# Could a “living planet” alter its own trajectory only by changing its shape?

In Stanislaw Lem's novel Solaris the planet is able to correct its own trajectory by some unspecified means. Assuming its momentum and angular momentum is conserved (it doesn't eject or absorb any mass), would this be possible (in Newtonian mechanics) and how? If not, can it be proven? The assumption is that the planet orbits a star (or perhaps a binary star) system.

Intuitively this seems possible to me. For example, tidal forces result in a planet losing its rotational energy, so it seems possible that by altering its shape, a body could alter at least its rotation speed.

My ideas go as follows: Assume we have an ideal rod consisting of two connected mass points. The rod rotates and orbits around a central mass. When one of the points moves towards the central body, we extend the rod, getting it closer to the center. thus increasing the overall gravitational force that acts on the rod. When one of the points is getting away from the center, we shrink the rod again, thus decreasing the combined gravitational force. I haven't run any simulations yet, but it seems this principle could work.

Update: An even more complex scenario (conserving momentum and angular momentum) would be if the planet ejected a piece of matter and absorbed it again after some time.

• Do you know that the Earth transfers its rotational kinetic energy slowly to the Moon through tidal effects? Orbit shifting effects such as these, particularly if enhanced by cyclic shape shifts, are indeed possible. Indeed, around a black hole there are unstable orbits whence things can get flung off through the presence of higher order terms in the effective potential and this effect can be very fast. I'd have to do detailed calculations or ask a question to work out how fast this can happen for more "normal" conditions. But it is certainly a real effect. – WetSavannaAnimal Jun 14 '16 at 13:42
• @AndriiMagalich No, this is not tidal locking, but transfer of kinetic energy / AM to the Moon from the Earth's spin AM and KE. See physics.ucsd.edu/~tmurphy/apollo/doc/Dickey.pdf, for example. The Moon drifts away from the Earth by about 4 centimeters a year. – WetSavannaAnimal Jun 14 '16 at 14:44
• @WetSavannaAnimalakaRodVance Hmm... Looks like you are correct, I did not think of that – Andrii Magalich Jun 14 '16 at 15:14
• @WetSavannaAnimalakaRodVance While I agree with you on the mechanism, I do believe that the mechanism you describe is precisely tidal locking. – Taemyr Jun 15 '16 at 11:07
• @Taemyr Thanks; I need to revise this stuff, but I've been holding off because this kind of problem really sucks me in deeply and I could see myself wasting several days that I really can't spare right now! – WetSavannaAnimal Jun 15 '16 at 11:09

## 8 Answers

If you allow for non-Newtonian gravity (i.e., general relativity), then an extended body can "swim" through spacetime using cyclic deformations. See the 2003 paper "Swimming in Spacetime: Motion by Cyclic Changes in Body Shape" (Science, vol. 299, p. 1865) and the 2007 paper "Extended-body effects in cosmological spacetimes" (Classical and Quantum Gravity, vol. 24, p. 5161).

Even in Newtonian gravity, it appears to be possible. The second paper above cited "Reactionless orbital propulsion using tether deployment" (Acta Astronautica, v. 26, p. 307 (1992).) Unfortunately, the paper is paywalled and I can't access the full text; but here's the abstract:

A satellite in orbit can propel itself by retracting and deploying a length of the tether, with an expenditure of energy but with no use of on-board reaction mass, as shown by Landis and Hrach in a previous paper. The orbit can be raised, lowered, or the orbital position changed, by reaction against the gravitational gradient. Energy is added to or removed from the orbit by pumping the tether length in the same way as pumping a swing. Examples of tether propulsion in orbit without use of reaction mass are discussed, including: (1) using tether extension to reposition a satellite in orbit without fuel expenditure by extending a mass on the end of a tether; (2) using a tether for eccentricity pumping to add energy to the orbit for boosting and orbital transfer; and (3) length modulation of a spinning tether to transfer angular momentum between the orbit and tether spin, thus allowing changes in orbital angular momentum.

If anyone wants to look at the article and edit this answer accordingly with a more detailed summary, feel free. As pointed out by Jules in the comments, the "previous paper" mentioned in the abstract appears to be this one, which is freely available.

The idea of "swimming in spacetime" was also discussed on StackExchange here and here.

• While that paper isn't available, the "previous paper" mentioned in your quote is probably this one: ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19890017506.pdf -- also knowing that Geoffrey A. Landis is involved, I'd imagine he has a laymans explanation somewhere (although it doesn't seem to be on his web site if there is one). – Jules Jun 14 '16 at 17:25
• @Jules: Good find. I've edited my answer to include it. – Michael Seifert Jun 14 '16 at 17:32
• Would using a tether in this way apply a net force on the satellite with a matching opposed force on the body it's orbiting? Also, it this theorized, or actually used in space right now? – Martin Carney Jun 14 '16 at 19:04
• Swimming in spacetime without background field is extremely inefficient, though... you wouldn't make it very far using that technique. – CuriousOne Jun 14 '16 at 19:29
• I found the original referenced paper: Satellite Relocation by Tether Deployment by G. A. Landis and F. J. Hrach, 1989. – Petr Pudlák Jun 15 '16 at 11:27

Conservation of angular momentum tells us that in an isolated system, total angular momentum remains constant in both magnitude and direction.

The key here is that the conserved quantity is the total angular momentum: spin+orbital angular momentum.

An example:

For a planet, angular momentum is distributed between the spin of the planet and its revolution in its orbit, and these are often exchanged by various mechanisms. The conservation of angular momentum in the Earth–Moon system results in the transfer of angular momentum from Earth to Moon, due to tidal torque the Moon exerts on the Earth. This in turn results in the slowing down of the rotation rate of Earth, at about 65.7 nanoseconds per day, and in gradual increase of the radius of Moon's orbit, at about 3.82 centimeters per year.

(source: Wikipedia)

Let's suppose that Solaris' sun does not rotate. If Solaris' spin axis' direction is $\vec n$, the total angular momentum will be

$$\vec L_\text{total} = \vec L_\text{spin} + \vec L_\text{orbital} = I \omega \ \vec n + M r^2 \Omega \ \vec k$$

Where $\omega$ is the spin angular velocity, $\Omega$ the orbital angular velocity and $r$ the distance between Solaris and its sun.

So if Solaris is able to change its moment of inertia $I$ by changing its mass distribution, we see that it is indeed possible for it to adjust its trajectory, because if $I$ changes then $\omega, \Omega$ and $r$ will have to change to conserve total angular momentum.

• This is substantially correct, although I believe incomplete: one needs to write down energy conservation conditions too. It would be good if one could come up with some numbers through a complete system of equations. – WetSavannaAnimal Jun 14 '16 at 14:48
• @WetSavannaAnimalakaRodVance You are right, we should include energy conservation and see what can really happen. As soon as I have some time I will update my answer for sure :-) – valerio Jun 14 '16 at 15:01
• Yes, it's not trivial; I'd like to have a go at it myself but I could see myself easily wasting many hours. Particularly if the irregular body is tumbling, so that its $I$ tensor is changing both from shape shifting and from the movement of its principal axes. This is possibly a problem where principal axes wouldn't be that helpful; I'm sure it must have been tackled in the literature somewhere. – WetSavannaAnimal Jun 14 '16 at 15:12

A different mechanism: On a long timescale, by increasing the surface area exposed to the sun (flattening the planet), the radiation pressure would increase, boosting to a higher orbit. Changing the albedo would be a more effective means to the same end but could allow assymetric force as well Either way it would be simpler in a tidally-locked planet. This has been proposed for deflecting asteroids. Extrapolating from figure 3 at that link, a perfectly reflective surface of the same scale as the asteroid/planet would take millenia for enough deflection to avoid an asteroid/coemt hitting Earth. There doesn't appear to be a limit to the timescale in the question, so assuming geological timescales this might be what you're looking for.

• I think a combination would be preferable. Use radiative pressure to gain orbital energy and rotation. Then exchange the rotation for more orbital energy by using tidal forces. – Taemyr Jun 15 '16 at 11:04

Thanks to Michael Seifert's answer, I found a paper he referenced: Satellite Relocation by Tether Deployment by G. A. Landis and F. J. Hrach, 1989.

By extending a tether radially, a satellite can increase or decrease its orbital speed (pictures below copied from the paper): The principle then can be used to pump an eccentric orbit: Similarly a planet like Solaris could then assume an elliptical shape, prolonging itself in a radial direction, to change its trajectory.

• ...though I daresay it's safe to assume this is not how Solaris stabilises its orbit. This would be purely simple Newtonian gravity, but the book does tell us that Solaris actually tweaks the Minkowski metric (in a way nobody understands). It's probably necessary, too; I doubt merely modulating the ellipticity a bit would be sufficient. These tethers need to have length that approaches the orbital scale, so the planet would pretty much have to spaghettify itself. – leftaroundabout Jun 15 '16 at 21:21

You can always use the process of tidal acceleration/deceleration. In nature this process might be very slow, such as for the system Earth/Moon. However, you can alway speed it up, by artificially increasing the frequency of the shape oscillations. In a natural system tidal acceleration will stop when the two objects are in tidal locking (both object always facing each other), but this can be overcomed. Tidal locking stops the acceleration because the objects no longer change their moment of inertia. If you do keep changing the shape artificially though, the process can continue indefinitely (but it will become faster as the bodies become closer or slower as they get farther away). the end product though, will be a huge change in the body rotational speed, which will be the end product of these changes in distance.

By conservation of momentum and energy, the only possible way to change a planet's trajectory is to eject some (large) mass at high velocity in specific direction, like rockets do. But you are also correct that by increasing the moment of inertia, the rotational speed can be changed. But this cannot influence the movement of center of mass.

Edit2: Other answers capture what I missed while looking for a fast solution. The interplay between rotational and orbital angular momentum can indeed produce some effect (credit to @WetSavannaAnimalakaRodVance and @valerio92).

Let's assume that the rotational axis of the planet and it's orbit are aligned. Then, we have 2 invariants:

$$E = \frac12 I \omega^2 + \frac12 m R^2 \Omega^2 - G \frac{m M}{R}$$ $$L = I \omega + m R^2 \Omega$$

where $I$ is a moment of inertia of a planet and $\omega$ is the rotational frequency while $\Omega$ is the orbiting frequency. $m$ and $M$ are masses of the planet and a star, respectively. Now, let's exclude $\omega$:

$$\omega = \frac{1}{I} (L-M R^2 \Omega)$$ $$E = \frac{1}{2 I} (L-M R^2 \Omega)^2 + \frac12 m R^2 \Omega^2 - G \frac{m M}{R}$$

For $\Omega$ we have a condition of staying on orbit:

$$\Omega^2 R = G \frac{M}{R^2}$$ $$\Omega^2 = G \frac{M}{R^3}$$

Then,

$$E = \frac{1}{2 I} \left(L-M R^2 \sqrt{G \frac{M}{R^3}} \right)^2 + \frac12 G \frac{M m}{R} - G \frac{m M}{R} = \frac{1}{2 I} \left(L-M R^2 \sqrt{G \frac{M}{R^3}} \right)^2 - \frac12 G \frac{M m}{R}$$

There might be a mistake somewhere, but we can solve this for $R$ and, keeping $L$ and $E$ constant, we can vary $I$ changing the orbit radius.

Edit: Not directly related to the question formulated in the title. Okay, among futuristic options would be destruction of some nearby objects like closest planets or the hosting star. If this won't destroy our planet, it's course will definitely change. But to do so, one needs to accurately disperse the mass comparable or much bigger than the planet.

Basically, everything boils down to changing distribution of the mass.

• Not even in the presence of a gravitational field of a star (or multiple stars) the planet orbits around? – Petr Pudlák Jun 14 '16 at 11:50
• Gravitational field conserves (at least, very, very well) the energy and momentum. Also, see my update – Andrii Magalich Jun 14 '16 at 11:58
• But it should be possible to use the gravitational maneuvering like the cosmic probes do to minimise the amount of things to blow up. – Andrii Magalich Jun 14 '16 at 12:01
• @AndriiMagalich The destruction would need rockets and momentum or some such, if it were done on purpose to control the orbit. – anna v Jun 14 '16 at 12:21
• On a pedantic point, all a star does is blow up continually. A single nuclear device to blow up the star is like using a firecracker to completely displace mount everest. – Neil Jun 14 '16 at 14:09

Fun question! Try this very simple answer (Newtonian as you asked)..

If a planet changes its shape from a round ball into a twine spool like shape (i.e. with a thicker center and elongated thinner ends) and assuming the elongation is done exactly along the radial line to the star (Sun), then for simplicity sake, the amount of mass that gets closer to the star is the same amount that gets farther from the star.

Preserving angular momentum in the above elongation scenario implies that the planet's spin rate along it's own axis will increase however for the same of brevity, we will assume that there is none or insignificant gyro effect at play and we'll look at only the gravitational forces... ...

Since the gravitational force is inversely proportional to the square of distance between the masses, it leads to an obvious/direct inference that the sum total gravitational force on the whole mass of the planet (part closer to star and the part farther) will increase, hence the planet will be pulled closer.

To make this a bit more clear.. Let's say a 1/4th of the planet (PartA) moved X km closer to its Star/Sun. At the same time 1/4th of the planet (PartB) moved X km away from the Star/Sun. The remaining 1/2 stayed at original distance and isn't part of the gravitational force change calculations. The original G force on PartA increases based on the standard Newtonian formula \begin{equation} \ F = G * (m1 * m2) / r^2 \end{equation} So this means that if "r" was originally Y km, now it is (Y - X) km. The means that F(on PartA) has increased inversely proportional to the reduction in "r". It also means that F(on PartB) has decreased inversely proportional to the increase in "r", however due to inverse square the increase is larger than the decrease, so as a whole the planet is experiencing more gravitational force. Which means the planet will start to move closer to the Star.

Hmm.. hey! that's pretty cool "a living planet altering its shape because it wants to be pulled itself closer to its Sun".

Similarly if the planet changes its shape to be more like a plate flattened along the planetary trajectory (i.e. perpendicular to the radial line), it will be able to reduce the gravitational pull and could get farther away from the star.

So yes, a living planet that was originally a round ball, by the above described fairly straightforward alteration of its shape, could effect a change in its own trajectory.

• Downvotes don't make sense. If you really think my answer is not valid, comment with your opinion instead of unproductive down votes. At least I'd have the option to clarify or explain. – LMSingh Jun 14 '16 at 19:51
• I simply can't understand your logic, it was the probable reason of the downs. For example, why will the spin rate exponentially increase? (I didn't vote down.) – peterh Jun 15 '16 at 0:03
• @peterh thanks. I'll edit the post to clarify. The reference to spin rate is a side effect but not material to the gravitational force. – LMSingh Jun 15 '16 at 7:41

One different aspect to this in case we are really talking about a living planet. Because gravity really kicks in at that scale, anything the size of a planet is so smoothly round that a polished billard ball feels ashamed for its own imperfection. So if this being is really planet-sized, it better be of really low density ...

## protected by Qmechanic♦Jun 15 '16 at 11:44

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