Start with the free Future of AI course—2 hours, no technical background needed, completely free. You can begin understanding AI and how it relates to you today: https://bluedot.org/minutephysics
At a major axis ratio of 1:15, you're already well in the territory where a double-elliptic transfer is more efficient. The reason is actually exactly what was described in this video.
There's also the fact that if you swing your orbit out far enough, you'll eventually find a second gravitation well you can swing off of for even more fuel savings.
The other problem with bi-elliptical transfers (and going to infinity) is that eventually you're going to hit the gravity well of another body and that'll screw up all your planning.
The really fun thing is that since everything is symmetric, it's also more efficient to use a bi-elliptic transfer to *lower* your orbit. The first step is shooting yourself into an even higher elliptical orbit. Going up in order to go down.
I imagine if you tried to overshoot by an infinite amount, you'd reach a point where you are so far away that the gravity of the object you are orbiting can't pull you back faster than the universe is expanding. So there must be some finite limit imposed by the fact that the universe is expanding.
2:52 I have actually used this in KSP before lmao. Just barely didn't have enough fuel to return to kerbin, so I ejected myself slightly further out and came back with 12m/s to spare.
Two approximations in this video:
First, the graphs use Δv as the cost, which is definitely not the same as fuel. In fact, a small saving in Δv can result in a huge saving in fuel.
Second, an infinite bi-elliptic transfer is, technically, a bi-parabolic transfer.
6:30 - There are also other mass out there, so if you go too far out, your elliptical orbits might have its trajectory altered by the gravity well of some other astronomical object and cause you to never reach the theoretical recircularization point. Even if you're overshooting by a large-but-finite distance.
theres another problem - when you go so far out you must be extremely precise on how much of a burn you do, so at these insane distances of overshooting thousands of times farther you must be very precise, or you will end up needing to make a course-correcting burn, and at the millions of times overshooting its simply impossible to be so precise so you will be needing to make a course-correcting burn
Start with the free Future of AI course—2 hours, no technical background needed, completely free. You can begin understanding AI and how it relates to you today: bluedot.org/minutephysics
We use cookies, device fingerprinting and cross-device tracking to personalise content, serve targeted advertising, and analyse how you use our site across your devices. By clicking Accept all you consent to the use of these tracking technologies. Your data may be used to build a profile of your interests and show you personalised ads on other sites.
Privacy policy
Comments (20)
Daily Junction discussion mixed with clearly attributed comments from the original video provider.
Join in — free. Comments on Daily Junction are for members, so real names stay rare and bots stay out.
One field. We email you a 6-digit code — no password needed. Your comment is kept while you do it.
Under 13? You’ll need a parent’s OK first — it takes them one click.
First, the graphs use Δv as the cost, which is definitely not the same as fuel. In fact, a small saving in Δv can result in a huge saving in fuel.
Second, an infinite bi-elliptic transfer is, technically, a bi-parabolic transfer.