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| Fig. 1: When an object is in orbit, it is constantly "falling" such that it always "lands" at the same distance r from the center of the planet it's orbiting. v is the orbital velocity in m/s. (Image source: D. Jensen) |
We wish to explain the amount of energy required to put an object in orbit, or in theory how much energy could be harnessed by taking an object out of orbit. For example, could we theoretically power society by harnessing the momentum of the moon?
We make a few approximations to simplify the math and explanations. For example, we assume the Earth is a sphere (with constant radius) and objects in orbit move circularly around the planet. We omit the effects of atmospheric drag and the kinetic energy introduced by the rotation of the earth. In this analysis of the moon's orbital energy we also neglect the effects of the spatial distribution of the moon's mass on its embodied potential and kinetic energy.
It takes energy to lift an object away from the Earth - or in general to separate any two objects, because of gravity. Newtonian physics define the force of gravity acting on the two objects by
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(1) |
where G = 6.67 × 10-11 m3sec-2kg-1 is the gravitational constant, m1 and m2 are the masses of the two objects in kg, and r is the distance in meters between the two objects' centers of mass. [1,2] The energy required to move an object is the integral of force over distance, so the minimum energy required to move an object of mass m (in kg) some altitude h (in meters) from the surface of the earth is
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(2) |
where R is the radius of the Earth (6.37 × 106 m), and M is the mass of the Earth (5.98 × 1024 kg) [2].
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| Fig. 2: An object of mass m kg is in orbit at an altitude of h meters above the surface of the Earth, with orbital velocity v. (Image source: D. Jensen) |
An object in orbit also has kinetic energy, because it is moving. When something is in orbit, it is technically in free fall, but it is moving fast enough that its "fall" misses the planet and thus it keeps falling rather than crashing and burning. We can calculate the speed required to stay in a circular orbit because we know that at any point in time the distance between the planet and the orbiting object is the same. Fig. 1 shows this diagrammatically. Let's call the distance between the center of the planet and the orbiting object r, and assume that the object's velocity tangent to the planet's gravitational pull is v, in m/s. Then at some very small time-step T, Δx = vT and Δy = 1/2 × Fg / m × T2.In order for the object to "fall" such that it "lands" on the circular orbit, it must be the case that
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(3) |
Plugging in for Δx and Δy as defined above and then taking the limit as T goes to zero results in the solution
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(4) |
Therefore kinetic energy of an object with mass m in orbit at an altitude h above the surface of the Earth is
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(5) |
where G, M, and R are the gravitational constant, the Earth's mass, and the Earth's radius respectively as we defined previously.
Suppose we have an object of mass m is in orbit at an altitude h above the Earth's surface, as shown in Fig. 2. The minimum amount of energy we would need to expend to get that object in orbit starting at the earth's surface is the sum of its kinetic energy and the energy expended pushing against gravity,
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(6) |
You will notice that as h gets really big, the final fraction term at the end of Eqs. (2) and (6) gets smaller and smaller, and both equations become approximately equal to GMm/R. Then the maximum orbital energy embodied in a 1 kg object in orbit around the Earth is ≈ 6.26 × 107 joules of energy. From Eq. (2) we can conclude that this is the minimum energy required to lift a 1 kg mass out of the reach of Earth's gravity altogether. Fig. 3 shows how Eorbital, Ug and kE are related to orbital altitude above the Earth.
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| Fig. 3: Orbital kinetic energy kE, gravitational energy Ug (or rather, the energy required to lift the object contrary to Earth's gravity), and total orbital energy Eorbital as a function of orbital altitude, per kilogram of orbiting object's mass. (Image source: D. Jensen) |
Suppose we had some means of harvesting orbital energy to power society. Most of our artificial satellites are too small to give us an appreciable amount of energy. We do, however, live on a planet with a large natural satellite—how much energy could we harvest from the moon?
The moon has a mass of roughly 7.36 × 1022 kg, and orbits the earth at an average altitude of 3.82 × 108 meters. Then from Eq. (6) we know that the moon embodies approximately 4.57 × 1030 J of orbital energy. If we were to harvest this energy in such a way that the moon stayed in a circular orbit, we could only extract about half this energy before the moon and Earth would inevitably collide. Even still, this is enough to fuel the current global energy use (≈ 6 × 1020 J) for more than 3.8 billion years [3]! Unfortunately for us, we don't as of yet have a means of harnessing orbital energy in a useful way.
Then again, maybe this is for the best: according to Eq. (4), bringing the moon into low-earth orbit would also require accelerating it to thousands of meters per second velocity relative to the surface of the Earth. Even if the moon and Earth didn't collide, the sweeping tidal effects would almost certainly destroy all life on the planet.
© Dillon Jensen. The author warrants that the work is the author's own and that Stanford University provided no input other than typesetting and referencing guidelines. The author grants permission to copy, distribute and display this work in unaltered form, with attribution to the author, for noncommercial purposes only. All other rights, including commercial rights, are reserved to the author.
[1] I. Newton, The Principia: Mathematical Principles of Natural Philosophy, translated by I. B. Cohen and A. Whitman (University of California Press, 1999).
[2] R. Halliday, R. Resnick, R., and J. Walker, Fundamentals of Physics, 10th ed. (Wiley, 2014).
[3] "BP Statistical Review of World Energy 2022, British Petroleum, June 2022.