Timeline for Could a satellite in LEO "pump" or change mass distribution to gain forward momentum?
Current License: CC BY-SA 4.0
33 events
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| Dec 22, 2018 at 18:14 | vote | accept | Muze | ||
| Nov 20, 2018 at 10:48 | comment | added | Litho | (cont) ... $v$ is the spacecraft's orbital speed and $r$ is the wheel's radius. | |
| Nov 20, 2018 at 10:45 | comment | added | Litho | @uhoh The spacecraft can start with a non-rotating wheel and then use a motor to spin it up over time in order to keep its attitude. But after a while, it will spin up to the maximal speed the motor can sustain, so if the spacecraft tries to use the spheres to raise its orbit after that, it won't be able to keep its attitude. And it will happen before a significant change in orbit: changing the orbit's radius by $\Delta R$ while keeping the attitude would mean that the edges of the wheel have to move with the speed on the order of magnitude of $v\Delta R/r$, where... | |
| Nov 19, 2018 at 20:11 | history | edited | Muze | CC BY-SA 4.0 |
picture was update or modified
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| Nov 19, 2018 at 15:04 | comment | added | uhoh | So your answer is about using some existing, previously imparted rotation of the spacecraft to slightly raise it's orbit. Once exhausted, raising would cease? That's different than how pumping a swing works. Both you and your swing are initially at rest. Energy is in the lunch you just ate, and momentum is zero. You pump to "pull" momentum out of the Earth. In this answer the spacecraft has to be "wound up" ahead of time or it couldn't raise if I understand correctly. | |
| Nov 17, 2018 at 18:24 | comment | added | Litho | Right, Earth's rotation is not affected. About the same center: there's a theorem that a system's total angular momentum w.r.t. some origin is equal to the sum of the a.m. of the system's CoM w.r.t. this origin (that is, the angular momentum the system would have if all its mass was concentrated at the CoM and it was moving with the same velocity) and the system's a.m. w.r.t. its CoM. So the total a.m. of the spacecratf w.r.t. Earth's center is the sum of its orbital a.m. w.r.t. Earth's center and its rotational a.m. w.r.t. its CoM. And this sum is what we apply the conservation law to. | |
| Nov 17, 2018 at 9:45 | comment | added | uhoh | @Litho okay thank you for the ping! I'll read through it and think some more as well. I think you are talking about an exchange between orbital angular momentum of the spacecraft (around the center of the Earth) and rotational angular momentum of the spacecraft (around the center of the spacecraft) and leaving rotational angular momentum of the Earth (around the center of the Earth) unchanged. I'm not sure it works that way. I think when you are conserving angular momentum, all components all have to be defined around one center of rotation. | |
| Nov 17, 2018 at 9:36 | comment | added | Litho | @uhoh I added expanation about angular momentum conservation. | |
| Nov 17, 2018 at 9:35 | history | edited | Litho | CC BY-SA 4.0 |
Explanation about angular momentum conservation for uhoh
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| Nov 16, 2018 at 22:59 | vote | accept | Muze | ||
| Nov 25, 2018 at 2:58 | |||||
| Nov 16, 2018 at 16:06 | comment | added | Muze | @uhoh I have been thinking on this. There is still a perigee and apogee that brings the satellite close to the Earth 2 times. As the satellite approaches the Earth the 2 weights would retract and then expand again at the pinnacle of the satellite's apogee and perigee. Energy is would be expanded and the natural orbit changed creating lateral momentum. I wish I could show some math. | |
| Nov 15, 2018 at 16:48 | comment | added | Muze | @Litho There would be a series weights on a .5 km pole made of solar vanes instead of massless balls. see linked questions. | |
| Nov 15, 2018 at 16:46 | vote | accept | Muze | ||
| Nov 16, 2018 at 16:10 | |||||
| Nov 8, 2018 at 8:24 | comment | added | Litho | @Chris Please show that math, then. | |
| Nov 7, 2018 at 17:04 | comment | added | Chris | @Litho It does have to apply equal forces to them, since the rod is by construction massless, and so by simple application of Newton's laws the forces are equal. If the rod is not massless, then you can move the center of gravity of the two masses, but only by moving the center of gravity of the rod in the opposite direction. The distance from the center of the earth at which the two spheres meet is, if you do the math correctly, independent of the magnitude of the force. | |
| Nov 7, 2018 at 11:24 | comment | added | Litho | @Chris Or one can look at it like this: the movement of system's CoM follows Newton's second law, $F=Ma$, where $M$ is the total mass of the system (spheres and rod) and $F$ is the total external force (i.e., gravity) acting on the system. Since $F$ is limited, if the time of the mechanism's work can be chosen to be arbitrarily small, the vertical displacement of the CoM during this time can be made arbitrarily small as well. | |
| Nov 7, 2018 at 11:24 | comment | added | Litho | @Chris Even if the mechanism does apply equal forces to the spheres, the difference of magnitudes of total forces acting on the spheres (gravity+mechanism's) does not depend on the magnitude of the mechanism's force, so if the mechanism works fast enough, then the point where the spheres meet can be arbitrarily close to the position of the system's CoM at the moment when the mechanism starts working. (As long as we apply Newtonian mechanics, of course. If we go into relativistic speeds, it will complicate things.) It doesn't have to be the system's original CoG. | |
| Nov 7, 2018 at 11:23 | comment | added | Litho | @Chris The mechanism which moves the spheres doesn't have to apply equal forces to them. It could control the speed of their movements relative to the rod. For example, the spheres have cogwheels inside, the rod has cogs, and the mechanism controls the speed of cogwheel's rotation. | |
| Nov 6, 2018 at 22:57 | comment | added | Chris | Your analysis is incorrect. The spheres can't move symmetrically when the rod is vertical, since each feels a different force of gravity, so an equivalent force on both will not move the two masses the same distance. You'll end up with both at the center of gravity, not at the center of mass, and the potential energy doesn't go up. | |
| Nov 3, 2018 at 20:04 | history | bounty ended | Muze | ||
| Nov 3, 2018 at 15:59 | comment | added | Muze | Also if you look at the linked question it has 5 weights, adjustable pole, and solar vanes. etc. I wrote this question to be a simple as possible. | |
| Nov 3, 2018 at 15:51 | comment | added | Muze | @uhoh Perpetual motion is often tagged to devises that are not fully understood and is often misused. In theory the energy to move the motors to move the weights inadvertently would propel the satellite There still would be a power source like solar. | |
| Nov 3, 2018 at 15:43 | comment | added | Muze | @Litho I think this is possible yes Earth isn't symmetric and neither is the orbit and the Moon is a driver not an inhibitor. I haven't seen any math opposing this model. In a symmetrical elliptical oblong orbit and/or using the Earth bulge on approach as the satellite as it comes closer to the Earth it would PULL the 2 weights together and PUSH as the satellite distance from the Earth may create lateral momentum, but I'm not sure that is why I asked the question. I'm sure there may be also a way to distribute the weight to use the Moon's gravity as an extra gradient as well. | |
| Nov 3, 2018 at 12:29 | comment | added | Litho | Let us continue this discussion in chat. | |
| Nov 3, 2018 at 11:40 | comment | added | uhoh | No, what is happening with the Moon is different than what's happening in your answer. As soon as you used the word "tides" you've moved to a deformed, non spherically symmetric Earth. It is the torque on that quadrupole moment thereby changing the rotation rate of the Earth about its axis that allows the Moon to move. Apples and oranges. | |
| Nov 3, 2018 at 11:33 | comment | added | Litho | @uhoh I.e., the angular momentum of Earth's "orbit" around the common CoM of Earth and the satellite changes similarly to how the angular momentum of Moon's orbit changes due to tides on Earth. (The difference, of course, is that the Moon causes the tides on Earth, while the satellite controls its own "tides".) | |
| Nov 3, 2018 at 11:30 | comment | added | Litho | @uhoh One need to take into account Earth's angular momentum not w.r.t. its center, since its center moves slightly due to satellite's attraction, but w.r.t. of the common center of mass of Earth and the satellite, and it does change. When the spheres are extended and the rod is at an angle (neither aligned nor perpendicular) to the direction to Earth, the resulting force the satellite acts on Earth with is not quite aligned to the direction between the centers of mass of Earth and the satellite, so it changes Earth's angular momentum w.r.t. their common center of mass. | |
| Nov 3, 2018 at 10:54 | comment | added | uhoh | With your spherically symmetric Earth, it is impossible to exchange angular momentum. In your scenario the satellite's angular momentum changes but the Earth's doesn't, which seems to violate conservation of angular momentum. So I don't see how your proposed mechanism can work. With $J_2$ you can exchange angular momentum with the Earth's rotation about its own axis. | |
| Nov 3, 2018 at 10:23 | comment | added | Litho | @uhoh It's not a perpetuum mobile, the extra energy comes from the mechanism which moves the spheres. It perfroms work againgst Earth's gravity. About using $J_2$ being more efficient: quite possible. As I said, the procedure I suggested is very slow, unless you build a huge spacecraft. | |
| Nov 3, 2018 at 10:18 | comment | added | Litho | @uhoh About center of gravity: does it matter? During rigid body's usual orbital movement, its total energy (potential+kinetic) is constant. During these maneuvres, it increases. So it will increase over time. Or do you mean that it may increase in such a way that, for example, a circular orbit doesn't stay circular; instead, its periapsis gets lower while its apoapsis gets higher at a higher rate? I would expect that since we move the spheres to the center at two opposite poins of the orbit, the effects on the orbit's shape would negate each other. But maybe not. | |
| Nov 3, 2018 at 9:35 | comment | added | uhoh | I still think that leveraging the Earth's strong $J_2$ with a modulated quadrupole moment may be the best mechanism, but I'm also not 100% sure of this yet... | |
| Nov 3, 2018 at 9:31 | comment | added | uhoh | This is an interesting solution for a spherical gravity field and doesn't require $J_2$, however there may be a problem here. Moving the spheres from the center to the edges keeps the center of mass at the center of the rod, but not the center of gravity because gravity isn't uniform. The trick that makes the $(\frac{1}{R+l} + \frac{1}{R-l}) \ne \frac{2}{R}$ thing worth mathematically neglects that it's the center of gravity rather than the center of mass that would remain at a given orbit. I'm afraid that this might be a perpetual motion machine-type solution. I'm not 100% sure though. | |
| Nov 3, 2018 at 9:08 | history | answered | Litho | CC BY-SA 4.0 |