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Being a newbie, I humbly submit the following question : How to minimize the energy (in terms of delta-v burns) to transfer a mass of 1 metric ton (1000kg) from Very Low Earth Orbit (say 200km above sea level) to Low Lunar Orbit (say 100km above the moon's surface) using a Hohmann transfer burn sequence (or Bi-elliptic which can supposedly be even more energy efficient)? The idea being to sacrifice the transit time to LLO in order to minimize propellant and engine power even if this requires a very large number of small burns (incrementally enlarging LEO to MEO to GEO and finally reach the equilibrium point where Earth's gravitational pull equals the moon's then slowly cycle down to LLO).

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These all look like real trajectory design concepts, so let's go through them in reverse order:

incrementally enlarging LEO to MEO to GEO and finally reach the equilibrium point where Earth's gravitational pull equals the moon's then slowly cycle down to LLO

This in inefficient, as it takes both more time and delta-v than going straight into a transfer orbit and then braking at the end. The only reason you would do this is if your engines aren't capable of rapid acceleration, which is the case for ion engines. This slow spiralling costs them delta-v, but is still a net win due to the much higher efficiency of the engines.

For a very gentle spiral, the delta-v cost between two circular orbits is simply the difference of the orbital velocities (fun thing to prove!). So, at least from a patched-conics perspective (we have "zero" velocity relative to the Moon when encountering it), we have:

$$\Delta v = v_{LEO} - v_{Moon} + v_{LLO}$$

or Bi-elliptic which can supposedly be even more energy efficient

A lunar transfer is really neither a Hohmann transfer nor a bi-elliptic transfer, as those apply to a target circular orbit, crucially with no Moon there. Nevertheless, the same idea for why bi-elliptic transfers are sometimes more efficient can be applied.

To scale: consider the difference lunar transfer orbit (blue) and the almost-escape trajectory (black). The cost difference from LEO (very small circle) is very low. We can see why from the vis-viva equation:

$$v = \sqrt{\mu \left(\frac{2}{r} - \frac{1}{a}\right)}$$

Both of these orbits have a large semi-major axis, so $\frac{1}{a}$ is almost zero.

biell1

Now suppose we do a very small burn "at infinity" (far away), again at close to zero cost, returning on the purple trajectory.

The savings come from the velocity difference between the Moon and the purple orbit is now smaller than the velocity difference between the Moon and the direct transfer orbit. Which saves some propellant when slowing down.

In reality, this gives very little benefit for the much longer time spent, and real world examples often combine this with subtle 3-body effects like multiple lunar flybys and having the Sun bend the trajectory slightly while far out from the Earth.

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    $\begingroup$ Absolutely spectacular response, so let's try to make it concrete...say NASA decided to test a small ion engine in real conditions by sending from the iss to LLO a tiny image sensor camera chip with the smallest transmitter possible (and appropriate solar panels or plutonium radioactivity power source) propelled and guided by a small ion engine. How many ion burns would be needed? How long would it take to receive images of the lunar surface from LLO? $\endgroup$
    – C. Burns
    Commented Dec 11, 2023 at 14:10

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