Through the use of chemical propulsion we have proven that it is possible for people to travel to space and even live there for awhile, so perhaps the time is ripe to use that knowledge and now prepare for a true spacefaring civilization. Chemical propulsion transportation systems are extrordinarily inefficient in putting payload into space, requiring incredible effort and expense to put small payloads into space, and the technology is not likely to reduce costs or increase payload size by more than an order of magnitude or so, not enough to produce a major highway between the earth surface and the vast resources of space. GEO, Geosynchronous Earth Orbit (the Clarke Belt) would be a high level waystation between earth and the rest of the solar system, being 91% up the earth's gravity well. A high efficiency, high capacity transportation link between the earth surface and GEO could provide a beachhead to the solar system.

An electrically powered transportation system from the ground to space will have the vast mass of the earth itself to thrust against, being attached to the ground. The transportation system could, in many ways, be analogous to a gigantic electric motor, which has its stationary stator part attached to the earth surface. The stator would provide inner magnetic levitation tracks for the armature segments to circle around in at high velocity in a high vacuum environment even where it goes through the earth's atmosphere, and would provide outer magnetic levitation tracks for payload carrying vehicles; electromagnetic coupling would transfer energy and momentum between the various parts of the transportation system. Electric motors can be made with 80% efficiency or more, although surely a ground to space system will have less efficiency, but let us do calculations based on 100% efficiency, as a starting point. A convenient upper transportation point would be the earth synchronous orbit, GEO, since it moves at the same angular velocity as the earth does, and so such a motor stator structure, rigidly fastened to the earth with its 24 hour a day rotation, would deliver its payload to its upper terminal requiring no additional energy to stay in orbit, the entire motor stator structure rotating with the planet as it delivers its cargo to GEO.

How much would be the minimum energy cost to put payload into space electrically? For a starting point, let us use the average cost of electrical power delivered to a home here in the US currently, about ten cents per killowatt hour: $0.10/KwHr. So for now let's use this figure in the following calculations.

Since an electrically powered launch system is reacting against the vast mass of the planet itself, essentially all the energy goes into the payload. Were it not for the impeding atmosphere, a payload mass launched from the earth surface at the well-known escape velocity of 25,000 mph would probably never return the earth, going far beyond the moon's distance. Destinations closer to earth would take less energy to reach, so such a calculation provides an outside value for electrically powered transportation to any earth orbit. So let's calculate that energy for escape velocity, the energy to head staight up and keep on going, never to come back (again, were it not for the earth's atmosphere, of course.)

The first calculation method will be based on the kinetic energy added to the launched mass; the second way is to use a simplification described by Arthur C. Clarke decades ago, and the third way is to use the gravitational equation.

The convention of using the symbol "E" followed by a number, to designate the number of zeroes to move the decimal point to the right, in a scientifically notated number, will be used here. For example, "1.0E6" would be 1,000,000; and "1.0E-2" would be 0.01.

First, the kinetic energy calculation: the energy required to accelerate one pound mass to the 25,000 mph escape velocity is:

Since it is easier to first use the metric system for such calculations, for 1 Kg accelerated to excape velocity:

Converting to metric system, 25,000 mph = (2.5E4 mph)(0.447 m/s per mph) = 1.1175E4 m / s

W = ((1 Kg)(1.1175E4)^{2}) / 2 = 6.24E7 Joules

Since 1 Joule = 2.78E-7 KwHr, the work needed to accelerate 1 Kg to 25,000 mph is 17.8 KwHr.

Since 1 Kg = 2.2 lbm, then it would theoretically take 7.89 KwHr to accelerate one pound mass to 25,000 mph, escape velocity.

The second calculation for the same energy required to leave the planet is based on a simplification described by Arthur C. Clarke decades ago.

Arthur C. Clarke many decades ago pointed out that the theoretical amount of energy needed to lift a mass up from the earth's surface out into far distant space is mathematicly equivalent to the energy needed to lift that mass up one planetary radius' altitude within a constant gravitational acceleration equal to that located at the planet's surface. (Ref "The Exploration of Space" by Arthur C. Clarke. Harper edition.) This makes it easy to make a calculation of the theoretical energy needed to do one thing: without vehicular overhead costs considered, lift a mass up from the ground and go to a theoretically infinite distance from the planet, not considering any other gravitational fields.

Note that the work, or energy, required to lift something is its weight ("weight" is the force of its mass times the gravitational accelleration where it's at) times the height through which it is lifted, the force exerted upward being equal in magnitude to its weight.

Energy, Work, Work = weight x height change

Clarke's calculation shows that, theoretically, the work required to lift something from ground to a point far distant above the Earth, is mathematically equivalent to that required to lift it from the ground to one planetary radius, as if the acceleration of gravity were to remain constant throughout the lift. Since the radius of the Earth is about 4000 miles, the work is equivalent to raising it 4000 miles up as if with a constant weight of that which it had on the surface. So for one pound of stuff, raised to a distance of one planetary radius under a constant surface gravitational field strength:

W_{1 lb} = 1 lbf x 4000 miles x 5280 ft / mile

W_{1 lb} = 2.1E7 ft lbf

Since 1 ft lbf equals 3.77E-7 KwHr,

W_{1 lb} = 2.1 x 10E7 ft lbf x 3.77E-7 KwHr,

W_{1 lb} = 7.9 KwHr

Stated otherwise, the energy cost to accelerate 1 pound of mass to escape velocity = 7.9 KwHr, nothing else considered.

The third way is to use the gravitational equation.

Given:

- Gravitational constant, G = 6.67E-11 m
^{3}/ Kg Sec^{2} - Earth mass = 5.983E24 Kg
- R
_{0}is the Earth equatorial radius = 6.378E6 meters - Destination R = infinity
- 1 joule = 1 Kg m
^{2}/ Sec^{2}= 2.78E-7 KwHr

where

R_{0} = radius of equatorial Earth surface = 6.378E6 m

R = infinity

W = GMm ((1 / R_{0}) - (1 / R_{infinity}))

W = (6.67E-11)(5.983E24)((1 / 6.378E6) - (1 / infinity))

W = (4E14)(1.57E-7) = 6.27E7 Joules / Kg to escape = 17.4 KwHr / Kg; @2.2 Kg / lbm = 7.9 KwHr per pound to GEO

Thus calculating all three ways yields the same amount of electrical energy to escape the gravitational influence of earth: 7.9 KwHr per pound mass.

At a cost of electric power of 10 cents per KwHr, 7.9 KwHr costs $0.79, which thus is the basic cost to move a pound of payload electrically from the ground to far distant space from earth. Any closer destination would cost less.

So how much would it cost to move payload electrically only to GEO, the earth-synchronous Clarke Belt orbit? Let's calculate it two ways, first using the gravitational equation again, then using Clarke's simplification, a bit expanded. These calculations only calculate the enerrgy needed to raise payload to an altitude, and the energy needed to give the payload orbital velocity to stay in orbit needs to be added to find the total cost.

Given:

- Gravitational constant, G = 6.67E-11 m
^{3}/ Kg Sec^{2} - g
_{0}= 9.807 meters / sec^{2} - Earth mass = 5.983E24 Kg
- Earth equatorial radius = 6.378E6 meters
- R
_{GEO}= 4.23E7 meters (22,300 mi above the equator) - 1 joule = 1 Kg m
^{2}/ Sec^{2}= 2.78E-7 KwHr

where R = radius of GEO altitude = 4.23E7 meters

R_{0} = radius of equatorial Earth surface = 6.378E6 m

W = GMm ((1 / R_{0}) - (1 / R_{GEO}))

W = (6.67E-11)(5.983E24)((1 / 6.378E6) - (1 / 4.23E7))

W = (4E14)(1.33E-7) = 5.31E7 Joules / Kg to GEO = 14.76 KwHr / Kg; @2.2 Kg / lbm = 6.71 KwHr per pound to GEO

That is the theoretical energy needed to lift one pound mass up to GEO; to stay there it also needs to be given orbital velocity:

Calculating orbital velocity at GEO, using the gravitational acceleration there as the centrepetal force's "a" as in F = m * a:

g_{GEO} = (GMm) / (R_{GEO})2 = (6.67E-11)(5.983E24) / (4.23E7)^{2}

g_{GEO}= 0.224 m / Sec^{2}

Since g_{GEO} = a = V^{2} / R_{GEO}, where R_{GEO} = 4.23E7 meters [this is the distance of GEO from center of the Earth]

Then V^{2} = a * R = (g_{GEO})(R_{GEO})

Therefore V = (a * R)^{1/2} = (0.224 * 4.23E7)^{1/2}

= 3.078E3 m / S = orbital velocity in GEO

Another way to calculate the orbital velocity in GEO is to calculate the circumference of the orbit then divide it be number of seconds in a day, one orbit:

V_{GEO} = (2 * pi * 4.226E7 Meters) / 8.64E4 seconds = 3.073E3 meters per second, about the same.

The Earth's equator, probably the starting point for payload to GEO, rolls along at 1,000 mph, or 4.47E2 m / S

So the delta V from equator to GEO, component at right angle to Earth's radius, is (3.073E3 - 4.47E2) m / S = 2626 m / S

The 1/2 mV ^{2} kinetic energy to be added from equator to GEO, horizontal component, then is 3.45E6 Joules per Kg; becoming an additional 0.96 KwHr/ Kg to orbit at GEO; at 2.2 lbm / Kg, that is 0.436 KwHr per pound mass to GEO to provide the orbital velocity component after getting up there.

So the total electrical energy to place one pound mass into geosynchronous earth orbit is the 6.71 KwHr to lift the mass up there plus another 0.44 KwHr to accelerate it to orbital velocity at that altitude, or **a total of about 7.15 KwHr to emplace one pound mass from earthsurface into GEO, Earth synchronous orbit, the Clarke Belt.**

We can also find the same result by using the same Clarke's simplification equivalency that we used earlier, to calculate the energy from GEO to far distant space, subtract this from the energy to lift from the ground to far distant space, thus the energy needed to go from the ground up to GEO. We already calculated above that it takes 7.9 KwHr of energy to escape the earth's gravitational well. And we calculated above that the gravitational acelleration ith the altitude of GEO is

g_{GEO}= 0.224 m / Sec^{2}

R_{GEO} = 4.23E7 meters [this is the distance of GEO from center of the Earth].

So using the Clarke simplification as we did above, only this time as if from a planet of the same mass as earth but with a radius the same as GEO, we find that the work to escape such a planet would be the work needed to lift one planetary radius height against a constant gravitational accelleration of the value at the planetary surface.

Work = energy = force * distance = mass * acceleration * distance

energy = 1 Kg * 0.224 m/Sec^{2} * 4.23E7 meters

energy = 9.48E6 Joules = 2.6 KwHr per Kg = 1.2 KwHr per pound mass

Subtracting this 1.2 KwHr per pound from the original 7.9 KwHr per pound from the earth surface, we get 6.7 KwHr to lift from the earth's surface to GEO, the same as we got using the gravitational equation calculation above. Once up there, we have to add more energy to give it the correct orbital velocity to stay there in orbit, of course, as we calculated above, bringing the total to 7.2 KwHr per pound moved up from the equator into earth synchronous orbit.

Just as taking a flight in an airliner costs more than just for the fuel energy to make the trip, so would the actual cost for electrically powered transportation between the earth and GEO. The cost of the development of the airplane, even of development of all aviation leading up to it, added to the cost of construction of the aircraft, its maintenance, the salary of those who operate and fly it for you... all this adds to the cost of each individuals trip in the airplane. So would also the cost of a trip to GEO on an electrically powered transportation system. But how much more?

An interesting form of electrically powered transportation from the ground to GEO, called KESTS (Kinetic Energy Supported Transportation Structures) is basically an electrical motor; electrical motors can be built for over 80 percent efficiency. But this motor must not only lift vehicles to GEO from the ground, it must also support its own weight and swing itself around with the planets 24 hour a day rotation. Explore more on the subject of KESTS and the potential for enabling large scale space colonization in GEO, even in the near future if mankind were to devote itself to the task.

Yes, certainly real world electrical transportation system inefficiencies would become high; but KESTS (Kinetic Energy Supported Electrically Powered Transportation Structures) space transportation systems need to be fully worked out so as to compare them with the exorbitant inefficiencies of conventional chemically fueled rocket launches which have to lift the fuel that lifts the fuel.

But once such relatively inexpensive electrically powered heavy lift capacity access to GEO becomes available, huge solar-electric powerplants can be built up there in the Clarke Belt to provide the transportation system with its own electrical power source, as well as to provide low pollution electrical power back to the Earth surface commercial electrical power grids.

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Copyright © 1999 James E. D. Cline