This tweet "from Hayabusa-2" via JAXA shows the diagram below, which seems to suggest that the Hayabusa-2 spacecraft tends to remain between the asteroid Ryugu and the Sun. Does Hayabusa-2 have to fight significant gravitational effects in order to stay there rather than orbit the asteroid, or at about 20 km is gravitation from the asteroid just a very small perturbation? Or is it more complex than that?

If they are just on similar heliocentric orbits, then how could the spacecraft stay between it and the Sun?

Could it be somewhat stabilized there due to gravity? Perhaps this is near the Sun-Ryugu L1 point?

Text and one of the images of the tweet:

While Hayabusa2 normally hovers at the home position, for the BOX-B “tour” operation, the spacecraft swung round to see Ryugu’s south pole and evening side. http://www.hayabusa2.jaxa.jp/en/topics/20180927e_BoxB/

enter image description here

  • $\begingroup$ I’ve had the impression that Ryugu’s gravity is pretty inconsequential at that distance and the home position is more to do with good positioning for observations, but I’m not sure on the details... $\endgroup$
    – Jack
    Commented Sep 29, 2018 at 16:57
  • $\begingroup$ @Jack a half-billion tons may not be inconsequential. For starters, try calculating the radius of the Hill sphere. If that returns a meaningful value, then Ryugu's gravity is probably consequential. $\endgroup$
    – uhoh
    Commented Sep 29, 2018 at 17:24
  • 1
    $\begingroup$ You’re right- I make the Hill radius to be ~70km. It varies quite a bit since Ryugu’s eccentricity is relatively high, but still always more than 20km... $\endgroup$
    – Jack
    Commented Sep 30, 2018 at 11:02
  • 1
    $\begingroup$ I don’t have a pen and paper to hand, but it might be worth seeing what the time-to-impact when dropped from 20km is - that may have been a strong consideration $\endgroup$
    – Jack
    Commented Sep 30, 2018 at 11:05
  • $\begingroup$ @Jack I think you should post that as an answer. It may not be conclusive but it certainly narrows things down quite a bit. Since it's well inside the Hill sphere, it would normally be in orbit. If it's not (and that's what the tweet and image suggest) then calculate say how much velocity it might accumulate in 1 hour in free-fall towards Ryugu. That's roughly how much delta-v per hour would be needed per hour to keep it there. $\endgroup$
    – uhoh
    Commented Sep 30, 2018 at 11:05

1 Answer 1


Hayabusa2's Home Position, also known as BOX-A, is a cubic region approximately 1km in size, 20km above Ryugu and directly on the Ryugu-Earth line.

We can show that the Home Position is not at any Lagrange points just by checking where they would be for Ryugu. The Hill radius of an object mass $m$ in orbit with semi-major axis $a$ around a body mass $M$ (the Sun) is given by:

$$r = a \sqrt[3]{\frac{m}{3M}}$$

which gives us ~75km. The L1 and L2 points are found approximately at the edge of the Hill sphere, so we can see that Hayabusa2 is dominated by Ryugu's gravity. Note - Ryugu has a relatively high eccentricity so the Hill radius varies significantly over time, but the mean is around 75km.

However, dominated is a strong word for Ryugu's influence on the craft - at the Home Position the radial acceleration from gravity is some $7.5\times10^{-8} ms^{-2}$ or just $0.28 mms^{-1}h^{-1}$. This makes conventional orbital manoeuvres impractical/unusable so Hayabusa2 can be considered to be co-orbiting with Ryugu rather than in orbit of it.

The reasons behind the positioning are all essentially down to this lack of a strong gravitational field and, perhaps more importantly, the fact that prior to the mission the strength and shape of the field wasn't known with much certainty. For example, in this presentation in the lead up to Hayabusa's arrival JAXA gives an 8-fold uncertainty in Ryugu's density and gives its mass as 1.7 × 1011 kg – 1.4×1012 kg!


For the above reasons, the initial approach and subsequent weeks were spent executing very careful manoeuvres to better map Ryugu's gravitational field in preparation for later surface activities. Since these manoeuvres couldn't be planned precisely in advance, the Ryugu-Earth line was used as the basis for the coordinate system for simplicity. The Ryugu-Earth and Ryugu-Sun pair define the Z-X plane - see here for more detail.

Other considerations include:

  • Ensuring that Hayabusa always has line of sight to Earth for communication
  • Ensuring Hayabusa's solar arrays are never eclipsed by Ryugu
  • Ensuring good illumination of the surface for tracking and observation (during BOX-A manoeuvres - the BOX-B regime moves laterally to provide different illumination angles)


The 20km separation essentially gives a 'safe' position from which to operate. We can show that the time-to-impact in freefall from this distance is approximately 6.5 days:

$$t = \frac{\pi r^{\frac{3}{2}}}{2\sqrt{2GM}} \approx 573000s$$

This uses the simplification that Ryugu's radius is 0 to get an approximation which is good enough for our purposes since the full formula is too messy for here! I also ignore the perturbation from the Sun which is small but not insignificant.

This of course means that ground teams have plenty of time to correct any drift due to the failure of the autonomous navigation system or signal drops. For reference this would give a drift of only 50cm over the course of an hour.


Hayabusa2 has a chemical propulsion system used for attitude control and manoeuvring at the asteroid. This consists of a set of 12 20N Hydrazine thrusters which can provide an acceleration of ~$0.1ms^{-2}$ - plenty enough to counteract any drift caused by Ryugu's gravity.

Further reading:

JAXA's website provides good up-to-date presentations from press releases on Hayabusa2 here at the bottom of the page.

I use 4.5E11kg for Ryugu's mass which is up-to-date, but the reported value still has a large relative uncertainty. My value for Ryugu's diameter is 850m which is approximately its mean diameter but the true value obviously varies substantially.


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