excellent job, minor point: the orbital velocity is reduced with altitude because earth gravity is reduced, but also because the orbit curvature is reduced!
Thanks! Years ago the SpaceX commentary mentioned that they were putting a satellite in a much higher transfer orbit and that it would save the satellite fuel, extending its lifetime. I couldn't figure out the search term to find out more about that orbit, but "bi-elliptic transfer" was what I needed!
@@gasdive Glad to help! There are a few cases where a bi-elliptic transfer can be helpful, but you have to be careful such that you don't lose all of your savings on the extra time it takes.
Great information. What about orbiting in a vacuum tube (such as Hyperlink) on or below the earth's surface. I assume orbiting the Lagrange points, the sun, and other bodies are another video. How about during lunar landings and the potential to fling rocks into a very low lunar orbit.
I might do another video on those topics... Note that you can't throw rocks into lunar orbit - any ballistic trajectory from a planet or moon comes and hits the moon unless it has escape velocity. You would either need something out there to catch it or a way to circularize the orbit.
I thought Baikonur was lower than 51.6 degrees N, but because a launch to its natural inclination would overfly China, something they didn’t want to do, they instead launch to a higher inclination to skirt around the country (or part of it). EDIT: also, is the diagram on how to get from GTO to GEO flawed? With the way you have it set up the satellite would be at its highest northern point on the ground track, and have to, at that point, shift its position downwards on the ground track until it’s at 0 (then correct), which would take a lot of dV. It would as such be far easier to have the apogee be at the equator on the ground track and instead have the points where the ground track is at the highest and lowest latitudes be more midway between the apogee and perigee on the orbit.
You are correct. Baikonur is at around 45 degrees north. And yes. You want to do the circularization and then do the inclination change if you are doing it in multiple burns.
Maybe I am interpreting the illustration with the inclanation change wrong, but for mee it looks like the apogee is 90° away from the intersection of the plane of the orbit and the 0° plane. Wouldn't you try to start into an orbit where the apogee is on the intersection of these planes? As you want to burn at apogee, as you said, you would "rotate" the orbit around this point. With the orbit shown in the illustration (at leas how I understood it) you would need to burn 90° away from the apogee to lower the inclation, which would be less efficent.
Good question... I did a little research, and I think my depiction is incorrect... You want to do the inclination change at the apogee because that is the lowest speed and it's a high-cost maneuver even then. So the solution is that you do the early circularization burns just to raise the perigee up, and then when you are circularized - or close to it - you have an orbit where your high point does cross the equator, thereby giving you the cheapest way to change the inclination.
Ya You're going to slaughter the explanation of and orbit then jump straight to equations? Damn I gotta so here, newton's great understanding about gravity and the moons orbit was in separating the travel path into two linear directions, down and to the side, in moving the moon along these seperatly he realized that if you put a point on the earth below the moon and move it down, then move it to the side, the point curves back away from the moon and the moons height from the point goes back to where it was before you moved it down Also I messed up the point doesn't go directly below the moon it stays as far back as it can from the moon before the earth's surface comes in between the moon and the point
@@EagerSpace ive commented before how underrated your channel is but this thread here is just the icing on the cake so cool of you take the time to clear up some confusion thank you dude
Because the speed you have to go sideways to be above the same point on the earth is higher up high than on Earth's surface, because the rotation speed is the same but the radius is larger. If you go straight up eventually you are going to fall behind Earth's rotation
You can reach the geostationary altitude by going straight up, but at that point you aren't in orbit and will just fall back to the earth. To get to orbit you need to both get to geostationary altitude and get enough sideways velocity to stay there. You can do that all at once - that's the second option I described - but you need a big rocket to do so, something like the Falcon Heavy. It's less efficient for two reasons. The first is that if you take that approach you need to put both the mass of the satellite and the mass of the rocket second stage into that orbit, and that roughly doubles the amount of mass. It's comparing what you get with a two-stage rocket to a three-stage rocket, where the satellite is the third stage. The second is because the electric ion thrusters are much more fuel efficient - they have a higher specific impulse - than the launch vehicles do, so your satellite can either be a lot lighter or it can carry more fuel for a longer lifetime. This is especially good for communications satellites because they already have to have large solar panels to power their electronics, so they can power the electric thrusters. Pretty much the only group still going straight to geosynchronous orbits is the military. I think it's mostly inertia and the fact that they are much less price sensitive.
@@chyza2012 well, definitely see your point, but maybe an escape trajectory can be considered an orbit? In that case you could reach an "orbit" that way 🤔. Will not do any revolutions but definitely not coming back 😅. Maybe the confusion stems from thinking a geostationary satellite has almost escaped Earth's gravity
