# 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 – Quietghost Sep 12 '19 at 13:47
• Assuming a massless, thousand kilometer long tether, is not a good decision in your analysis... – Michael Stachowsky 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. – Hohmannfan 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. – Magic Octopus Urn Sep 15 '19 at 2:25