### 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: calculate this with me to prove it to yourself:

Energy, in technical terms, is the capacity to do work. It takes energy, doing work, to lift a physical thing up to a higher place. Energy has a price, in our culture. Here are some easy calculations on how much work is required to lift something from the ground up to space. One doesn't have to be a "rocket scientist" to do such calculations, and they produce some very interesting and extremely hopeful numbers regarding the limits to human civilization's potential expansion into the vast resources of space around us. The amount of energy, work, required to do something, often controls whether or not it gets done.

### 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 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 foot-pound: 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 3.77E-7 KwHr,

Work = 2.1 x 10E7 ft lbf x 3.77E-7 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 = 6.67E-11 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 = 2.78E-7 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 = (6.67E-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 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 added-energy 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.

#### Calculating theoretical work (energy) required to go from the ground up to GEO, the Clarke Belt:

Given:

• Gravitational constant, G = 6.67E-11 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 = 2.78E-7 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 = (6.67E-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:

#### Calculating the gravitational acceleration at GEO altitude, gGEO:

gGEO = (GMm) / (RGEO)2 = (6.67E-11)(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, 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 cross-country hop, for comparison.