:
*  R = mass of spaceship / mass of projectile      6.4-6
   Let us define the following variables:
   M = mass of projectile
   V = velocity of projectile
   P = momentum of projectile
   m = mass of spaceship
   v = velocity of spaceship
   p = momentum of spaceship
   m' =  new mass of spaceship
   v' =  new velocity of spaceship
   The momentum of the projectile as it approaches the spaceship will
   be  given  by the product of its relative velocity, (V-v), and its
   mass, 'M':
*        P = ( V - v ) * M           6.4-7
   The  momentum  of  the  projectile  will  be  transferred  to  the
   spaceship  (  plus  the  projectile  )  when  the  electromagnetic
   projectile launcher on board the spaceship catches the projectile.
   We  can  calculate  the velocity change for the spaceship from the
   law  of  conservation of momentum. Let 'dv' be the velocity change
   of the spaceship. Then:
*        dv = P / m'                       6.4-8   or
         dv = ( V - v ) * M / ( m + M )    6.4-9   or
         dv = ( V - v ) / ( R + 1 )        6.4-10
   Notice  that  the  mass of the projectile cancells out in equation
   6.4-10 leaving us with the ratio. Now we have the following:
*        m' = m + M                        6.4-11
         dv = ( V - v ) * M / m'           6.4-12
         v' = v + dv                       6.4-13
   By iteratively recalculating equations  6.4-11, 6.4-12, and 6.4-13
   we  can find a number 'N' such that the final mass and velocity of
   the spaceship are given by:
*        m' =  m + N * M                   6.4-14
         v' >= v + 500                     6.4-15
   The  spaceship  has  now  caught  'N' projectiles and next we will
   throw them back. In a similar manner we find the following:
*        dv = V * M / m'                   6.4-16
         m' = m - M                        6.4-17
         v' = v + dv                       6.4-18
   Notice  that  in  equation  6.4-16 there  is no subtraction of the
   velocity  of  the  spaceship.  This is because the velocity of the
   spaceship relative to itself is zero. After we have thrown all the
   projectiles, we find that the mass of the spaceship is back to its
   original  value  and the velocity of the spaceship is at least one
   kilometer  per  second  faster than when we started. The following
   table  was  calculated by iterating equations  6.4-11, 6.4-12, and
   6.4-13 until  the  ship's  velocity  was  at  least 500 meters per
   second  faster  than its initial velocity.  Then equations 6.4-16,
   6.4-17, and 6.4-18  were  iterated  until all 'N' projectiles were
   thrown away.  The  final  velocity,  'Vel', is given in the table.
   These numbers are shown for various values of 'R', the mass ratio.
*        Table 6.4-2   Number of Projectiles Required
             for 1 Km/sec Increase in Ship Velocity
        R = 1000     R = 2000     R = 3000     R = 4000     R = 5000
    V    N  Vel       N  Vel       N  Vel       N  Vel       N  Vel
    5  112 1004.25  223 1000.09  334 1000.37  445 1000.50  556 1000.58
   10   53 1009.51  106 1004.64  158 1000.17  211 1000.07  264 1000.01
   15   35 1008.02   69 1001.39  104 1003.68  138 1001.45  173 1002.82
   20   26 1019.92   52 1000.06   77 1000.59  103 1000.49  129 1000.43
   25   21 1008.53   41 1009.39   62 1000.86   82 1003.21  103 1004.32
   30   17 1006.93   34 1007.06   51 1007.10   68 1007.12   85 1001.13
   35   15 1003.11   29 1003.97   44 1003.77   58 1004.04   73 1003.90
   40   13 1029.71   26 1009.85   38 1003.73   51 1000.29   64 1006.23
   45   12 1025.16   23 1003.54   34 1011.33   45 1003.97   57 1008.26
   50   11 1040.79   21 1016.69   31 1008.65   41 1004.63   51 1002.21
   N - Number of projectiles
   R - ratio of mass of spaceship to mass of one projectile
   V - velocity of projectiles in kilometers per second
   Vel - spaceship velocity increase in meters per second
        The  table shows that for a spaceship which weighs 3000 times
   as much as the projectiles, it will require 158 projectiles moving
   at 10 kilometers per second to increase the ship's velocity by one
   kilometer  per  second. Notice that the required number is only 77
   at 20 kilometers per second.
   6.4.3  Spaceship acceleration
         Now  we  wish to calculate the acceleration of the spaceship
   during  the  catching  and  throwing  of  the projectiles. This is
   important because humans are fragile and cannot tolerate very high
   accelerations.  For  this  calculation we have selected a electro-
   magnetic  projectile launcher 6 kilometers in length. We assume it
   is  capable  of  launching  projectiles  according to table 6.4-1.
   Therefore  the  time to catch (or launch) a projectile will be the
   same  as in table 6.4-1. Again 'V' is in kilometers per second and
   'T'  in  seconds. It is clear that the more massive the spaceship,
   the  less  will be its acceleration. The deceleration (or acceler-
   ation)  of  the  projectile  is  given  by  equation  6.4-4.   The
   following derivation applies only to throwing projectiles from the
   spaceship.   Acceleration is generally expressed as a small change
   in velocity, dv, during a small change in time, dt, - or:
*        a = dV / dt                       6.4-19
   We now substitute for dV from equation 6.4-16, giving:
         a = V * m / M / dt                6.4-20
   Notice  that this equation is only approximate because the mass of
   the  spaceship  (denoted by 'M') is changing with each projectile,
   but, since the  mass of all projectiles is still small compared to
   the  mass of the spaceship, the total mass is roughly  equal to M.
   We now replace 'm / M' by its equivalent, '1 / R', giving:
*        a = V / R / dt                    6.4-21  or
         a = V / ( R * dt )                6.4-22
   At this point 'dt' is the  time  required to launch the projectile
   (given by equation 6.4-5).  So we substitute '2*D/V' for 'dt'.
*        a = V / ( R * 2 * D / V )         6.4-23  or
         a = V * V / ( 2 * D * R )         6.4-24
   Finally, we can replace 'V*V/2*D' with 'A' from equation 6.4-4.
*        a = A / R                         6.4-25
        Notice  that the acceleration experienced by the spaceship is
   quite  acceptable  for  velocities  up to 25 kilometers per second
   when the mass ratio is two or three thousand. The columns labelled
   'G' are in Earth gravities.
*        Table 6.4-3 - Ship's Acceleration for EMPL of ** 6 ** Kilometers
         R = 1000     R = 2000     R = 3000     R = 4000     R = 5000
    V      G   T        G   T        G   T        G   T        G   T
    5    0.21 2.40    0.11 2.40    0.07 2.40    0.05 2.40    0.04 2.40
   10    0.85 1.20    0.42 1.20    0.28 1.20    0.21 1.20    0.17 1.20
   15    1.91 0.80    0.96 0.80    0.64 0.80    0.48 0.80    0.38 0.80
   20    3.40 0.60    1.70 0.60    1.13 0.60    0.85 0.60    0.68 0.60
   25    5.31 0.48    2.66 0.48    1.77 0.48    1.33 0.48    1.06 0.48
   30    7.65 0.40    3.82 0.40    2.55 0.40    1.91 0.40    1.53 0.40
   35   10.41 0.34    5.21 0.34    3.47 0.34    2.60 0.34    2.08 0.34
   40   13.59 0.30    6.80 0.30    4.53 0.30    3.40 0.30    2.72 0.30
   45   17.20 0.27    8.61 0.27    5.74 0.27    4.30 0.27    3.44 0.27
   50   21.24 0.24   10.62 0.24    7.08 0.24    5.31 0.24    4.25 0.24
        The following table gives the same data as table 6.4-3 except
   that  the  length  of  the  on  board  electromagnetic  projectile
   launcher is assumed to be 10 kilometers instead of 6 kilometers as
   in table 6.4-3.
*        Table 6.4-4 -  Ship's Acceleration for EMPL of ** 10 ** Kilometers
         R = 1000     R = 2000     R = 3000     R = 4000     R = 5000
    V      G   T        G   T        G   T        G   T        G   T
    5    0.13 4.00    0.06 4.00    0.04 4.00    0.03 4.00    0.03 4.00
   10    0.51 2.00    0.25 2.00    0.17 2.00    0.13 2.00    0.10 2.00
   15    1.15 1.33    0.57 1.33    0.38 1.33    0.29 1.33    0.23 1.33
   20    2.04 1.00    1.02 1.00    0.68 1.00    0.51 1.00    0.41 1.00
   25    3.19 0.80    1.59 0.80    1.06 0.80    0.80 0.80    0.64 0.80
   30    4.59 0.67    2.29 0.67    1.53 0.67    1.15 0.67    0.92 0.67
   35    6.24 0.57    3.12 0.57    2.08 0.57    1.56 0.57    1.25 0.57
   40    8.16 0.50    4.08 0.50    2.72 0.50    2.04 0.50    1.63 0.50
   45   10.32 0.44    5.16 0.44    3.44 0.44    2.58 0.44    2.07 0.44
   50   12.74 0.40    6.37 0.40    4.25 0.40    3.19 0.40    2.55 0.40
   6.4.4  Spaceship power requirements
        The next item of interest is the amount of power necessary to
   operate  these  electromagnetic  projectile launchers - especially
   the  ones  on  board spaceships. The kinetic energy of any mass is
   given by one half the product of that mass times the square of its
*        E = 0.5 * m * V * V               6.4-26
   If the launchers could operate at 100% efficiency, this would give
   us a good idea of the power required. Table 6.4-5 shows the energy
   of a one kilogram projectile travelling at the specified velocity,
   'V',  in  kilometers  per  second.  The energy, 'E', from equation
   6.4-26 is in megajoules. The power, 'P', is given in megawatts for
   three  different lengths of electromagnetic  projectile launchers.
   'P' is calculated by dividing the required energy 'E' by the  time
   'T' from the corresponding entry of table 6.4-1.  Of course, these
   launchers  will not achieve 100% efficiency, but Henry Kolm states
   that EMPLs can achieve "electrical to mechanical conversion
   efficiencies as high as 90%" [SM 2, p.300].
*        P = E / T                   6.4-27   or from 6.4-5
         P = 0.5 * E * V / D         6.4-28
   Table 6.4-5  Power (MW) Required for a 1 Kg mass at velocity 'V'
                     Length of EMPL in Kilometers
                          6        8        10
       V       E          P        P         P
       5      12.5       5.21      3.91      3.13
      10      50.0      41.67     31.25     25.00
      15     112.5     127.84    105.14     84.59
      20     200.0     333.33    250.00    200.00
      25     312.5     651.04    488.28    390.63
      30     450.0    1125.      849.      671.
      35     612.5    1801.     1331.     1074.
      40     800.0    2666.     2000.     1600.
      45    1012.5    3750.     2812.     2301.
      50    1250.0    5208.     3906.     3125.
   V - Velocity of projectile in kilometers per second
   E - Energy of 1 kg mass at velocity, 'V', in megajoules
   P - Required power of EMPL in megawatts
        In order to propel a spaceship of the expected size and mass,
   it  will  be necessary to use much more massive projectiles than 1
   kilogram.  Projectiles  of  about 1000 kilograms will be required.
   Fortunately  this  does  not mean that we will need 1000 times the
   power.  The  power requirements of table 6.4-5 were for continuous
   operation  but  such  is  not required. Electromagnetic projectile
   launchers  can  be  operated or powered by capacitors which we may
   take  as  long as we wish to charge up. This is clearly a tradeoff
   wherein  we can manage with less power if we are willing to accept
   the  penalty  of  lengthy  capacitor charging periods between each
   projectile  launch.  To  achieve  1000  times  the energy of table
   6.4-5,  we  can  use 10 times the power applied over 100 times the
   time.  In other words, we will take about 100 seconds to charge up
   the  capacitors  of the EMPL fully, using 10 times the power given
   in  table  6.4-5  in  order  to launch (or catch) each projectile.
   Table  6.4-6  shows the time, 'T'(in days), required to accelerate
   the  spaceship  by  one  kilometer  per  second.  The  number   of
   projectiles,  'N',  was  calculated exactly as in table 6.4-2. The
   velocity  of  the projectiles relative to the spaceship is 'V', in
   kilometers  per  second.  The  power  of  the  on  board  EMPL (in
   megawatts)  is  indicated  by the column headers in the table. The
   mass  ratio,  'R',  used was 2000, the mass of the projectiles was
   1000  kilograms,  and the length of the ship's EMPL was assumed to
   be 6 kilometers. The energy of each projectile was calculated from
   equation  6.4-26 -  with  'm' set to 1000 kilograms. The capacitor
   charging time (in minutes) is then given by:
*        CT = E / P / 60.0                  6.4-29
        Where P is the power of the spaceship's EMPL in megawatts.
   We have allowed 10% extra  time  between  each  projectile launch.
   Therefore,  the  time  between  projectiles, 'DT' (in minutes), is
   given by:
*        DT = 1.1 * CT                      6.4-30
   The total time, T(in days), is therefore given by:
         T = 2 * DT * N / 60.0 / 24.0       6.4-31
   Table 6.4-6 - Time (T) to Accelerate Spaceship 1 km/sec (in days)
           EMPL = 100 MW        EMPL = 200 MW        EMPL = 400 MW
    V      N     DT    T        N     DT    T        N     DT    T
    5     223   2.08  0.71     223   1.04  0.35     223   0.52  0.18
    6     182   3.00  0.83     182   1.50  0.42     182   0.75  0.21
    7     154   4.08  0.96     154   2.04  0.48     154   1.02  0.24
    8     134   5.33  1.09     134   2.67  0.55     134   1.33  0.27
    9     118   6.75  1.22     118   3.38  0.61     118   1.69  0.30
   10     106   8.33  1.35     106   4.17  0.67     106   2.08  0.34
   11      96  10.08  1.48      96   5.04  0.74      96   2.52  0.37
   12      87  12.00  1.59      87   6.00  0.80      87   3.00  0.40
   13      81  14.08  1.74      81   7.04  0.87      81   3.52  0.44
   14      75  16.33  1.87      75   8.17  0.94      75   4.08  0.47
   15      69  18.75  1.98      69   9.38  0.99      69   4.69  0.49
   V - Velocity of projectiles in kilometers per second
   N - Number of projectiles in each group
   T - Time in days to catch and launch one group
   R - Mass ratio is 2000 in this example
   DT - Time between projectiles in minutes
   6.4.5  Flight profile to Mars
         With  today's  technology  we can build an EMPL at the North
   pole  of  the moon. The recommended length is 10 kilometers and it
   will  be  powered  by a 2500 megawatt nuclear power facility. This
   EMPL  should  be capable of launching 1000 kilogram projectiles at
   velocities of up to 20 kilometers per second. From equation 6.4-26
   and  6.4-29 you  can see that the capacitor charging time would be
   about 2 minutes for each launch. The EMPL will be constructed on a
   series  of  circular tracks so that it can be rotated between each
   launch  to compensate for the orbital motions of the Earth and the
   moon.  The  orbital  velocity of the Earth around the sun is about
   29.8  kilometers  per second and that of the moon around the earth
   is about 1.02 kilometers per second.
         The spaceship will be assembled in orbit, most likely at one
   of  the  Lagrangian  points known as L4 or L5. It would then orbit
   the  Earth  at  the  same  velocity and period as the moon itself.
   Escape  velocity  at  such  an  orbit is about 1.44 kilometers per
   second.  Thus we would need only about 0.420 kilometers per second
   of additional velocity in order to escape Earth's gravity.
         The  spaceship  will  be  powered  by a 500 megawatt nuclear
   powered  electricity  generating  facility which will drive the on
   board  EMPL. The on board EMPL will be about 6 kilometers long and
   will  be  able  to  launch  or  catch 1000 kilogram projectiles at
   velocities  of  up  to  10 kilometers per second. Both the nuclear
   power facility and the on board EMPL will be constructed using the
   lightest  (lowest mass) materials possible. The target mass of the
   entire spaceship will be 3000 metric tons so that the ratio of the
   mass of the spaceship to the mass of the projectiles will be 3000.
   The  crew's  quarters will be roughly the shape of a ring with the
   EMPL passing through its central axis. The crew themselves will be
   lifted  from  Earth by one or more of the various aerospace planes
   now  being  developed - such as NASP(US) or Sanger(Germany) or the
   Space  Van(US)  or Hotol(UK/CIS). The crew need not go to the moon
   but  instead  could  be  moved  directly  from LEO to the orbiting
   spaceship  thus saving much time, money, and effort. A crew of one
   thousand  persons  can  be accommodated. They will be supported by
   crops grown by an on board hydroponic food production facility.
        The  spaceship  will  also  be  stocked  with several hundred
   projectiles.  These  will  be  launched  from the on board EMPL to
   begin the voyage to Mars. Once the spaceship has left lunar orbit,
   the  lunar EMPL can begin launching groups of projectiles to build
   up  the  spaceship's  velocity.  Each  group  will  have a uniform
   velocity  but each group will also be launched about one kilometer
   per  second  faster  than the previous group so that the spaceship
   will  receive  each  group  at  a  relative  velocity  of about 10
   kilometers  per second. Within each group, the projectiles will be
   separated  by the time and distance corresponding to the capacitor
   charging time of the on board EMPL.
        Table  6.4-7  shows  the duration of a trip to Mars using the
   lunar EMPL and spaceship described above. The column labelled 'GR'
   is the number of the group of projectiles launched from the lunar
   EMPL.  Group 0 is actually not launched from the lunar EMPL but is
   instead a group launched from the spaceship (its original stock of
   projectiles).  The  column  labelled  'VP'  is the velocity of the
   projectiles  in  kilometers per second. Again, group 0 is launched
   from  the  spaceship  while the other groups are launched from the
   lunar  EMPL.  The  column  labelled  'VS'  is  the velocity of the
   spaceship at the start of the launching of the corresponding group
   (in  kilometers  per  second).  The  column  labelled  'VE' is the
   velocity  of the spaceship at the end of catching and launching of
   the  incoming group of projectiles (in kilometers per second). The
   column  labelled  'TIME' is the elapsed time from the start of the
   mission  -  in  days.  The column labelled 'LEFT' is the number of
   days  to  reach  Mars  if  no  further  groups  of projectiles are
   received.  The  column  labelled  'TRIP' is the total trip time in
   days  if no further groups of projectiles are received. The column
   labelled  'PROJ'  is  the  total number of projectiles required if
   none  is reused. The column labelled 'DT' shows the number of days
   that  the trip is shortened by the receipt of the current group of
*  Table 6.4-7 -  Flight profile to Mars   ( R = 3000 )
   GP     VP      VS      VE     TIME   LEFT    TRIP    PROJ    DT
    0   10.00    1.15    2.15    0.44  422.48  422.92    316    0.00
    1   12.15    2.15    3.16    0.66  286.60  287.25    474  135.67
    2   13.16    3.16    4.17    0.88  216.60  217.48    632   69.77
    3   14.17    4.17    5.19    1.10  173.88  174.98    790   42.51
    4   15.19    5.19    6.20    1.32  145.05  146.37    948   28.61
    5   16.20    6.20    7.21    1.54  124.26  125.80   1106   20.57
    6   17.21    7.21    8.23    1.76  108.54  110.29   1264   15.50
    7   18.23    8.23    9.24    1.97   96.22   98.19   1422   12.10
    8   19.24    9.24   10.25    2.19   86.29   88.48   1580    9.71
    9   20.00   10.25   11.26    2.41   78.19   80.61   1738    7.88
   10   20.00   11.26   12.21    2.63   71.67   74.30   1896    6.31
   11   20.00   12.21   13.11    2.85   66.28   69.14   2054    5.16
   12   20.00   13.11   13.97    3.07   61.76   64.83   2212    4.31
   13   20.00   13.97   14.78    3.29   57.89   61.18   2370    3.65
   14   20.00   14.78   15.56    3.51   54.54   58.05   2528    3.13
   15   20.00   15.56   16.29    3.73   51.60   55.33   2686    2.72
   16   20.00   16.29   16.99    3.95   49.00   52.95   2844    2.39
   17   20.00   16.99   17.65    4.17   46.67   50.84   3002    2.11
   18   20.00   17.65   18.28    4.39   44.56   48.95   3160    1.88
   19   20.00   18.28   18.88    4.61   42.65   47.26   3318    1.69
   20   20.00   18.88   19.45    4.83   40.90   45.73   3476    1.53
   The distance to Mars is assumed to be 0.524 A.U. = 7.8385E+7 km.
    6.5 Specific Impulse comparisons

          Specific  impulse is defined to be the ratio of the thrust of
     the  rocket to the weight flow rate of the propellant [62, p.3-5].
     When  stated  in  this  manner,  it  is difficult to understand. A
     simpler  way to express it is the number of seconds that one pound
     of  propellant  would burn while producing one pound of thrust. It
     is a measure of the energy in the propellant.
          The  importance  of  higher  specific  impulse cannot be over
     emphasized.  Imagine that you have two rockets, one of which has a
     specific  impulse  which is twice that of the other. By looking at
     equation  6.1-1  it  can be seen that the mass ratio of the higher
     specific  impulse will be the square root of the mass ratio of the
     lower  specific impulse. The most immediate consequence of this is
     an EXPONENTIAL DECREASE in the cost of payload in orbit - assuming
     only  a  moderate  increase in the unit cost of the more energetic
          Consider the following real-life example. We plan to launch a
     rocket  into  low  earth  orbit - a trip which requires a velocity
     change  of  about  9300 meters per second. Suppose we use a liquid
     oxygen  and  liquid hydrogen engine such as the space shuttle main
     engine (SSME) with a specific impulse of 475 seconds. Plugging the
     numbers into equation 6.1-1 we get:

         M = m * exp( dv/g*Isp )               6.1-1
         M = m * exp( 9300/9.8*475 )            or
         M = m * exp( 1.99785 )                 or
         M = m * 7.3732

          The  mass  ratio  is  7.3732. Now we consider using a nuclear
     thermal  rocket  with  a specific impulse of 950 seconds to do the
     same job. Plugging the numbers in we get:

         M = m * exp( 9300/9.8*950 )            or
         M = m * exp( 0.998926 )                or
         M = m * 2.7154

          Now  the  mass  ratio  is  2.7154 which is the square root of
     7.3732.  In  the  first  case, the engine plus fuel tanks plus the
     payload  are  1/7.3732 = 13.6% of the launch weight. In the second
     case,  the  engine plus fuel tanks plus the payload are 1/2.7154 =
     36.8% of the launch weight. But, in both cases the engine plus the
     fuel  tanks are about 8-10% of the weight. So we really have about
     5%  payload  in the first case and about 28% payload in the second
     case.  Using  equation  6.1-5  we calculate the amount of fuel per
     kilogram of payload.

     Table 6.5-1   Comparison of Specific Impulse
         Isp                             475         950
         mass ratio ( for dv = 9300 )    7.3732      2.7154
         kgs fuel per kg payload         13.0        1.99
         payload mass (%)                5.6         28.8
         fuel mass (%)                   86.4        63.2
         engines and tanks mass (%)      8.0         8.0
         cost per kg fuel                x           2x
         total fuel cost                 86.4x       126.4x
         fuel cost per kg payload        15.5x       4.39x
         fuel cost ratio                 1           0.28
         payload ratio                   1           5.14

          Assuming  the more energetic propellant is twice as expensive
     leads  to  only  a  46.2%  increase  in  total  fuel cost. So, for
     roughly  the  same  cost,  we can put more than five times as much
     payload in orbit using the nuclear thermal rocket as can be put in
     orbit  with  the space shuttle main engine. Notice that the amount
     of  fuel  is  actually less for the nuclear thermal rocket ( 63.2%
     vs. 86.4% ) than for the space shuttle main engine.
         The following table shows the specific impulses of some of the
     various types of propulsion systems reviewed in sections 6.1 - 6.4

     Table 6.5-2  Specific Impulse of Various Propulsion Systems
         Propulsion system     Specific impulse    Source
         LOX-LH2                 450 - 475         many
         O-BeH                   705               [66, p.43]
         Augustine engine        900 - 1,000       [68, p.171]
         Nuclear thermal (fission)
            solid core           500 - 1,100       [66, p.46]
            liquid core        1,300 - 1,600       [66, p.46]
            gas core           3,000 - 7,000       [66, p.46]
         Momentum exchange     1,000 - 10,000      author
         Nuclear electric        800 - 30,000      [AW 32, p.24]
         Ion drive             5,000 - 25,000      [50, p.213]
         Nuclear fusion        1,000,000(.033c)    [53, p.3]
         Antimatter rocket     28,775,000(.94c)    [68, p.239]
         Photon drives         30,000,000(1.0c)    [50, p.238]

          Sadly,  all  is  not  wonderful in the world of high specific
     impulse.  The  problem  is that often when the specific impulse is
     high,  the  thrust  is  low!  This  is true of ion drives, nuclear
     electric  systems, nuclear fusion systems, antimatter rockets, and
     photon  drives.  "So  what?",  you may ask. The answer is that low
     thrust means very slow acceleration and therefore very long trips.
     Take  an ion drive for example. A million pound vehicle may have a
     thrust  of  only  100  pounds  and a peak acceleration of 0.0001 g
     (gravities)  [14,  p.143].  A  trip to Mars at 0.0001 g would take
     about 644 days or 1.76 years - each way!
          Fortunately,  the  momentum transfer propulsion system offers
     the  answer.  Since this scheme does not use on board propellants,
     the  specific  impulse  is  calculated  by  dividing  the  exhaust
     velocity by "g", the acceleration of gravity. The exhaust velocity
     of  this  system is dependent only on how powerful your EMPLs are.
     Therefore, the same applies to the specific impulse of the system.
     In  other  words,  you  can have as high a specific impulse as you
          In  summary, only LOX-LH2, nuclear electric propulsion (NEP),
     nuclear thermal propulsion (NTP), and momentum transfer are viable
     candidates  for propulsion systems. Perhaps in 25 years we can add
     fusion propulsion to the list and in 50 years, antimatter rockets.
     6.6 Travel time comparisons
          One  destination  of great interest is Mars. How long will it
     take  to  fly to Mars? The following table was calculated assuming
     that  the distance to Mars was 0.52 astronomical units - i.e. that
     Mars  was  at  its  maximum orbital distance from the sun which is
     1.52  A.U. This table shows the time in days that it would take to
     reach  Mars  if the spaceship were under the constant acceleration
     given  in  the first column. By the way, 1.0 g is normal gravity -
     i.e. what we all experience every day.

     Table 6.6-1  Travel Time to Mars and Jupiter
                                 Time (in days)
         Acceleration      Earth to Mars    Earth to Jupiter
                            (0.52 A.U.)      (4.20 A.U.)
         1.0 g                  6.4             18.3
         0.1 g                 20.4             57.9
         0.01 g                64.4            183.1
         0.001 g              203.7            579.1
         0.0001 g             644.3           1831.2

          Of  course,  this is not a strictly realistic example for two
     reasons.  First,  the  above  table  is  calculated for a constant
     acceleration  which  most  propulsion  systems  are not capable of
     doing  (exceptions are the low thrust systems: ion rocket, nuclear
     electric,  nuclear  fusion, antimatter rocket, and photon drives).
     Second, Mars and Jupiter are orbiting the sun just as is the earth
     and thus the distances given represents only the distances between
     orbits  and  ignores the distances around the orbits - although it
     is  a  reasonable  approximation  because  the  existing   orbital
     velocity  will carry you around the orbit as the propulsion system
     moves you between the orbits.

          In  real-life, most spacecraft use Hohmann transfer orbits to
     fly  from  one planet to another. In 1925, a German engineer named
     Walter  Hohmann  discovered  the technique now named after him for
     changing  orbits  with  a minimum amount of energy - the so-called
     least-energy  transfer orbit. This requires a power thrust tangent
     to  the starting orbit, then an unpowered coast out (or in) to the
     destination  orbit,  and  finally a second power thrust tangent to
     the  final  orbit which causes the spaceship to assume the desired
     final  orbit.  (Recently it has been shown that the Hohmann scheme
     is  not  always  the lowest energy. See "An Estimate of the Global
     Minimum  Delta  V  needed  for Earth Moon Transfer" by T. Sweetser
     [AAS 91-101]).
         The problem with the Hohmann transfer orbit is that it is very
     slow  -  you  have to pay the piper. You must coast around half an
     orbit  of  the  transfer ellipse; which in the case of a trip from
     earth  to  Mars is 257 days or 8.5 months. Thus if you want to get
     there  faster,  you  must  use  more  fuel  or  find  a  different
     propulsion system.
         One of those alternatives is nuclear thermal propulsion (NTP).
     Although  the  idea  has been around for at least 30 years, it has
     recently  received  special  attention  due  to  the report of the
     Synthesis  Group. In an article by J.R. Asker in Aviation Week [AW
     32,  p.24-5,147-9]  of 3/18/91, the Synthesis Group emphasized NTP
     saying  "nuclear  rockets  are seen as the most practical means of
     sending  explorers  to  Mars  and  the  only possible vehicles for
     visiting  the  other planets" [AW 32, p.24]. The latter portion of
     this  statement  is  clearly  false. A number of other articles on
     nuclear  rockets  may  be  of  interest  [AW 33, p.18-20], [AW 41,
     p.54], [AW 51, p.38]. The report of the Synthesis Group [61, p.21]
     states  that  a  NTP  rocket can make the trip to Mars in 160 days
     (either  way).  This  is  a  reduction  of  97 days or about 38% -
     clearly a dramatic improvement.
          The other alternative propulsion system is momentum transfer.
     It  can  transport a 3000 MT spaceship with a crew of 1000 to Mars
     in less than 2 months. In summary, we have:

     Table 6.6-2  Travel Time of Candidate Propulsion Systems
         Propulsion system        Travel time to Mars
         Chemical rocket             257 days
         Nuclear rocket              160 days
         Momentum transfer           <60 days By the way, light makes the trip from earth to Mars in about 258 seconds when earth and Mars are closest to each other and about 1252 seconds when they are furthest apart. Thus, the round trip radio transmission delay time vary from about 8.6 minutes to 41.7 minutes at worst. 
     6.7 Hazards of space travel
          There  are  many  hazards associated with extended periods of
     exposure  to  micro-gravity.  Among  the  most  dangerous  are the
     following:  (1)  bone calcium loss or decalcification of the bones
     which  is similar to osteoporosis, (2) atrophy of the body muscles
     including the heart, and (3) deterioration of the blood.
          Bone  calcium loss is very dangerous because not only does it
     make  the  bones  weaker and more susceptable to being broken, but
     the  calcium  loss  continues as long as one is in a micro-gravity
     environment.  Stine  reports  a  bone  calcium loss of 1 to 2% per
     month  on  the  Skylab missions [101, p.107]. This agrees with the
     report  by  John  Billingham  of NASA's Ames Research Center, that
     astronauts experienced a 10% bone loss during 8 months in orbit on
     Skylab  [AW  1, p.48]. Booth has stated that two Soviet Cosmonauts
     Musa  Munarov  and  Vladimir Titov, who spent a year (1988) in the
     Mir  space  station,  lost  5%  of their bone mass in spite of two
     hours of exercise daily and drugs to combat the losses [71, p.56].
         The muscles of the body atrophy in a micro-gravity environment
     due  to  the  absence of stress. Cosmonauts Berezovoi and Lebelov,
     who  spent 211 days on board Salyut 7, could barely walk when they
     returned  to  Earth and required weeks of intensive rehabilitation
     [22,  p.25].  Cosmonaut  Yuri  Romanenko,  who was 43 at the time,
     spent  326  days  in  orbit  on  board  the Mir space station from
     February 6, 1987 to December 21, 1987. Upon landing back on Earth,
     he  could neither stand nor walk [15, p.96]. But he did apparently
     grow  one  half  inch  in height during the mission [15, p.96]. It
     took him months to recover. Miles and Booth have reported that the
     heart  shrinks  as  much as 10% on long micro-gravity flights [23,
          Micro-gravity  also  affects  the  blood. Astronauts on board
     Skylab  suffered both loss of red blood cell mass and plasma loss.
     Red  blood  cell mass loss averaged 14% on a 28 day mission, 12.3%
     on  a 59 day mission, and 6.8% on a 84 day mission [15, p.92]. The
     plasma  loss  averaged 10% on the 59 day mission and 16% on the 84
     day mission [15, p.92].
         For a discussion of the radiation hazard see section 9.0.
     6.8 Interstellar travel
          The human race may be forever limited to the Milky Way galaxy
     unless one of the various science fiction techniques can be turned
     into reality.  These include diving through black holes or folding
     space  as  in  "Dune". The nearest large spiral galaxy outside the
     Milky  Way  is  Andromeda,  about 2,200,000 light years away [123,
     p.76]. At 10% of the speed of light it would take 22,000,000 years
     to  get there. Several dwarf galaxies such as Leo I and Leo II are
     closer  at  750,000  light  years  [123, p.69]. With today's crude
     technology,  the  fastest we can now go is about 30 kilometers per
     second  which  is  only about 0.01% of the speed of light. This is
     very  likely  the  reason why we do not see any aliens walking our
     streets - notwithstanding UFO reports or pictures in the "National
          Relativistic  effects  are  quite insignificant for very high
     velocities. For example, 10% of the speed of light (which is about
     1000  times  as fast as we can go now) yields only about 0.5% time
     dilation;  25%  of  the speed of light yields only about 3.2% time
     dilation; and it requires 86.6% of the speed of light to get a 50%
     time dilation.
          Travel  within  our  galaxy  is  entirely  possible  and will
     probably  begin  within  100  years.  There are two basic types of
     interstellar  spaceships:  the  G.K.  O'Neill type, i.e. Island 1,
     Island  2,  Island  10  or  whatever  he is up to, and second, the
     embryonic  type  in which "the crew" exist in cryogenic storage in
     the  form  of human eggs and sperm. O'Neill has been promoting his
     giant  spaceships  for many years from his Space Studies Institute
     in  Princeton,  New  Jersey.  But he has not convinced Congress to
     finance  them.  Indeed, Senator William Proxmire had the following
     comment  about  O'Neill's  proposal,  "Not  a penny for this nutty
     fantasy!"  [71,  p.104].  The  interested space cadet can read all
     about O'Neill's ideas in "The High Frontier" [95].
         The embryonic ship would be operated by robots. There would be
     no  crew  until  the  ship was within 15 years of its destination.
     Then the robots would use some eggs and sperm to conceive and grow
     a  crew.  Clearly  this  scheme  awaits  the medical breakthroughs
     necessary to grow a human embyro to full term outside the mother's
     body. Perhaps the crucial question about this type of spaceship is
     if  it  is  ethical to play God and create humans in the middle of
     space  billions  of  kilometers  from  earth where they will never
     experience or even see life on earth. One must try to imagine what
     it would be like to become conscious of the fact that you had been
     marooned  in  space  for  your  entire  lifetime by an advanced(?)
     civilization  that  you  never  knew  or  saw and which may now be
     entirely  dead. If our civilization were then dead, then one might
     say  that we have avoided the death of our species - especially if
     numerous  such  spaceships  have  been  seeded to numerous distant
     stars.  However, our feelings of guilt over putting them in such a
     situation  might  not  be  shared by those "children of the stars"
     because, having not experienced life on earth, how could they miss
     it  unless  we  arranged  for  them  to know what it was they were
     missing.  Their  reaction might be to commit suicide by destroying
     the  integrity  of  their  spaceship  -  perhaps  as  we  now  are
     destroying  spaceship  Earth  -  but far more easily. Very likely,
     they  would  simply try to make the best of their situation. After
     all,  they  would  be  well  provided for by our spaceship and the
     robots. They might find their existence quite pleasant.