# Could a complex system of reaction wheels be used to propel a spacecraft?

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.

• 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. Commented Aug 29, 2019 at 17:16
• Check out 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" Commented Aug 29, 2019 at 17:26
• Angular momentum is additive. Any complex system of reaction wheels can be replaced with a simpler, larger system.
– Mark
Commented Aug 30, 2019 at 0:33
• 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.) Commented Aug 30, 2019 at 20:22
• This idea sounds a bit like the Dean Drive Commented Aug 31, 2019 at 8:06

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.

• 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
Commented Aug 29, 2019 at 23:47
• @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). Commented Aug 30, 2019 at 7:09
• @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. :) Commented Aug 30, 2019 at 15:04
• @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°... Commented Aug 31, 2019 at 1:48
• @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. Commented Aug 31, 2019 at 1:55

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.