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

Could a satellite in LEO “pump” or change mass distribution to gain forward momentum?Could antimatter be used for spacecraft propulsion?Could radiated heat propel space-craft in outer space?Linear motion reaction wheels?What is the maximum speed an ion engine can propel a spacecraft at?Would Elon Musk's roadster be able to rotate itself using its wheels as reaction control?Why aren't magnetic bearings used more frequently in reaction wheels?Could a cubesat propel itself to Mars?Would a grinding machine be a simple and workable propulsion system for an interplanetary spacecraft?Could a CubeSat be propelled by this low-tech electromagnetic propulsion system?

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4

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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.

propulsion spacecraft physics

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asked 8 hours ago

SentiCarterSentiCarter

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$endgroup$

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  $begingroup$
  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.
  $endgroup$
  – zeta-band
  8 hours ago
  

  
  
  
  


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  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"
  $endgroup$
  – Organic Marble
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Angular momentum is additive. Any complex system of reaction wheels can be replaced with a simpler, larger system.
  $endgroup$
  – Mark
  1 hour ago
  

  
  
  
  

add a comment | 

4

$begingroup$

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.

propulsion spacecraft physics

share|improve this question

asked 8 hours ago

SentiCarterSentiCarter

2111 bronze badge

New contributor

SentiCarter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  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.
  $endgroup$
  – zeta-band
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  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"
  $endgroup$
  – Organic Marble
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Angular momentum is additive. Any complex system of reaction wheels can be replaced with a simpler, larger system.
  $endgroup$
  – Mark
  1 hour ago
  

  
  
  
  

add a comment | 

$begingroup$

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.

propulsion spacecraft physics

share|improve this question

asked 8 hours ago

SentiCarterSentiCarter

2111 bronze badge

New contributor

SentiCarter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|improve this question

asked 8 hours ago

SentiCarterSentiCarter

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SentiCarter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

asked 8 hours ago

SentiCarterSentiCarter

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New contributor

SentiCarter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 8 hours ago

SentiCarterSentiCarter

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New contributor

SentiCarter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 8 hours ago

SentiCarterSentiCarter

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2 Answers
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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.

share|improve this answer

answered 8 hours ago

Tom SpilkerTom Spilker

12.4k3535 silver badges6060 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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...
  $endgroup$
  – uhoh
  2 hours ago
  
  

  
  
  
  

add a comment | 

1

$begingroup$

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.

share|improve this answer

answered 4 hours ago

Roko MijicRoko Mijic

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9

$begingroup$

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.

share|improve this answer

answered 8 hours ago

Tom SpilkerTom Spilker

12.4k3535 silver badges6060 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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...
  $endgroup$
  – uhoh
  2 hours ago
  
  

  
  
  
  

add a comment | 

9

$begingroup$

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.

share|improve this answer

answered 8 hours ago

Tom SpilkerTom Spilker

12.4k3535 silver badges6060 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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...
  $endgroup$
  – uhoh
  2 hours ago
  
  

  
  
  
  

add a comment | 

$begingroup$

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.

share|improve this answer

answered 8 hours ago

Tom SpilkerTom Spilker

12.4k3535 silver badges6060 bronze badges

$endgroup$

share|improve this answer

answered 8 hours ago

Tom SpilkerTom Spilker

12.4k3535 silver badges6060 bronze badges

answered 8 hours ago

Tom SpilkerTom Spilker

12.4k3535 silver badges6060 bronze badges

answered 8 hours ago

Tom SpilkerTom Spilker

12.4k3535 silver badges6060 bronze badges

1

$begingroup$

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.

share|improve this answer

answered 4 hours ago

Roko MijicRoko Mijic

13522 silver badges1010 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

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.

share|improve this answer

answered 4 hours ago

Roko MijicRoko Mijic

13522 silver badges1010 bronze badges

$endgroup$

add a comment | 

$begingroup$

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.

share|improve this answer

answered 4 hours ago

Roko MijicRoko Mijic

13522 silver badges1010 bronze badges

$endgroup$

share|improve this answer

answered 4 hours ago

Roko MijicRoko Mijic

13522 silver badges1010 bronze badges

answered 4 hours ago

Roko MijicRoko Mijic

13522 silver badges1010 bronze badges

answered 4 hours ago

Roko MijicRoko Mijic

13522 silver badges1010 bronze badges