The higher altitude you wanna reach the more delta-v you need. E.g. the SpaceShipTwo never went higher than 90 km (56 mi), so it could fly well below even 1/10th of orbital speed. However, at some certain targeted apogee, a spacecraft will need to reach at least 7.7 km/s (4.8 mi/s) around engine cut-off to reach it. And around the engine cut-off altitude the orbital velocity will probably be considerabely lower anyway. I wonder what's the minimum apogee in order for a spacecraft to have to go at orbital speed or faster, even if your trajectory is suborbital in that the perigee is within Earth.
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4$\begingroup$ If you fly straight up with a long burn with low thrust you could get to a very high altitude with zero final speed. I don't think this question is really answerable in the general case, you can only answer it based on a specific vehicles performance data $\endgroup$– Alan BirtlesCommented Aug 12 at 18:08
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$\begingroup$ I can't tell if this is about SpaceShipTwo or something else. what do you mean, exactly? $\endgroup$– Austin OvertonCommented Aug 12 at 21:00
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$\begingroup$ @AustinOverton It's about any spacecraft (or anything that has been propelled into a suborbital trajectory) that reaches orbital velocity at engine cut-off altitude. So the question is, what targeted altitude (i.e. apogee) requires orbital velocity even if your trajectory intersects the Earth, that is, you'd fall back onto Earth eventually. $\endgroup$– Star CaptainCommented Aug 13 at 5:37
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tl;dr: You have to do TWO things to get into an actual safe orbit that will go around the Earth without hitting it.
The expression "orbital velocity" is the minimum speed you would need to make a complete orbit around the Earth without burning up. At that minimum velocity, the only shape that the orbit can be is very nearly a circle (very slightly elliptical if you're following the Karman line in a polar orbit instead of an equatorial orbit.
To get into this minimum, very temporarily safe orbit, you need to do two things at the same time:
- achieve a speed of ~7.7 km/s at about 300 km or ~7.84 km/s at 100 km (if you only want to go around roughly once before reentry and death)
- make sure the direction of your motion at that speed is tangential and not radial.
If you pass 300 km above Earth at a speed of 7.7 km/s but you are moving in a direction that deviates up by more than a few degrees from tangential, you will go higher for a while, then you'll turn around and die.
If it deviates lower than tangential, you'll "die harder" (and quicker).
If you have "orbital velocity" at the desired altitude, but you find yourself moving straight up, then you are in a [hyperskinny elliptical orbit with eccentricity ~ 1 who's apogee will go much higher, but alas, whose perigee is near the center of the Earth.
The escape velocity at 300 km is 10.9 km/s. If you are going straight up at 7.7 km/s instead of sideways for an orbit, you'll peak out at about 7100 km apogee (above Earth's surface), then fall straight back.
Terminology:
Once all of those "sub-orbital" tourist (and sounding rocket) flights are truly in Earth orbit once they're outside most of the atmosphere and traveling ballistically. It's just that those orbits intersect the earth. So we call them "suborbital".
Basically "orbit" can be any ballistic trajectory outside of substantial atmospheric drag, or it can mean "revolve completely and safely around the Earth at least once".
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$\begingroup$ My question is about what you describe in your latter paragraphs: not reaching an orbit around the Earth, but staying at a trajectory that intersects the Earth, yet having to go at orbital velocity in order to reach a certain apogee. So you say, if I wanted to reach 7100 km (4400 mi) at eccentricity 1, I'll be at 7.7 km/s at the engine cut-off altitude? $\endgroup$ Commented Aug 13 at 5:44
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$\begingroup$ @StarCaptain if I understand it correctly, you are asking what your apogee would be if you flew straight upwards at an orbital velocity? $\endgroup$ Commented Aug 13 at 7:43
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$\begingroup$ @StarCaptain Why don't you consider revising the title of your question and the text in the main body as well. I think that you are asking something like "If a spacecraft shuts its engines of when it attains orbital velocity at say 400 km, but it's going straight up instead of sideways for orbit, how high would it go? What would its apogee be?" I'm pretty sure the answer is easy: 2 x (6378 + 400) km, but I'm not sure if that's your question yet. $\endgroup$– uhohCommented Aug 13 at 7:57
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1$\begingroup$ @uhoh No, but you answered my question anyway. The "altitude" in my question is only the apogee I mean. My question is "if you shut down the engines when going straight up, what's the minimum targeted apogee for you to be orbital velocity during the cut-off?". You answered it: 7100 km. $\endgroup$ Commented Aug 13 at 8:20
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1$\begingroup$ uhoh - in Jonathan McDowell's 2018 paper on the Karman line he says "Empty rocket stages are frequently left in low orbits of under 200 km and reenter after several days; the final tracked orbit is often between 130 and 140 km". Presumably many of those were in circular orbits. But stages are different than satellites, it would be interesting to find out what the lowest recorded single orbit of a satellite was. $\endgroup$ Commented Aug 14 at 0:18