How much delta-v would Hayabusa-2 need per day to remain 20 km sunward of Ryugu?

Question

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?

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.