5
$\begingroup$

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

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

$\endgroup$
1
  • 1
    $\begingroup$ 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
4
$\begingroup$

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 ;)

$\endgroup$
9
  • 1
    $\begingroup$ 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 ? $\endgroup$
    – Cornelis
    Apr 17 '17 at 16:19
  • 1
    $\begingroup$ @Conelisinspace "m" is the Phobos' mass, "v1" is the initial orbital velocity and "v2" is final. "r1" is the initial orbital radius and "r2" is final. "G" is the universal gravitational constant. "M" is the mass of Mars. And no, mv^2/2 is not all the energy terms of the Phobos, it is just kinetic and you need to consider the potential term (GM/r) too... $\endgroup$
    – AliRD
    Apr 18 '17 at 17:26
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @uhoh It's been some time since, but still I would very much like tto know he right answer from you. Indeed, with the formula in this answer one still doesn't know the values of v2 and r2. $\endgroup$
    – Cornelis
    Jul 7 at 10:35
  • 1
    $\begingroup$ @Cornelis I think AliRD explained those in a comment You can get the velocities from the vis-viva equation and you'll see that that equation is related to this expression for energy. It's been a few years, but I think my objection was only on the use of the word "energy" since going to a lower orbit means the spacecraft loses energy. I am not 100% sure if that's what I meant but I think so. In some old answer I remember putting vis-viva into this equation, I'll look for it. $\endgroup$
    – uhoh
    Jul 7 at 11:59
7
$\begingroup$

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 μm/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.

$\endgroup$
10
  • $\begingroup$ 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? $\endgroup$ 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. $\endgroup$ Apr 17 '17 at 16:30
  • 1
    $\begingroup$ 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. $\endgroup$ Apr 17 '17 at 17:51
  • 1
    $\begingroup$ Applying a low constant magnitude thrust that always points directly toward at Mars would have almost exactly the same effect as would a low constant magnitude thrust that always points directly away from Mars. Both are incredibly inefficient mechanisms for slowly raising Phobos's orbit. $\endgroup$ Aug 19 '18 at 11:30
3
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ What about orbital changes ? en.wikipedia.org/wiki/Tidal_locking#Orbital_changes And since huge amounts of energy are involved, the change in orbit will be very slow,slow enough for the tidal locking to compensate in time i think. $\endgroup$
    – Cornelis
    Aug 19 '18 at 11:31
  • $\begingroup$ Or just make your rocket aimable by a fraction of a fraction of a fraction of a degree. $\endgroup$
    – PcMan
    Jul 8 at 0:02
3
+100
$\begingroup$

First, one assumptions:

The acceleration is so low that instantaneous impulse solutions are out of the question, and the trajectory can be modelled as a very gentle spiral.

This is quite reasonable, as an absolutely enormous amount of thrust would be necessary to provide high acceleration to a $1.0659×10^{16} kg$ rock.

So let's get started then. First, we need the delta-v.

For gentle continuous thrust spirals between circular orbits, the equation is surprisingly simple:

$$\Delta v = v_0 - v_1$$

Yes, just the difference between the initial and final orbital velocities.

To bring Phobos down to say, half the orbital radius, we need to supply 885 m/s (orbital velocity scales with the inverse square root or radius, so the velocity difference for half the radius is $\sqrt{2} - 1$ times the orbital velocity of Phobos, which is 2.14 km/s). We can then see what exhaust velocities are needed to spend only 1% of the mass of Phobos.

We take the rocket equation...

$$\Delta v = v_e \cdot ln\left(\frac{m_1}{m_0}\right)$$

... and turn it around!

$$v_e = \frac{\Delta v}{ln\left(\frac{100}{99}\right)} = 88.1 km/s$$

The energy required for this would be:

$$E = \frac{m_{propellant} \cdot v_e^2}{2}$$

For the half orbital radius, 1% Phobos mass example, that would be $4.14 ×10^{23} J$, or 4000x the World's yearly energy production.

Note that you do not want to use a higher specific impulse than necessary, as at small propellant mass fractions, the energy requirements go up linearly with the exhaust velocity.

As a general equation:

$$E = \frac{m_{propellant} \cdot \left(\frac{\Delta v}{R_{mass}}\right)^2}{2}$$

Were the mass ration $R_{mass}$ is:

$$R_{mass} = ln\left(\frac{1.0659 \cdot 10^{16} kg}{1.0659 \cdot 10^{16} kg - m_{propellant}}\right)$$

And the $\Delta v$ is:

$$\Delta v = \sqrt{\frac{\mu}{r_{final}}}- 2138 m/s$$

In the range 0 m/s (no change) to 1400 m/s (gracing the surface of Mars).

$\endgroup$
5
  • $\begingroup$ Thank you, I asked for formula, now I got them ! But to learn from this answer could you explain where you got them from, in particular those with the ln expression ? And how did you calculate the 885 m/sec ? $\endgroup$
    – Cornelis
    Jul 7 at 12:32
  • 1
    $\begingroup$ @Cornelis Both added. $\endgroup$ Jul 7 at 16:13
  • $\begingroup$ I think an exhaust velocity of 88 km/sec is not realistic, is it ? Could a much lower one not bring down the required energy ? $\endgroup$
    – Cornelis
    Jul 7 at 16:53
  • $\begingroup$ @Cornelis The energy would indeed go down, but the mass would go up! If the exhaust velocity is less than 88 km/s you have to use more than 1% of the mass of Phobos as propellant. That's a lot of mass! $\endgroup$ Jul 7 at 20:02
  • $\begingroup$ Clear, what a waste of energy ! $\endgroup$
    – Cornelis
    Jul 11 at 13:13
1
$\begingroup$

Goal: lower Phobos's orbit.

Status: ACHIEVED!

By the time you read this line, Phobos' orbital altitude is already lower than when you started reading this question.
Tidal deceleration is dropping Phobos by about 2cm per year, and in less than 50 million years Phobos will impact Mars.
Well, actually it won't. In only about 20-25 million years, Phobos will descend below its Roche's limit, and turn itself into a nice rocky ring for Mars.

You should be worrying about how much energy is needed to keep Phobos UP, and prevent armageddon from raining fiery death down on your newly-terraformed Mars!
(spoiler: you need about 60N of continuous thrust to counteract the Tidal deceleration)

$\endgroup$
2
  • $\begingroup$ We don't have the time to wait a million years ! And by the time Phobos is rurned into a "nice rocky ring" will there still be tidal deceleration to "rain it down" ? $\endgroup$
    – Cornelis
    Jul 7 at 22:13
  • $\begingroup$ When Phobos starts falling apart, probably valuable material could be extracted.from the inside. $\endgroup$
    – Cornelis
    Jul 8 at 13:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.