Hayabusa-2 is near the astroid Ryugu, and both are essentially in nearly-matching heliocentric orbits around the Sun. Hayabusa-2's "hangout" is about 20 km sunward of Ryugu.

For the purposes of this question, it might be easiest to assume they are both in circular orbits, 1 AU or 150 million km from the Sun, and Ryugu's mass is 450 million metric tons, though it's not required to do so.

This answer describes the situation quite nicely.

If Ryugu wanted to remain there for months, how much delta-v is required per day to do so, and in what direction? You're welcome to use impulses or continuous thrust, but m/s per day is the target unit for the final answer.

"Bonus points" for what it's motion relative to Ryugu would be if the propulsion suddenly stopped and the spacecraft were allowed to drift under the gravitational attraction of Ryugu and the Sun. Which way would it start moving, and after 3 months where would it end up?


1 Answer 1


Let's first have a look at the acceleration due to gravity. $$a = G \frac{m}{r^2} = 6.7\cdot10^{-11} \cdot \frac{450\cdot10^9}{20000^2} \rm\frac{m}{s^2} = 7.5\cdot10^{-8}\rm\frac{m}{s^2}$$ Within a day this sums up to a $$\Delta v = a \cdot 86400~\rm s = 6.5~\rm\frac{mm}{s}$$ or $2.4~\rm\frac{m}{s}$ per year.

Now, the probe is stated to have the same orbital period as the asteroid, but a distance to the Sun that is 20 km less. Kepler doesn't like this. How our orbit looks like depends on how we define the starting conditions. If we start from a trajectory that keeps us exactly sunward of the asteroid for some time, we are in an orbit that has a smaller semi-major axis and a shorter period because we are slower than required. After a year we're several dozen kilometer (~ 100 km) in front of the asteroid and have to compensate for that. This slowing down to let the asteroid catch up is on the order of 3 mm/s if done constantly over the year. A more realistic assumption is that we do the adjustment say once per month, taking us more fuel, but still substantially below 1 m/s.

We can optimize our starting position by increasing the initial speed slightly and being on an elliptical orbit with the same semi-major axis and the same period as the asteroid. Unfortunately we are now moving between a larger and smaller distance to the Sun, sometimes being on the "dark side" of the asteroid and risk colliding with it four times a year. So, we have to use some fuel to re-adjust the distance. As the difference in orbits is so tiny, we can assume that this takes about the same amount of fuel.

There is actually one possible orbit that always stays sunward of the asteroid and doesn't require station keeping at all (despite correcting some perturbations): L1 - the Lagrange point between Sun and asteroid where their gravitational pull just cancels out. This point is about 70 km from the asteroid and doesn't fit the 20 km requirement.

There is one more aspect we shouldn't forget about: Solar radiation pressure. It amounts to a force of about $10\frac{\mu N}{m^2}$ in Earth orbit (according to wikipedia). Hayabusa has a surface area of about 10 m² and is accelerated by $$a = \frac{F}{m} = \frac{100~\rm N}{500~\rm kg} = 2 \cdot 10^{-7} \rm\frac{m}{s^2}$$ - three times as large as the gravitational acceleration. Again a convincing hint on how weak gravity actually is (your scale will disagree).

So, in total one can expect some 10 m/s per year for station keeping, but this is for sure increased by some factor due to the irregular shape of the asteroid and other sources of gravity.

To add to the bonus question: It's simple, you'll end up on the surface of the asteroid after a few days. Lateral forces are not high enough to make you miss it.

  • 2
    $\begingroup$ "or 2.4 m per year". Should this read m/s? $\endgroup$
    – DrSheldon
    Oct 2, 2018 at 21:32
  • $\begingroup$ Two orbits w/ same period (return to the same configuration after one year) will have the same semi-major axis. In the sentence "...a slightly elliptical trajectory that takes us further away from the asteroid ever so slightly over the course of a quarter of a year, then brings us back." do you mean further away from the Sun instead? A slightly elliptical orbit would intersect the circular orbit of the same period twice a year, so if the spacecraft is closer to the Sun than the asteroid now, it intersects in 3 months and is farther from Sun in three more. $\endgroup$
    – uhoh
    Oct 3, 2018 at 2:25
  • 1
    $\begingroup$ @uhoh If your current state is "closer to the Sun" and "same speed (actually lower as you specified 'between Sun and asteroid') as the asteroid", you're not on an orbit with the same semi-major axis. For that you would need to be faster than the asteroid. But that's all minuscule contributions because the difference is so small. $\endgroup$
    – asdfex
    Oct 3, 2018 at 11:29
  • $\begingroup$ I was just trying to understand the shape, parameters, and purpose of the elliptical orbit to which you refer, right now I don't understand the nature of this elliptical orbit. The reason I thought you might be talking about one is that a circular orbit 20km closer will move faster, so in 30 days it will advance in its orbit by about 15 km from its "home position". I had thought you'd matched the semi-major axes as a way to reduce that sizable walk-off rate. Right now I'm still not clear what your proposed orbit is, is it possible to add a summary separate from the paragraphs of text? Thanks! $\endgroup$
    – uhoh
    Oct 3, 2018 at 12:29
  • 1
    $\begingroup$ @asdfex the spacecraft is absolutely in a heliocentric orbit without question. That it may make parts-per-billion adjustments to that orbit does not make it a non-orbit. Thanks for the edit btw. $\endgroup$
    – uhoh
    Oct 3, 2018 at 13:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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