# Orbital insertion with "inertial parachute"?

Inspired by Is there any way to land a rover on the Moon without using any thrusters?, but off-topic as a solution as that question asks specifically about crumbling zones.

The idea:

• A spacecraft heading for the Moon will remain linked to the spent upper stage with a tether.
• With a small initial nudge, they will drift several thousand kilometres apart while coasting.
• When approaching the Moon, the trailing upper stage is accelerating slower than the spacecraft, thus pulling the tether and acting like a brake.

(the fact that the dry mass of the upper stage is usually significantly lower than the spacecraft is less relevant, this is about the orbital mechanics involved)

I'm able to model the scenario where the spacecraft is heading straight towards the Moon and impacting in the following way:

The acceleration with respect to the Moon is the arithmetic mean of the two objects (I'm assuming a massless tether).

$$a(r) = \frac{m_1a_1(r) + m_2a_2(r)}{m_1 + m_2}$$

$$a(r) = \frac{m_1 \cdot \frac{\mu}{r^2} + m_2 \cdot \frac{\mu}{(r + l)^2}}{m_1 + m_2}$$

That gives an integral, which yields a formula for the impact velocity

$$v_{impact} = \sqrt{2\int_{\infty}^{r_0}a(r)dr + v_{\infty}^2} = \sqrt{\frac{2\left(m_1 \cdot \frac{\mu}{r_0} + m_2 \cdot \frac{\mu}{r_0 + l}\right)}{m_1 + m_2} + v_{\infty}^2}$$

But that's just a special case. What I would rather use this for is orbital insertion.

How do I calculate the velocity at periapsis, given two linked masses arriving from a hyperbolic orbit?

The periapsis height can obviously be picked arbitrarily, and if the altitude of the trailing object was known at that moment too, a similar integral to the above can be used. However, I find it difficult to approach, since that quantity is non-trivial. Unlike the face-on trajectory, the acceleration vectors are in slightly different directions, which gives the whole system a spin, which makes the orbit non-keplerian. The maximum tension on the tether would also be of interest.

• It looks like you are asking about the dynamics of tether orbits. I have only an inkling of knowledge about them myself, but there are some articles out there concerning their instabilities. They may not be analytically solvable except for the very simplest of cases like the one you described. Search for "Ultra Long Orbital Tethers Behave Highly Non-Keplerian and Unstable" by Rugescu and Mortari Sep 12 '19 at 13:47
• Assuming a massless, thousand kilometer long tether, is not a good decision in your analysis... Sep 12 '19 at 18:09
• @Michael Stachowsky provided large enough masses at the end and a thin enough tether, it is. If the mass of the tether was significant, I would rather spend that on rocket fuel. Sep 12 '19 at 21:35
• I kind of want a ballpark on how much 1000 km of cable having the strength for this would weigh too! But the theoretical calculation is much more interesting than that. Sep 15 '19 at 2:25

### Partial result:

I've come to realise that cutting the tether at periapsis is not the optimal way of going about this.

Because, as long as the angle between the tether and the velocity vector of the payload is larger than 90 degrees, the payload would still be bleeding energy, and thus end up in a tighter capture orbit.

$$\angle\vec{A B}\vec{v_{A}} \geq \frac{\pi}{2}$$

The form of the question going "How do I calculate the velocity at periapsis, given two linked masses arriving from a hyperbolic orbit?" is thus not exactly correct, as what one is really looking for is the full set of 2D orbital elements.

This probably makes the problem even more intractable...

On the flip side, this isn't too hard to simulate: I think you're overthinking this. Won't it behave almost as if it were a single mass at it's center of mass? Set your periapsis to the distance between the center of mass of the spacecraft and set the rotational velocity to what your velocity will be at periapsis, you can cut the tether when it gently deposits you on the surface.

However, there are two problems. First, as already mentioned, your tether is made of unobtainium. A sane tether for this purpose requires a bond strength far in excess of what is theoretically possible. (Look at momentum transfer tethers and skyhooks for more on this--in a non-expended form and possible with current tech for the lesser bodies.)

Second, how are you going to spin your tether? I'm pretty sure that's going to take as much as landing would, although since it's done in space it could be done with a low-thrust engine.

• Behaving like as it were a single mass at it's centre of mass can not be the case, as one can see by linking two satellites in the same orbit: i.stack.imgur.com/3chHn.png Jul 12 at 22:16