Similar to pumping your feet on a child's swing can a satellite run up and down a pole to produce a forward momentum?

Red dots is the satellite's position running up and down the pole on an elliptical orbit around Earth. The pole uses solar vanes at each end to keep vertical to Earth. The pole could use magnetic entrapment to suspend the satellite not to have touching or moving parts when moving up and down the pole similar to a mono rail.

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    $\begingroup$ Hmm... a spacecraft with a dynamic quadrupole moment... how interesting! The whole thing together (dot plus rod) is "the satellite", and the whole thing has one center of mass, but the dot-plus-rod satellite will now vary its distribution of mass around that center. Certainly there are some interesting possibilities using tidal effects and/or Earth's large J2 in some clever way; after all, Earth's one natural satellite is slowly boosting itself over time. Hmm... $\endgroup$
    – uhoh
    Oct 26, 2018 at 0:13
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    $\begingroup$ @uhoh are we looking to engineer a complete spacecraft, or to prove a theoretical concept ? If we are just assuming that the spacecraft can control his position on the rod, I don't think the implementation details have ANY consequences on the question for that particular mater. $\endgroup$
    – Antzi
    Oct 26, 2018 at 6:22
  • $\begingroup$ @uhoh The Moon's slow escape from Earth is a consequence of tidal forces; the Moon is massive enough to raise a tide; due to the phasing of the tidal peaks (mass redistribution on Earth's surface) there is a torque acting to transfer angular momentum from Earth's rotation to the Moon's orbit. It only works because of the Moon's large mass, and even then, only over astronomical time scales. Not an effect usefully transferable to an artificial satellite. $\endgroup$
    – Anthony X
    Oct 29, 2018 at 1:33
  • $\begingroup$ @uhoh my comment was in reference to yours of 2018-10-26 0:13:01Z. I may have misunderstood your point; the point I was seeking to make is that no artificial satellite is going to do what the Moon does i.e. raise its own orbit by raising a tide on Earth. $\endgroup$
    – Anthony X
    Oct 30, 2018 at 0:04
  • $\begingroup$ @AnthonyX I agree with you that the Moon's orbit and raising mechanism is a poor or at least inadequate model for this problem for sure. This one needs to be treated as a separate problem for several reasons. $\endgroup$
    – uhoh
    Oct 30, 2018 at 1:27

2 Answers 2


In theory, you can change your spacecraft's orbit without expending reactive mass, just by moving its parts relative to one another. Instead of a rod and a sphere, let's consider a rod and two identical spheres which can move along it, and let us assume that the spheres always move symmetrically to each other. Instead of keeping the rod pointed to Earth, let it keep its direction in an inertial frame.

When the rod is perpendicular to the direction to Earth's center, move the spheres from the rod's center to its edges. This way, the spheres end up slightly further from Earth's center than they were, so their potential energy increases. Since the spheres move symmetrically, the rod's position does not change, so its potential energy does not change either.

Later, when the rod is pointed to Earth's center, move the spheres from the rod's edges back to its center. If each sphere's mass is $m$, the rod's length is $2l$, and the distance from the rod's center to the Earth's center is $R$ at the moment, the total potential energy of the spheres changes from $-\mu m(\frac{1}{R+l} + \frac{1}{R-l})$ to $-\frac{2\mu m}{R}$, where $\mu$ is Earth's gravitational parameter. And we can see that $$ -\mu m(\frac{1}{R+l} + \frac{1}{R-l}) = -\frac{2\mu m R}{R^2-l^2} < -\frac{2\mu m}{R}, $$ so the potential energy increases again.

By repeating this again and again, you can move your spacecraft to a higher orbit. Of course, it's going to be very slow, unless the rod's length is comparable to the orbit's radius.

Edit: uhoh has pointed out that as spacecraft's orbit rises, its orbital angular momentum increases, so this answer seems to break the law of conservation of angular momentum.

The answer assumes that the spacecraft's orientation in an inertial frame stays constant. However, the spacecraft is not spherically symmetrical, and Earth's gravity applies torque to it. For example, when the spacecraft is in the upper left or in the bottom right positions in the picture below (not to scale), the torque is in the direction opposite to the direction of the orbital rotation, since the force acting on the forward-pointing (relative to the orbital motion) sphere is smaller than the force acting on the backward-pointing sphere.

enter image description here

And this torque's effect is not negated by the opposite-directed torque in other parts of the orbit: when the spacecraft is in the upper right or in the bottom left positions and the torque is in the same direction as the orbital rotation, its magnitude is smaller, since the spheres are pulled close to the center. So the torque's effect accumulates over time, and in order to keep its orientation constant in an inertial frame, the spacecraft has to have some way to compensate this torque. This compensating torque is what explains the increase in the spacecraft's total angular momentum. (Or, if there is no compensation, this increase in the orbital angular momentum happens together with the opposite change in the spacecraft's rotation around its center, so the total a.m. stays constant. I mean, the procedure described here doesn't require the spacecraft's orientation to stay constant, it just requires that sometimes the spacecraft is "horizontal", and sometimes it's "vertical". But I guess that if we don't try to compensate the torque, the spacecraft will end up always pointing to Earth, so the procedure won't be applicable anymore. On the other hand, moving the spheres along the rod changes the spacecraft's moment of inertia, and therefore changes its rotation speed, so some additional analysis is needed to figure out what happens in this case.)

How can the spacecraft compensate the torque? Well, in theory, it can do it with reaction wheels. Of course, any realistic reaction wheels would get saturated quickly, before a significant change in orbit, but I said from the beginning that this whole approach is not practical. The answer's purpose was to show that raising/lowering orbit without expending reaction mass is possible in principle, not that it's doable in practice.

Or, as Muze suggests, one can use solar vanes to keep the orientation. But in this approach one needs to make sure that the solar pressure does not negate the change of orbit.

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    $\begingroup$ 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. $\endgroup$
    – uhoh
    Nov 3, 2018 at 9:31
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    $\begingroup$ @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. $\endgroup$
    – Litho
    Nov 3, 2018 at 10:23
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    $\begingroup$ @Chris Please show that math, then. $\endgroup$
    – Litho
    Nov 8, 2018 at 8:24
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    $\begingroup$ @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. $\endgroup$
    – Muze
    Nov 16, 2018 at 16:06
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    $\begingroup$ @uhoh I added expanation about angular momentum conservation. $\endgroup$
    – Litho
    Nov 17, 2018 at 9:36

I am going to answer the question as asked, ignoring the comments.

The system in question is a sphere pierced by a rod. Some internal mechanism acts to move the two parts relative to each other.

Let's look at the two limiting cases - massless rod and massless sphere.

enter image description here

For the massless sphere, the c.g. (red star) is in the center of the rod, it stays where it is, and the sphere slides up and down relative to the c.g. (this is what you have drawn in the question) For the massless rod, the c.g. is in the center of the sphere, and the rod slides up and down relative to the c.g.

You have provided no info about the relative masses of the objects. But any combination of massive objects will result in a case between these two. The c.g. will stay where it is, and the objects will move about it; how much depends on their relative masses.

It is the c.g. that is in orbit, and moving two objects about their c.g. can do nothing to change that orbit. Only the application of an external force can do that.

What about the swinging kid? He is pulling on the chains. That is an external force.

enter image description here

I urge you to read this article about swinging (where the picture came from), it's very interesting and explains it well.

The comments talk about quadrupole moments, tidal forces, etc, but I don't think that is what you were really asking about. At least, that is not what the question itself says.

  • $\begingroup$ "Only the application of an external force can do that." Can you address why the Moon's orbit is moving farther from Earth all the time without an external force? I think this answer is an oversimplification. Also "quadrupole moment" is just a way to mention that the ball plus rod has moments. It's not something added to the problem, it's just a way to describe the problem. $\endgroup$
    – uhoh
    Oct 29, 2018 at 0:52
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    $\begingroup$ In both your limit cases, the mass distribution of the system (the rod and the sphere) relative to its center of mass is fixed and cannot be changed. But it can be changed when both the rod and the sphere have mass, and that is the whole point of moving them relative to one another. So the fact that the orbit cannot be changed in your limit cases does not mean that it cannot be changed in the general case. $\endgroup$
    – Litho
    Nov 20, 2018 at 11:01
  • $\begingroup$ @Litho is right, but still this answers the original question. I think everyone here is OK with generalizing it, in order to address all possible mechanisms that might achieve the goal of acceleration without reaction mass. $\endgroup$ Apr 4, 2019 at 20:19

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