This question already has an answer here:

Edit: Although this question may look similar to this one, here I am asking about a constant force in the direction of Mars, and no one of the answers there have answered this question.

Because Phobos has always the same face turned toward Mars, a constant force in the direction of Mars could be achieved by placing an electric propulsion system on the opposite side of Phobos.

How much will Phobos go toward Mars as a consequence of a that force?

And which are the formula to calculate this change in distance between Phobos and Mars ?


marked as duplicate by Organic Marble, Rory Alsop, peterh, Hohmannfan Aug 25 '18 at 8:09

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ But it is a duplicate. The physics remains the same, it doesn't matter how or when you apply the force. $\endgroup$ – Hobbes Aug 19 '18 at 9:12
  • 1
    $\begingroup$ But you can’t have a ‘constant force’ towards Mars because force is a vector and you’ve just described that its direction is always changing. I think this is the root of your confusion and why orbital manoeuvres can be unintuitive at first. $\endgroup$ – Jack Aug 19 '18 at 10:05
  • $\begingroup$ @Hobbes If you think it doesn't matter how or when you apply the force, then you haven't well read the answers ! $\endgroup$ – Conelisinspace Aug 19 '18 at 11:19
  • 2
    $\begingroup$ I just provided the comment that follows to one of the answers at the proposed duplicate: 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. $\endgroup$ – David Hammen Aug 19 '18 at 11:31
  • 3
    $\begingroup$ Since this question represents a fundamental misunderstanding of orbital maneuvers it is not quite a duplicate, IMO. Even impulsive maneuvers can be rather counterintuitive ("Forward takes you out, out takes you back, back takes you in, and in takes you forward." What???) Low thrust can be even more counterintuitive. $\endgroup$ – David Hammen Aug 19 '18 at 11:40

I think this can be explained much more simply. Suppose we add a force $F$ pointing towards Mars at all times to Phobos. This will not change the angular momentum $h$ of Phobos. So if it is distance $r$ from Mars, it will be moving around Mars at a velocity of $h/r$ (it may also be moving towards or away from Mars).

So there is a stable circular orbit at radius $r$ where $$GM_{mars}/r^2 + F/M_{phob} = h^2/r^3$$ When $F=0$ we have the original orbit at radius $R$, so we get $$h^2 = GM_{mars}R$$ Combining those, we get $$GM_{mars}/r^2 + F/M_{phob} = GM_{mars} R/r^3$$ or

$$R/r = 1 + Fr^2/{GM_{phob}M_{mars}}$$

so provided $F$ is small compared to the gravitational force of Mars on Phobos we simply move the circular orbit a little closer to Mars, and then all our thrust is needed to keep it there, without making any further progress.

Putting in numbers from Wikipedia, we get that $$GM_{phob}M_{mars}/R^2 \sim 4\times 10^{15}N$$ so to lower the orbit by 1m (current radius is close to 10000km) would take a force of about $400 MN$, equivalent to that produced by about 5 billion of the ion thrusters used on Dawn (consuming about a terawatt of power), or 60 of the giant F1 rocket engines used on the Saturn V (consuming about 200 000 gallons per second of liquid fuel and oxidizer).

  • $\begingroup$ Impressive calculations, i will have to think about them for a while ! Meanwhile i've changed the question somewhat, but i don't think this will be disadvantageous for your answer. $\endgroup$ – Conelisinspace Aug 19 '18 at 22:05
  • $\begingroup$ Your answer is exactly what i was looking for, thank you ! $\endgroup$ – Conelisinspace Aug 20 '18 at 7:41

In the (non-inertial) rotating frame of reference bound to Phobos, with nadir/prograde directions setting the axis, the constant force towards Mars is the weight of Phobos.

Phobos semi-major axis (or orbital radius; it has very low eccentricity): 9376 km
Phobos mass: 1.06 x 10$^1$$^6$ kg, Mars mass: 6.39 x 10$^2$$^3$ kg

The force of gravity: $F = G {{m_1 m_2}\over r^2}$

Result: 5.22 x 10$^1$$^5$ N, force equivalent to weight of 532 billion tonne on Earth.

That force, though, is constantly offset by the centrifugal force (remember: non-inertial frame of reference!) - that's why Phobos isn't falling down onto Mars immediately. So its speed towards Mars is pretty much zero - save for minuscule movement as tidal forces brake it and centrifugal force drops slightly - resulting in Phobos orbital radius dropping, orbital speed rising and equilibrium between weight and centrifugal force restored.

Accelerating towards Mars will achieve nothing, as the resulting reduced orbital radius immediately converts to increased orbital speed, and equivalent centrifugal force ejecting it back to prior orbit or even further as soon as your pushing force vanishes.

Due to tidal forces, Phobos loses altitude - moves towards Mars at 1.8 centimeters per year; but that's a pretty much constant speed - if there was an isolated, unbalanced force, it would be accelerating its descent, as per Newton's second law of motion. Meanwhile, it sticks to the same average altitude loss rate, meaning the forces inwards/outwards are perfectly balanced.

  • $\begingroup$ Thank you for the answer, but can you mention the formula that you use for your calculations ? ( If you don't know how to write formula, you could use the one in one of the answers of the possible duplicate as an example. ) $\endgroup$ – Conelisinspace Aug 19 '18 at 14:46
  • $\begingroup$ @Conelisinspace: edited. $\endgroup$ – SF. Aug 19 '18 at 18:55
  • $\begingroup$ I've changed the question somewhat. If you have objections, please let me know. $\endgroup$ – Conelisinspace Aug 19 '18 at 21:59

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