Back to TechnologyArenas Back to home page Index How much of the energy that is consumed by rocket launch vehicles going to GEO is actually given to the lifted payload itself? Here we find, through three different paths of reasoning, that it is 15.7 KWhr/kg, 7.2 KWhr per pound mass, about 72 cents of electricity per pound lifted up from ground into GEO. Much greater efficiency is needed to enable large scale projects such as SPS in GEO Energy is the ability to do work, and it is indeed work to lift something from the ground all the way up into the Geostationary Earth Orbit, GEO. 22,300 miles above the Earth's equator. Just how much work it is, however, limits what is worthwhile to lift up there. Current transportation modes are extraordinarily wasteful of energy, requiring far more work expended in the rumble hustle to space than is actually useful energy spent. but it is the best we have gotten so far; can we now look toward doing much better? It is very possible, as is shown on these pages. So let's look at how much work is involved that is actually supplied to the payload; that is the useful part; all else is very expensive wasted fluff. We will find that the limiting values of energy are quite low, enabling quite a new range of possible projects to be achieved up there, given a new and sufficiently efficient transportation mode if we choose to do so (such as explored in many of these pages); some new large scale projects begin to look feasible then, that could enormously help our civilization along with the planetary ecosystem we live within. There are several aspects involved. One is the actual energy added to the payload mass by having been lifted up there, and that is a very specific value. Other aspects involved concern the actual kind of grunt work done to lift it up there, and current techniques are extremely wasteful of energy, enormously adding to the cost of lifting payload up there into GEO. So first let's determine the energy given to payload by lifting it up there; and then next look at the many forms of grunt work involved in techniques such as rocket launch vehicles, tether space elevators, and carousel space escalators; each of these can do a lot of wasteful energy grunting under load while supporting the weight during the time it is being carried up to GEO, if we are not careful. Rockets have no choice but to do it the hard way, grunting the whole way mostly under its own weight instead of the weight of the payload itself. The tether and the carousel have the potential to provide "rest stops" along the way, where the structure briefly supports the load while the lifter gathers its strength; the less time of actual input of lift energy, the higher potential efficiency involved, and resonant techniques can make the actual lift proceed fairly gently. The energy given to mass by having lifted it to a higher location, is calculated differently from lifts which are of a very tiny proportion of the planet's radius. If you are going very high, like to GEO, the gravitational field reduces quite a lot along the way, and so the higher you get, the less energy is involved to lift a given distance up more. It's an "inverse square" thing. And the equations are not in common use. So let's figure it out for ouselves here. For a reality check, we know that the value cannot be greater that the amount of energy needed to give a mass "escape velocity" to hurl it to the furthest reaches of space (if there were nothing else up there, like the Sun's gravitational field, however) so we will calculate that here. Then we can use a technique provided by Arthur C. Clarke back in the mid1950's to calculate that escape energy, and then modify that technique to provide a lift only to GEO. And then we will do an actual calculation. What we will find is that the energy actually added to payload in the process of emplacing it up in Earth orbit has a surprisingly very low cost, expressed in electrical energy terms, of 7.15KWh; or about $0.72 per pound lifted up to GEO. Let's get started: some lift energy calculation basics for beginners. To calculate the amount of work (energy) done to an object when lifting it to a higher place, basically is a simple calculation: multiply the weight of the object and multiply it by the height it is lifted. Note that the weight of something is its mass multiplied by the force of gravity existing at its location. So we assume here that the force of gravity does not significantly vary over the height being raised, limited to, say, the height of existing tallest buildings. So if we lift something off the ground which weighs (in English units) one pound, to a position which is one foot higher, the work done on the object is one footpound: Energy equals Weight times height lifted: E = W x h Shifting from English units of measurement to the simpler International System of Units of measurement However, when one gets into more complex calculations, it has been found that the use of a system of units known as the International System of Units (SI) is clearer and simpler in the long run. instead of a "foot", it uses a meter (about 3 feet) as its basic unit of length. And for its unit of weight ... well, to make things clearer overall, SI separates out the mass of the object from the force of gravity at its location at the moment, to give it more adjustable values for use at different locations, more universally usable. In SI units, the "kilogram" is the basic amount of mass; and the gravitational acceleration (which, when multiplied by the physical mass of something gives the weight of that something where it is at) on the ground (on the surface of the Earth) is 980 cm per second per second ("cm" stands for a hundredth of a meter, or "centimeter"). Note that the notion of "time" is embedded into the notion of "acceleration", which could be thought of as how fast something is speeded up per unit of time... therefore it may be of interest that the notion of "time" is embedded into the notion of the "weight" of something. The force of gravity becomes less the higher up we go, so the higher we go the less work it takes to go even higher The value of gravitational acceleration varies with the height of the object. The further one gets from the center of the mass of the earth, the lower the gravitational acceleration one experiences. And this force of acceleration decreases rapidly, inversely proportional to the square of the distance from the center of the Earth, as the fixed gravitational pull is spread out in all directions from the planet ever more thinly to include ever larger encompassed spherical surface high above the planet. This means that the higher one gets above the ground, the easier it is to lift the object higher yet. And the amount drops off as the square of the distance from the center of the Earth. (Note that referring to the "square" of something simply means the length of line moved sideways through a distance equal to its own same length, thus covers an area which is its "square".) And it means that the equation we used above (E = W x h), to determine the energy given to an object by lifting it off the ground, is only effectively valid for use near the ground, since the value of gravitational acceleration used to determine the object's Weight "W" changes when we get much higher above the ground. There are slightly more sophisticated ways to calculate the work, energy, needed to lift payload up through a varying strength gravitational field. Arthur Clarke's simplified calculation for the energy given to a payload's mass when lifted from the ground up out of the gravity well 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: Work = 1 lbf x 4000 miles x 5280 ft / mile Work = 2.1E7 ft lbf Since 1 ft lbf equals 3.77E7 KwHr, Work = 2.1 x 10E7 ft lbf x 3.77E7 KwHr, Work = 7.9 KwHr Stated otherwise, the energy cost to accelerate 1 pound of mass to escape velocity = 7.9 KwHr, nothing else considered. Note that, at current American electric power costs of ten cents per KHr, this is 79 cents. (And yes, of course there is a lot more expense to be considered for the process of lifting payload to space!) Another way to make this calculation is to use the gravitational equation. Calculating theoretical work (energy) required to go from the ground up to essentially beyond the influence of the planet's gravitational field: Given: * Gravitational constant, G = 6.67E11 m3 / Kg Sec2 * Earth mass = 5.983E24 Kg * R0 is the Earth equatorial radius = 6.378E6 meters * Destination R = infinity * 1 joule = 1 Kg m2 / Sec2 = 2.78E7 KwHr Then, calculating the work (energy) from equatorial Earth surface to essentially beyond the influence of the planet's gravitational field: where R0 = radius of equatorial Earth surface = 6.378E6 m R = infinity W = GMm ((1 / R0)  (1 / Rinfinity)) W = (6.67E11)(5.983E24)((1 / 6.378E6)  (1 / infinity)) W = (4E14)(1.57E7) = 6.27E7 Joules / Kg to escape = 17.4 KwHr / Kg; @2.2 Kg / lbm = 7.9 KwHr per pound to GEO Thus calculating both ways yields the same amount of electrical energy to escape the gravitational influence of earth: 7.9 KwHr per pound mass. Note that this is the same amount of energy which would be given up by the one pound mass if it returns to rest again on the equator. At a cost of electric power of 10 cents per KwHr, 7.9 KwHr costs $0.79, which thus is the basic addedenergy 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 earthsynchronous 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. First, calculating the lesser theoretical energy cost of going from Earth's equator into geosynchronous Earth orbit, using the gravitational equation. Calculating theoretical work (energy) required to go from the ground up to GEO, the Clarke Belt: Given: * Gravitational constant, G = 6.67E11 m3 / Kg Sec2 * g0 = 9.807 meters / sec2 * Earth mass = 5.983E24 Kg * Earth equatorial radius = 6.378E6 meters * RGEO = 4.23E7 meters (22,300 mi above the equator) * 1 joule = 1 Kg m2 / Sec2 = 2.78E7 KwHr Then, calculating the work (energy) from equatorial Earth surface to Geosynchronous Earth Orbit (the Clarke Belt): where R = radius of GEO altitude = 4.23E7 meters R0 = radius of equatorial Earth surface = 6.378E6 m W = GMm ((1 / R0)  (1 / RGEO)) W = (6.67E11)(5.983E24)((1 / 6.378E6)  (1 / 4.23E7)) W = (4E14)(1.33E7) = 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: Calculating the gravitational acceleration at GEO altitude, gGEO: gGEO = (GMm) / (RGEO)2 = (6.67E11)(5.983E24) / (4.23E7)2 gGEO= 0.224 m / Sec2 Since gGEO = a = V2 / RGEO, where RGEO = 4.23E7 meters [this is the distance of GEO from center of the Earth] Then V2 = a * R = (gGEO)(RGEO) 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: VGEO = (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. At $0.10 per KwHr average cost of electrical power in the US currently, that equals $0.72 per pound from equatorial Earth surface up to GEO. We can also use the Clarke equivalency, slightly modified, to calculate the energy needed to lift from the ground to GEO: 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 gGEO= 0.224 m / Sec2 RGEO = 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/Sec2 * 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. Again, that is theoretically only $0.72 (72 cents) electrical equivalent energy cost added per pound, starting from the ground at the Earth's equator when being lifted up to Geostationary Earth Orbit! In more personal terms, for a 165 pound person to be lifted similarly, from the ground up into GEO, the energy added to the person, expresed in electrical power cost terms, would theoretically be $119. This is in the range of today's airline ticket price for a crosscountry hop, for comparison. Copyright © 2008 James E. D. Cline. Permission granted to reproduce providing inclusion of a link back to this site and acknowledgment of the author and concept designer James E. D. Cline. 
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