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Because Phobos has always the same face turned toward Mars, an electric propulsion system could be placed there to slow down the orbital speed.

But what would be the energy needed to supply the force of the propulsion system ?

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    $\begingroup$ If you would allow the engine to be put in a different location on Phobos, consider momentum instead of energy, and don't care about anything ON the Martian surface, then... $\endgroup$ – uhoh Apr 17 '17 at 9:33
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It is possible, but too costly!

And the orbital energy reduction is done best by retro-thrust, and that means, the exhaust of the propulsion device must be on the face in the direction of orbital motion (so that thrust is retrograde).

The energy needed would be:

$m \left(\frac{v_1^2}{2}-\frac{v_2^2}{2}-GM\left(\frac{1}{r_1}-\frac{1}{r_2}\right)\right)$

And this is in the best scenario, the propulsive energy that is not the same as solar energy input (considering solar-electric propulsion), because propulsion efficiency is not 100%.

For mining, It is indeed better to keep the Phobos up! Because it is much easier to loft materials off the Phobos' weak gravity and use it for construction in Martian orbit. And also, the debris of Phobos falling on Mars would wreak havoc in Martian atmosphere and climate and might pose danger to Martian colonies. Because when the Phobos falls apart, it's orbital height would be much lower and decays much faster. Also, after falling apart, Kessler syndrome happens and makes low Mars orbit a very dangerous place. And Mars gets its own ring after all! cool ;)

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    $\begingroup$ Could you specify m,v1,v2,G,M,r1,r2 ? I have read that Phobos is 'falling' 2 meters in a hudred years. Would it be too costly to keep Phobos up ? $\endgroup$ – Cornelisinspace Apr 17 '17 at 16:19
  • $\begingroup$ If the exhaust of the propulsion device is yet on the far side of Phobos can one use then, E= 1/2.m.v2 ? $\endgroup$ – Cornelisinspace Apr 17 '17 at 16:32
  • $\begingroup$ @AliRD no, you need to accept energy, not use it! Define the terms in your equation carefully and you'll see that Phobos actually gives kinetic energy to the molecules of the thrust. Right now your answer is... not right. As I mentioned here you need to talk about momentum, not energy. Flag me when you improve the equation and I'll reverse my temporary down vote :) $\endgroup$ – uhoh Apr 17 '17 at 16:47
  • $\begingroup$ @uhoh It is correct, by retro thrusting, both momentum and energy are reduced. I don't follow what you're saying? The specific energy of orbital motion equals {v^2/2-mu/r} and the work done by the propulsive force is approximately Tv.dt... this is subtracted from the orbital energy that is the specific orbital energy times mass. $\endgroup$ – AliRD Apr 18 '17 at 17:20
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    $\begingroup$ Would it be enough to, instead of using an engine, to paint half of Phobos black and the other hemisphere white, so that the sunlight heats it when approaching the Sun, braking its speed by light reflection and evaporation from heating? Is it straight off clear that such effects cannot compete with Mars' tidal forces? $\endgroup$ – LocalFluff Sep 15 '17 at 7:13
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First of all, the best place to place any such rockets would be at the point where the rocket is facing the direction that Phobos is rotating around Mars. That would give you about 4 x more bang for your buck to deorbiting Phobos than pointing directly at Mars, due to orbital mechanics.

Secondly, Phobos is actually being slowly deorbited naturally. It will take about 30 million years, but eventually it will land.

If you really wanted to deorbit Phobos, there is one major issue. Phobos is believed to not really be a cohesive body. If one put too much strain on it via a propulsion maneuver, it would probably break apart. Also, bringing it closer to Mars will almost certainly have the same effect.

Lastly, anything is in fact possible, if you get a large enough engine. To bring it from it's current 6,000 km orbit to a 3,000 km orbit would require a delta-v of around 660 m/s (I can't find an exact value, but that should be pretty close) The mass is about 10$^1$$^6$ kg. To have that much change in velocity over 10 years (315360000 s), it would require an average acceleration of 2.08 um/s$^2$. That would require a thrust of about $2.1*10^{10} N$, which is a considerable amount. All that really needs to be done is to hook up an engine that can maintain that thrust for 10 years, and you would accomplish your goal.

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  • $\begingroup$ Assume that because Phobos is more like a large asteroid than a "celestial body" that it could ... become enclosed in a material that prevented it from breaking apart. Now assume that I want it at 160 km and I want to have it be stationary over a single position such that I could build a Space Elevator. Would that same engine have the ability to keep it there? $\endgroup$ – Enigma Maitreya Apr 17 '17 at 14:50
  • $\begingroup$ Stationary at low altitude requires an ENORMOUS amount of energy. Far better would be to raise it up. But if you wanted to do that, then use Deimos. $\endgroup$ – PearsonArtPhoto Apr 17 '17 at 14:53
  • $\begingroup$ Ok, I assume your saying Deimos because it is what roughly 20% of the mass of Phobos and 67% the size so it becomes easier to package and less mass to manipulate and at 8 Miles in diameter, it would be sufficiently large to work as a habitat for the space end of the Elevator. Besides Deimos is leaving Mars and why let that resource just ... leave. So you convinced me so now it is the same question would that same engine be sufficient to keep Deimos there IF it is needed at all. I did not originally take into consideration the density of the Martian Atmosphere. $\endgroup$ – Enigma Maitreya Apr 17 '17 at 16:30
  • $\begingroup$ Also, it is much closer to the elevator space, and is more than large enough. Read about the physics of a space elevator, basically you have to be above the stationary orbit of the planet in order for it to work. $\endgroup$ – PearsonArtPhoto Apr 17 '17 at 16:31
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    $\begingroup$ The deltaV for a hohmann transfer to a 3000km orbit would be approximately 210m/s + 260m/s so a total dV of ~470m/s, you could add a bit to that for the spiralling trajectory but I doubt it'd be much, 500m/s would probably be a fair estimate. $\endgroup$ – Blake Walsh Apr 17 '17 at 17:51
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Phobos is tidally locked now, but if you started changing Phobos' orbit, you'd be changing its orbital period. Tidal locking is a very slow process, so pretty soon Phobos will lose tidal lock and will start to spin relative to Mars. That means your rocket engine will be in the wrong position to provide retrograde thrust for most of Phobos' rotation period. So you end up either taking a very long time to deorbit Phobos or having to build a rocket engine that can travel across Phobos' surface.

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