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I was think about different ways spacecraft could feasibly move around and I came across Reaction Wheels as a way they can rotate. So I was curious if a complex system of them could be used to propel a large spacecraft forward and backwards instead of spinning on the spot? It could be a way of maneuvering within a small area (around a POI or adjusting it’s orbit) while conserving more energy than higher tech propulsion systems.

*I’m not very well versed in physics so please forgive me if this is very out of the park.

SentiCarter
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    Short answer: No. Longer answer is that in space, the center of mass keeps moving with the velocity it has. To change that velocity (speed or direction) you have to eject some mass or use the momentum of light in some fashion. So reactions wheels can spin you, but they can't change your velocity. – zeta-band Aug 29 '19 at 17:16
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    Check out https://en.wikipedia.org/wiki/Reactionless_drive "no rotating (or any other) mechanical device has ever been found to produce unidirectional reactionless thrust in free space" – Organic Marble Aug 29 '19 at 17:26
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    Angular momentum is additive. Any complex system of reaction wheels can be replaced with a simpler, larger system. – Mark Aug 30 '19 at 00:33
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    A cellphone's vibrator moves the cellphone backwards and forwards slightly. In principle you could do a little better than that, and move the object almost as far as the object is large - but no more than that, because you can't move the center of mass this way, and the center of mass always has to be somewhere inside the object. (In practice, from an engineering standpoint, I think the best you could realistically do is to make the spacecraft vibrate.) – Harry Johnston Aug 30 '19 at 20:22
  • You could use a rotating wheel to accelerate some propellant and eject it. That avoids the problems associated with the high temperatures of the usual rocket engines. – Peter - Reinstate Monica Aug 31 '19 at 07:56
  • @HarryJohnston A vibrating phone in space would merely rotate around its centre of mass. When sitting on a table, it has other forces acting on it (i.e. the imperfect elastic reaction of the table) which allow its centre of mass to move. – CJ Dennis Aug 31 '19 at 07:57
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    This idea sounds a bit like the Dean Drive – Michael Harvey Aug 31 '19 at 08:06
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    @CJDennis, yes, but the fact that the center of mass can't move doesn't mean that the rest of the object can't move around it. As an extreme example, consider a lightweight spherical shell connected by retractable lines to a much heavier and much smaller sphere - the object as a whole could move away from its original position in any direction, almost as far as its radius. Not a useful design from an engineering standpoint, but the physics is sound. – Harry Johnston Aug 31 '19 at 11:50
  • @HarryJohnston That doesn't fit into any definition of "propel" that I know of. – CJ Dennis Aug 31 '19 at 12:03
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    @CJDennis, I didn't intend to suggest that it did. Apparently my original comment wasn't as clear as I'd hoped. The point is that while you can move this way, the movement is extremely circumscribed, and not at all useful. – Harry Johnston Aug 31 '19 at 12:20
  • @zeta-band Why do you write your answers in the comment section? – pipe Aug 31 '19 at 22:31
  • Didn't the FBI or was it the CIA attempt to make a flying saucer this way using propulsion? – TheRealChx101 Sep 01 '19 at 01:13
  • By the way, if you consider the fuel of a conventional rocket part of the rocket, the center of mass of that rocket does not move at all when you start the engine, in free fall. It can't. – Peter - Reinstate Monica Sep 01 '19 at 03:25

2 Answers2

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Previously posted comments are correct: in free space (assumed free of any other bodies' gravity fields) there is no way to convert the reaction wheels' angular motion to translational motion.

There is one tongue-in-cheek way: throw a reaction wheel off the spacecraft in the direction opposite the direction of the desired delta-V! ;-)

If you abandon the free-space assumption and allow non-spherical gravitating bodies in the vicinity of the spacecraft, then it is possible, by turning the spacecraft at the right time and the right rate, to have tidal forces from the gravitating body wind up imparting a truly tiny delta-V on the spacecraft.

Tom Spilker
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    I've been long fascinated by the question Could a satellite in LEO “pump” or change mass distribution to gain forward momentum? and I think your mention "truly tiny delta-V" addresses this in some way. Do you know of any place I could read further about how to turn the spacecraft "at the right time and the right rate" to maximize this? Something with some basic fundamental equations? There was a time when I knew how to write down a Lagrangian for a dynamical system, but that time has long-since passed... – uhoh Aug 29 '19 at 23:47
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    @uhoh This is exactly how the Moon is moving away from the Earth. As the moon orbits, tidal forces disturb the shape of each body thus moving their centres of mass. Since the Earth-Moon system is in a tidally-locked, resonant system, the effect is not random or chaotic but acts to transfer rotational energy from the Earth into orbital energy in the Moon: Earth's rotation slows down and the Moon's orbit speeds up (which means it moves away). – Oscar Bravo Aug 30 '19 at 07:09
  • @OscarBravo It's definitely in the same general set of equations ballpark, but I'm more interested in finding out how an "articulated Moon" that could modify its own shape, or at least its own quadrupole moment, would optimize the timing and waveform of said modulation in order to maximize its rate of climb using Earth's quadrupole moment (J2). – uhoh Aug 30 '19 at 10:19
  • @OscarBravo: I assume (since it's a consistent influence) that the moon's orbit increased concentrically (i.e. it doesn't become more elliptic). Does that mean that this effect eventually tapers off when the moon orbits higher than it did before? Or will it always continue to climb until it finally leaves the Earth behind? – Flater Aug 30 '19 at 13:38
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    @uhoh Scott Manly has an older vid on Kerbal Space Program where transferring fuel from one large tank to another equally large tank allowed it to translate. He joked that this would be a cheap, albeit slow and tedious, way to travel to other planets, but I'm pretty sure limited to KSP's... cavalier approach to physics. :) – Lux Claridge Aug 30 '19 at 15:04
  • @OscarBravo rules definitely change at the Macro level compared to the micro – Magic Octopus Urn Aug 30 '19 at 17:49
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    @uhoh Here's an example: a roughly dumbbell-shaped spacecraft, with its mass concentrated in the spheres of the dumbbell, flies by an oblate planet on a polar orbit, with periapsis at the planet's equator. On the inbound leg, the spacecraft turns such that the longitudinal axis (i.e., the axis that goes through the centers of the spheres) points at the equatorial bulge. In this orientation, the gravitational attraction force is slightly greater than if the axis were pointed 90° from that orientation. It keeps pointing that sphere at the bulge until it is directly over it, then turns 90°... – Tom Spilker Aug 31 '19 at 01:48
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    @uhoh ...to put the two spheres equidistant from the equatorial bulge. Now, for the outbound leg, the gravitational force is slightly less than in the inbound orientation, so the outbound deceleration will be slightly less than the inbound acceleration, and the spacecraft will have picked up a tiny amount of delta-V. Momentum is conserved, though: the asymmetrical pass also imparted some delta-V to the planet. But if the spacecraft's delta-V is tiny, the planet's delta-V is miniscule! Unlike a standard, symmetrical hyperbolic flyby, this flyby changes the v-infinity. – Tom Spilker Aug 31 '19 at 01:55
  • @TomSpilker okay I will work this on paper (or Python) after my 2nd cup of coffee. I'm going to need both brain cells for this one. Thanks for the scenario! – uhoh Aug 31 '19 at 01:57
  • @LuxClaridge Yeah; the problem is that in KSP, pumping fuel around changes the center of mass without actually moving the vehicle. In real-life, the center of mass doesn't move. – Luaan Aug 31 '19 at 09:09
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No, it's a textbook case of conservation of the linear momentum vector in the absence of any external forces.

Linear momentum of a system Sum(mv) is a conserved quantity even if individual parts are allowed to change their momentum vectors.

Actually reaction wheels also conserve angular momentum of the total system (ship + wheel) as well! But that's Okay because you can keep the spinning reaction wheel inside the ship. It's also the case that angular position (facing) isn't conserved, so you can turn around in space using a reaction wheel, and at the end of the process the reaction wheel isn't spinning.

Roko Mijic
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