
"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 12 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 reallife 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 ) 6.11
.
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
changes.
* Table 6.11 Earth to Moon delta velocities
Earth to low earth orbit (LEO) 9300 m/s
TransLunar 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 nonpolluting 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.5323]. 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 6.12 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; 6.12
final mass = tanks + engines + payload; or
mf = t + e + p; 6.13
.
We want to solve for "fuel" in terms of "payload". Equation
6.11 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 ); 6.14
.
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 ); 6.15
.
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 6.11
and 6.15 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 6.11. 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 6.11.
* M = m * exp( dv/g*Isp ) 6.11
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 ); 6.15
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 6.15 is
that certain missions are impossible with single stage rockets. If
* mr * x >= 1 + x (mission impossible) 6.16
.
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 6.17
.
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.43145].
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.153161]. 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.447]. The specific
impulse of these systems are 5001100 seconds for the solid core,
13001600 seconds for the liquid core, and 30007000 seconds for the
gas core type [66, p.447].
The second is nuclear electric propulsion [AW 32, p.245]. 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.173182] 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 Helium3
as the fuel in [7, p.46]. A more recent article by Edward Teller
et al (including J.F. Santarius) describes nuclear propulsion by
fusion in a magnetic dipole [53, p.134]. That proposal also
requires Helium3. Corliss also briefly mentions nuclear fusion
[50, p.169170], but not using Helium3. 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 5e5 for fusion propulsion, which is in agreement with
figures of 1e4 to 1e5 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
222223 [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 matterantimatter 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.195215], or [68, p.221,2256]
or [66, p.434].
Typical ion rockets have specific impulses between 5000 and
25,000 seconds and thrust to weight ratios of 1e4 to 1e5
[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.616], or [50, p.1679] or
[68, p.226228]. Project Daedalus was very similar except that
nuclear fusion "bombs", actually tiny fuel pellets, were to be
used instead of fission bombs [66, p.679].
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.24754], and [66, p.89105].
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 ultathin 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.656], or [66, p.108113]. 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 matterantimatter 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
antielectron although it was named the positron. The antiproton
was found at Lawrence Berkeley Laboratory in October of 1955.
These antiparticles 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 antiproton
are electrically charged particles, but they could be combined to
produce antihydrogen 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; 6.31
.
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
matterantimatter rocket has a mass ratio (as described in
equation 6.11) 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
return.
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
1e12 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 (6.31) 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 tradeoffs 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
cost.
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 6.41
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 6.41
D = 0.5 * A * T * T 6.42
.
By squaring the first equation and substituting into the second
equation we get:
* D = V * V / ( 2 * A ) 6.43 or
A = V * V / ( 2 * D ) 6.44
.
Equation 6.44 was used to calculate the results of table 6.41
which were converted to Kgs by dividing the result by 9800 meters
per second per second. By substituting for 'A' in equation 6.41
from equation 6.44 we get:
* T = 2 * D / V 6.45
Table 6.41 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 6.42 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 6.46
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, (Vv), and its
mass, 'M':
* P = ( V  v ) * M 6.47
.
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' 6.48 or
dv = ( V  v ) * M / ( m + M ) 6.49 or
dv = ( V  v ) / ( R + 1 ) 6.410
.
Notice that the mass of the projectile cancells out in equation
6.410 leaving us with the ratio. Now we have the following:
* m' = m + M 6.411
dv = ( V  v ) * M / m' 6.412
v' = v + dv 6.413
.
By iteratively recalculating equations 6.411, 6.412, and 6.413
we can find a number 'N' such that the final mass and velocity of
the spaceship are given by:
* m' = m + N * M 6.414
v' >= v + 500 6.415
.
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' 6.416
m' = m  M 6.417
v' = v + dv 6.418
.
Notice that in equation 6.416 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 6.411, 6.412, and
6.413 until the ship's velocity was at least 500 meters per
second faster than its initial velocity. Then equations 6.416,
6.417, and 6.418 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 6.42 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 6.41.
Therefore the time to catch (or launch) a projectile will be the
same as in table 6.41. 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 6.44. 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 6.419
We now substitute for dV from equation 6.416, giving:
a = V * m / M / dt 6.420
.
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 6.421 or
a = V / ( R * dt ) 6.422
.
At this point 'dt' is the time required to launch the projectile
(given by equation 6.45). So we substitute '2*D/V' for 'dt'.
* a = V / ( R * 2 * D / V ) 6.423 or
a = V * V / ( 2 * D * R ) 6.424
.
Finally, we can replace 'V*V/2*D' with 'A' from equation 6.44.
* a = A / R 6.425
.
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 6.43  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 6.43 except
that the length of the on board electromagnetic projectile
launcher is assumed to be 10 kilometers instead of 6 kilometers as
in table 6.43.
* Table 6.44  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
velocity.
* E = 0.5 * m * V * V 6.426
.
If the launchers could operate at 100% efficiency, this would give
us a good idea of the power required. Table 6.45 shows the energy
of a one kilogram projectile travelling at the specified velocity,
'V', in kilometers per second. The energy, 'E', from equation
6.426 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 6.41. 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 6.427 or from 6.45
P = 0.5 * E * V / D 6.428
Table 6.45 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 6.45 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
6.45, 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 6.45 in order to launch (or catch) each projectile.
Table 6.46 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 6.42. 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 6.426  with 'm' set to 1000 kilograms. The capacitor
charging time (in minutes) is then given by:
* CT = E / P / 60.0 6.429
.
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 6.430
The total time, T(in days), is therefore given by:
T = 2 * DT * N / 60.0 / 24.0 6.431
Table 6.46  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 6.426
and 6.429 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 6.47 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
projectiles.
* Table 6.47  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.35].
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 6.11 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
propellant.
Consider the following reallife 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 6.11 we get:
M = m * exp( dv/g*Isp ) 6.11
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 810% of the weight. So we really have about
5% payload in the first case and about 28% payload in the second
case. Using equation 6.15 we calculate the amount of fuel per
kilogram of payload.
Table 6.51 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
above.
Table 6.52 Specific Impulse of Various Propulsion Systems
Propulsion system Specific impulse Source
LOXLH2 450  475 many
OBeH 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
want.
In summary, only LOXLH2, 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 6.61 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 reallife, 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 socalled
leastenergy 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 91101]).
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.245,1479] 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.1820], [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 6.62 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 microgravity. 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 microgravity
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 microgravity 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 microgravity flights [23,
p.57].
Microgravity 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
Inquirer".
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.