Edit: Since i've learned a lot from the excellent answer below, i've changed the content of my question totally to make it more realistic.

The orbital radius of Phobos is gradually decreasing by 2 meters every one hundred years by tidal deceleration.Scientists estimate that Phobos will be destroyed in approximately 30-50 million years.

Could an electrical propulsion system on the side of Phobos facing Mars powered by solar energy on the sunny sides be powerful enough to prevent this from happening ?

  • $\begingroup$ You should mention you just asked a very similar question $\endgroup$ – J. Chomel Apr 18 '17 at 11:08
  • $\begingroup$ It is a totally other question ! This question is about saving Phobos, the other question was about destroying Phobos ! $\endgroup$ – Cornelisinspace Apr 18 '17 at 16:57
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    $\begingroup$ Physics doesn't care about your goals. Physics does care about getting the formulas right. You're entirely ignoring the fact that you're moving Phobos against gravity. $\endgroup$ – MSalters Apr 19 '17 at 14:31
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    $\begingroup$ You should rather consider talking with historic architecture restoration people. They do manage to keep structures the size of Phobos from falling apart. Proper concrete injections, steel braces, binding infusions, and with their care Phobos would remain whole and unbroken right until impact into Mars. $\endgroup$ – SF. Aug 13 '18 at 17:28
  • $\begingroup$ @SF And create space for living as well ! One of the best environments near Earth to get practical experience i think, and with plenty water. $\endgroup$ – Cornelisinspace Aug 14 '18 at 7:42

Something is very badly wrong with your calculations. Even putting aside that if you want to raise the orbit of something, you need to thrust in the direction of travel.

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But to actually calculate the energy requirements you need to first treat the problem in terms of momentum

  • You want to change the velocity of Phobos by 1mm/s
  • That requires sending some mass in the opposite direction
  • The required change in momentum is $0.001m/s\times10^{16}kg$ = $1\times10^{13}kg\cdot m/s$
  • Lets say the electric drive has an exhaust velocity of 20km/s
  • The amount of propellant required is thus $\frac{10^{13}kg\cdot m/s}{2\times10^4m/s}=5\times10^8kg$

So a change in velocity of 1mm/s, would require firing 500000t of propellant at 20km/s in the other direction. This happens to be about equal to the mass of a supertanker or the Empire State Building.

Now that we know we need to fire out 500000t/s at 20km/s we can calculate the energy requirements using $E=\frac{1}{2}mv^2$ (hint: it's going to be bad). It is simply equal to $$0.5\times5\times10^8\times20000^2 = 10^{16}J$$

This amount of energy is roughly equal to the energy unleashed in the detonation of a fairly large thermonuclear warhead. So we actually could impart a change in velocity of 1mm/s on Phobos, altough we would need to nuke it, which would not save it from falling apart.

Phobos is huge. It's tiny on the scale of moons and planets, but it's still huge relative to things humans build and manipulate. If you take all the coal which is mined every year - which is about 5 billion tonnes, Phobos is 2000 times bigger than that. So even dismantling it into pieces to be flung out of a high powered electric mass driver would be a monumentally huge task.

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  • $\begingroup$ Thank you for your clear explanation ! It is said by NASA that the orbital radius of Phobos is decreasing with 2 cm every year. If the velocity of Phobos could be changed with 1 nm per sec., this would be 3.1 cm every year. The amount of propellant required would then become 500 kg ! And the required amount of energy would become 10,000 MWs. A solar PV system with a size of 100m by 100 m could have the power of about 1 MW. So within one year the required amount of energy could be easily reached. Or am i wrong again this time ? Could the 500 kg propellant be found on Phobos ? $\endgroup$ – Cornelisinspace Apr 18 '17 at 16:48
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    $\begingroup$ @Conelisinspace Actually the way orbital decay works, a reduction in orbital altitude of 2cm would be caused by a reduction in orbital velocity of like ~0.002mm/s per year, this could be counteracted by burning 0.2kg of rocket propellant per second. Any kind of matter can be used as propellant, rocks can be flung by mass driver or vaporized by a laser, water can be superheated into a plasma and so on. You'd be looking at about 3MW to counteract the orbital decay. However I suspect in the short-mid term humanity will be much more interested in pillaging Phobos than preserving it. $\endgroup$ – Blake Walsh Apr 18 '17 at 17:29
  • $\begingroup$ Pillaging could become a rather dusty business with a layer of regolith 100 meters thick ! $\endgroup$ – Cornelisinspace Apr 19 '17 at 8:00
  • $\begingroup$ Could you specify how you calculate the reduction in orbital velocity and the amount of propellant to be burned per second ? $\endgroup$ – Cornelisinspace Apr 19 '17 at 8:12
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    $\begingroup$ @Conelisinspace Consider an object in orbit returns to it's original location, so after doing a thrust the maximum change in altitude is on the far side of the planet, for Phobos that would be 4 hours later (half the orbital period). So if the change in velocity is say 0.001m/s, the change in altitude at the far side of the planet will be 0.001m/s * 4 * 3600 = 14.4m - (that's just a first order approximation - actual calculations much more complicated). Calculating the propellant required to achieve that change in velocity is as per my answer. $\endgroup$ – Blake Walsh Apr 19 '17 at 9:27

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