Energy Given to Mass By Lifting From Ground Up Into GEO

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Calculating the energy given to mass by lifting if from the ground and putting it into

Geostationary Earth Orbit

By Jim Cline

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, 7.3 KWhr per

pound mass,

That is, about 73 cents of electricity per pound lifted up from ground into GEO.

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Much greater efficiency than rockets provide, is needed to enable large scale projects such as

SPS in GEO. Yet the energy given to mass by putting it into GEO is quite small, so there is

the potential for economically building 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 rumblehustle 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


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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 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 useat 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


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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.


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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!)


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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


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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-synchronousClarke 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:


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* 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


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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


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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 KWh 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


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Copyright 2008 James E. D. Cline

Energy Given to Mass By Lifting From Ground Up Into GEO

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