How much energy is needed to bring Phobos closer to Mars?

Question

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 ?

Answer 1

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 ;)

Answer 2

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.

Answer 3

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.

Answer 4

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).

Answer 5

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)