In case of a StarShip (but it could be any other lander), it would arrive a Mars with minimal fuel remaining and it would need to slowdown for reentry, how can it be slowed down for free? Maybe using Phobos?

I am curious if it’s possible to lose a lot of speed by entering into a parallel orbit of Mars' inner moon Phobos and using its pull as a brake but it probably requires advance computer simulations.

For some reason this parallel braking via gravity assist evokes gyroscope force ideas. May be it would take thousands of iterations but the speed should eventually come down, right?

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    $\begingroup$ Phobos is absolutely tiny. Its escape velocity varies with location because it's sort of a potato shape, but at the highest its escape velocity is a little lower than an olympic sprinter's top speed. If Usain Bolt could stay stuck to the surface long enough to get a running start, he could potentially long-jump entirely out of Phobos's gravity well. At the ends, you could leave it forever by taking a brisk walk (~3 mph). I think the only real way to use Phobos as a brake is by crashing into it. $\endgroup$ Commented May 9 at 15:40
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    $\begingroup$ Think about what it would take to get into an orbit around Phobos. $\endgroup$ Commented May 9 at 17:09
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    $\begingroup$ What is a parallel orbit? $\endgroup$ Commented May 9 at 17:44
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    $\begingroup$ Phobos might be useful for lithobraking. $\endgroup$
    – Mark
    Commented May 10 at 7:27
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    $\begingroup$ @Mark how many times? $\endgroup$
    – fraxinus
    Commented May 10 at 7:32

2 Answers 2


This wouldn't really work.

Phobos is tiny, and its gravity is proportionally weak (in some areas, the escape velocity is around a slow jog), so first off, any sort of gravity assist you do is going to have a very weak effect.

This question digs into the math of how much velocity change you can get from a gravity assist maneuver: How much delta-v can we squeeze out of a gravitational slingshot? But the bottom line is the Δv you can get is more or less controlled by the angle of the change in direction your craft experiences, with 60 degrees being the optimum.

The maximum angle change you can achieve on a flyby is limited by the mass of the object, the altitude of your closest approach, and the relative speeds of the vehicle compared to the object. I hope it's intuitive that the smaller the mass of the object and the faster you're passing it, the less effect the pass will have on your ship's direction and thus the smaller the momentum transfer will be. You can increase the effect by passing closer to the target. If you had an infinitely dense point-source of mass (a black hole), you could theoretically attain maximum momentum transfer regardless of speed by getting closer to the mass -- but that doesn't work for three dimensional objects because you can only get so close before you're lithobraking instead of getting a gravity assist.

Since each pass changes your orbit, it's unlikely that you can arrange to have many gravity assist passes to slowly lower your speed; you'd have to spend fuel to adjust after each pass to hit the next one. You'd be spending a lot of time and effort for not much benefit.

Since it's Mars, there's a much easier answer, though: aerobraking would use one or more passes through the atmosphere of Mars to slow down instead of trying to use gravity assists to do it. If you can handle the heat, this can provide virtually unlimited braking capacity, up to and including coming to a complete halt by landing on the surface in a single atmospheric entry (like, say, how a rover gets onto Mars's surface).

On some other planets, a gravity assist can be a useful way to cut down the cost of fuel -- for example, JUICE is intended to brake into Jupiter orbit by making a close pass by Ganymede that slows it just enough to stay in a long elliptical path where it'll adjust to hit the rest of its flyby targets -- but Mars just doesn't have enough mass in orbit to be worth messing with.

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    $\begingroup$ Thanks. In that flyby of Ganymede would it matter if spacecraft was going on the same direction or the opposite of it? Just wondering if traveling in opposite direction from Phobos would increase the drag. $\endgroup$
    – estinamir
    Commented May 9 at 19:32
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    $\begingroup$ Never mind, there practically would be zero drag either way. $\endgroup$
    – estinamir
    Commented May 9 at 20:37
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    $\begingroup$ Approaching from the front doesn't actually change things except that it increases the approach speed (head-on instead of tail-chase) which reduces the potential amount of deflection by making you spend less time close to the body. $\endgroup$ Commented May 9 at 21:04
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    $\begingroup$ +1 for the "lithobraking" double entendre. $\endgroup$
    – phil1008
    Commented May 12 at 6:47
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    $\begingroup$ It's a euphemism rather than a double entendre (which would imply there's two distinct meanings), but yeah, it's a pretty common joke. I'm not sure who first coined it; I first ran into it on the Kerbal Space Program boards, where it was pretty popular, and in the decade since, I've seen it become a commonly accepted term alongside "rapid unscheduled/unintentional disassembly" (i.e. a rocket explosion). (I know RUD originates from the US military, where it referred to a rifle falling apart when you try to load it, usually due to failing to seat the parts properly after cleaning.) $\endgroup$ Commented May 12 at 14:29

Phobos isn't massive enough to gravitationally affect a spacecraft's trajectory. The gravity on Phobos is only 0.006℅ of what it is on Earth.

Phobos would be useful in the settlement of Mars as a kind of "way station" to prepare for landing on Mars itself. Also, a couple O'Neill cylinder spacesteads could be embedded into Phobos. This little Martian moon is about 27 kilometres long and 18 km wide. Thus an O'Neill cylinder with dimensions of up to eight kilometers in diameter and a length of up to 25 kilometers. A second O'Neill cylinder with a diameter of up to 5 kilometres could also be built within Phobos. The main Phobos O'Neill cylinder would have up to 200 square kilometres of land inside, an area akin to a small country. The smaller one would have an area of up to 100 square kilometres, still nothing to scoff at. Multiple cities and towns could be placed in each cylinder habitat and depending upon the population density, tens of millions of people could end up calling these spacesteads home! Alternatively, they could be partially (or entirely) outfitted as nature reserves or farms.

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    $\begingroup$ The bulk of this answer is irrelevant to the question, and the part that is relevant is not supported. $\endgroup$ Commented May 11 at 23:38

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