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If you launch a rocket 101 km high and it doesn't reach orbital velocity then it will return to Earth without completing an orbit. Even though it reached outer space.

But if you launch it 10 million km high then most likely it will not fall back to Earth because the gravity is too low to bring it back.

Then, where is the limit?

If you launch it above N km then it will not fall back for sure. But if you launch it below N km without it reaching orbital velocity, then it comes back. That implies, I think, that it has to reach orbital velocity in order to go above N km.

Glorfindel
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Joe Jobs
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    asteroid crashes to earths all the time. – user3528438 Aug 07 '20 at 19:04
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    If you launch a rocket 101 km high and it reaches orbital velocity then it will return to Earth without completing an orbit too.. – Uwe Aug 07 '20 at 19:57
  • Sorry I though reaching orbital velocity means you will make at least one orbit. What am I missing? – Joe Jobs Aug 07 '20 at 20:32
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    @JoeJobs atmospheric drag. It varies roughly exponentially with altitude, doubling every 5 or 10 km. At 200 km a spacecraft might be able to stay for a few weeks but at 100 km it may not even last one orbit; as soon as it loses a little velocity it drops lower where the density is higher and so it loses velocity even more quickly. – uhoh Aug 08 '20 at 02:48
  • It would have been perfectly possible for the Apollo lunar missions to have been sub-orbital, at least of Earth. (Although they did do several Earth orbits for systems checkouts before trans-lunar injection.) Launch on the correct trajectory, loop around, the moon, and return to Earth. Indeed, SpaceX appears to be planning just such a mission, though details are scant: https://en.wikipedia.org/wiki/DearMoon_project – jamesqf Aug 08 '20 at 04:05
  • @uhoh that means orbital velocity is not constant and it is greater when the altitude gets lower - if there is any atmosphere in that low altitude. – Joe Jobs Aug 08 '20 at 10:38
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    @JoeJobs yes indeed; "loses velocity" are the wrong words there. We can't edit our comments so I can't fix it. The drag force pushes backwards against the direction of motion. The altitude drops but velocity actually increases. It's counterintuitive – uhoh Aug 08 '20 at 10:56
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    @JoeJobs "What am I missing": Orbit is about going fast sideways. "Up" happens automatically. You can go straight up to almost infinity and the cartload will return when the out fuel error happens (assuming no other bodies). To launch to orbit (and not "up") rockets folow a trajectory called "gravity turn". –  Aug 08 '20 at 11:19
  • So if it reached 101 km and the speed is equal to the orbital velocity for that altitude (calculated keeping count of the atmospheric drag) then it will make a complete orbit. With atmosphere, the orbital speed is greater than without. I can't find "loses velocity" in my comments – Joe Jobs Aug 08 '20 at 18:18
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    @JoeJobs FYI uhoh was referring to their own comment regarding "losing velocity" and was restating their wording. – fyrepenguin Aug 08 '20 at 20:06
  • Then whats wrong with it? Atmospheric drag indeed makes the spacecraft lose velocity. I think his/her wording is correct. The required velocity (to stay in orbit) is increasing indeed at lower altitudes but thats something else – Joe Jobs Aug 08 '20 at 20:10
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    I highly recommend this xkcd what-if. It won't answer the question (that's what the answers below are for), but it does help dispel some intuitions about how "high" space is. And, as always Munroe's wordings are so easy to read! – Cort Ammon Aug 09 '20 at 16:58
  • Did this make it to HNQ? Can't understand the upvote-quantity. The OP should ask specific questions about how to achieve a ballistic orbit (as opposed to , e.g., calling a world-circling jet plane trip an orbit). – Carl Witthoft Aug 10 '20 at 12:27
  • A jet plane can't reach 101 km. Hardly highest altitude. – Joe Jobs Aug 10 '20 at 12:35
  • "Orbital velocity" needs to be in the right direction. If you're going at orbital velocity parallel to earth's surface (sideways) you're circling around earth and are in orbit. If you're going straight up with a speed equal to "orbital velocity" you'll eventually fall back down and hit the ground. while some answers here are technically correct and explain where the limit is between falling back down and escaping earth, they don't help explain the obvious misconceptions in the question. – Vincent B Aug 10 '20 at 15:26
  • So if I go vertical at 500 km per second for 100 seconds and then I stop the engines then I still fall back to Earth? That's really interesting – Joe Jobs Aug 10 '20 at 15:31
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    In basic concept, yes. But 500 km/s is considerably faster than escape-velocity so you wouldn't come back at all. (this means you wouldn't be in Earth orbit, but rather solar orbit) This speed is also actually pretty close to escaping the solar system. As long as you launch slower than escape-velocity it's possible to be on a sub-orbital flight that returns to Earth, even though escape-velocity is faster than you need to go to achieve orbit. The top answer here explores where this limit is between sub-orbital and escaping Earth. – Vincent B Aug 11 '20 at 08:18

6 Answers6

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Hill sphere is the radius at which gravity of the body (a planet) becomes too weak to pull an object back in, and the object enters a solar orbit. In case of Earth, that is 1.5 million km. In reality this will be a bit less, because at the very edge of the Hill sphere the body moves quite slowly, is susceptible to other influences (like Moon gravity) and it takes very little disturbance to make it enter Earth orbit or escape into Sun orbit.

The thing with orbital velocity is it's a component directed parallel to Earth surface (at given location) - how much of velocity in the direction 'away' you have doesn't matter as long as it won't bring you out of the Hill sphere. So, unlike most launches that shortly after climbing some distance start increasing the horizontal velocity, this one would go straight up, with possibly some lateral component west to cancel out the initial velocity coming from Earth spin.

SF.
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    Thought experiment: if I launched something straight up, and several years later it collided with Earth again, was that a suborbital flight? After all, it never entered Earth orbit, just went from a heliocentric orbit matching Earth's to one different, and eventually back again.... – Anton Hengst Aug 08 '20 at 00:20
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    @AntonHengst Well, it was orbital, just not around the earth, but the sun. – Polygnome Aug 08 '20 at 06:54
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    @Polygnome what if I slowly drift n% of the way to L4, thrust a bit and slowly drift back to Earth? At what point does it begin to count as solar orbit? – John Dvorak Aug 08 '20 at 09:44
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    @JohnDvorak: Are you a lawyer or a politician? – SF. Aug 08 '20 at 09:49
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    @SF. I'm a mathematician (at heart; not actually paid for it) – John Dvorak Aug 08 '20 at 09:50
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    @JohnDvorak I'd put the limit at where patched conics would switch to Solar orbit, and consider venturing to L4 and back just going into solar orbit unstable due to Earth's perturbance, but I guess it would take a professional theoretical physicist to firmly draw the line, and probably only after being paid to do so by a lawyer or a politician. – SF. Aug 08 '20 at 09:58
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    @JohnDvorak How do you suppose you "drift to L4" without entering a heliocentric orbit? But lets say you do not leave the hill sphere. In that case, you are still on an hyperbolic escape trajectory. By convention, these aren't usually considered suborbital. You then slow down, entering a suborbital trajectory that falls back to earth. Was your whole flight suborbital? I'd say no, since you were on an escape trajectory. But you can certainly argue you were suborbital the whole time. But if thats a constructive discussion while those terms rely on convention anyways is on another page. – Polygnome Aug 08 '20 at 10:00
  • @Polygnome: If we include propulsive maneuvers, then a Martian sample return mission is suborbital. – SF. Aug 08 '20 at 10:04
  • @SF. - I thought there are three main categories: suborbital, orbital and leaving orbit. A Mars mission implies leaving orbit (escape velocity) – Joe Jobs Aug 08 '20 at 10:27
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    @SF. Its kinda hart to return samples from mars to Earth without escaping Mars, entering a heliocentric orbit and then entering the earth system. I'm not sure "suborbital" is a sensible description for that kind of trajectory, its commonly understood quite differently. But again, since its not dependent on a rigorous definition, you can certainly make up meanings that can't be proven wrong, but aren't well suited for communicating with others. – Polygnome Aug 08 '20 at 10:41
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    @Polygnome: It starts suborbital, ends suborbital, and has "momentary" other segments achieved through propulsive maneuvers. Of course it's absurd, my point was if you have segments of flight that are not suborbital, especially achieved through propulsive maneuvers, then the flight is not suborbital. – SF. Aug 08 '20 at 10:50
  • @JoeJobs That's my point too. No reasonable person would consider this a suborbital flight - but which could be the conclusion if going by the rules where it's still a suborbital flight where propulsive maneuvers lead to escape trajectory "for a time" but then another maneuver brings it back to suborbital. – SF. Aug 08 '20 at 10:55
  • But here, sitting in my chair, I'm very suborbital but I'm still on in a heliospheric orbit (with some silly non-inertial stuff happening) – Anton Hengst Aug 08 '20 at 18:45
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    @AntonHengst: You are landed on Earth. That's a fourth state to complete Joe Jobs' three main categories, and if you absolutely demand to categorize it as a form of orbital motion, you're a geostationary statite of Earth, station-keeping in the unstable orbit through continuous propulsive maneuver of lithobraking. Also, with your butt providing you with a constant 1g acceleration pushing against the chair and Earth gravity, you're a torch ship. – SF. Aug 08 '20 at 20:20
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    Accurate assessment. I am a geostationary statite of Earth. – Anton Hengst Aug 08 '20 at 20:48
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    Expanding upon Anton Hengst's thought experiment: The rocket burns 18 km/sec (neglecting drag and gravity loss during the burn) directly outward from the sun. This will take it out beyond the planets but it will eventually fall back. Time it perfectly and it hits Earth on the return. Since it carries Earth's orbital velocity you could say it's in solar orbit (it would go round and round if it didn't hit Earth) but in this scenario it doesn't complete a whole orbit. Suborbital?? – Loren Pechtel Aug 08 '20 at 23:57
  • And an even more extreme case: In addition to the outward velocity it burns to kill Earth's orbital velocity. It's now suborbital with regard to the sun. Is this a suborbital flight? – Loren Pechtel Aug 08 '20 at 23:59
  • @LorenPechtel It isn't, because at one point you were in a stable heliocentric orbit. Now, if you launched from a comet hurtling into the Sun, and ended up hurtling into the Sun, you would be suborbital. Your first example is quite interesting though. – John Dvorak Aug 09 '20 at 10:52
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    ...and this is the kind of trouble you keep getting by overthinking it and trying to reclassify odd edge cases of otherwise clear situations. Just use the patched conics approach; if at any point your conic indicates orbit or escape from the current SOI, it's no longer suborbital. – SF. Aug 09 '20 at 11:33
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Impossible to say without heavy modelling.

If the Earth was a solitary, stationary object, then the highest possible altitude of a sub-orbital flight would be nigh-infinite, since it would always be possible to construct an elliptical orbit which has its perihelion underneath the Earth's surface; the higher the aphelion of the orbit is, the more eccentric the orbit would need to be.

However, the Earth is not a solitary, stationary object. It orbits the sun, and is orbited by the moon, and that introduces an entirely new level of complexity to the problem. This additional complexity is known as the n-body problem; for any n beyond 3, it is almost impossible to predict the orbits of objects without computational modelling since they are chaotic systems and even a small change in initial conditions can lead to significant differences in the orbits of those objects.

nick012000
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Are you asking theoretically, or in the real world?

Theoretically, there is no upper limit, because gravity extends forever. Gravity does get weaker the farther you travel, but it never reaches zero. "Escaping" Earth, therefore, is based on speed, not distance. If you keep the speed below escape velocity (about 11 km/s at the surface, discounting air resistance) then your probe will inevitably succumb to gravity at some finite time in the future, after having reached a finite altitude. The closer to escape velocity you get, the longer it will take, and the higher altitude the probe will reach. There is no limit to this, however: the peak can be literally any (positive) altitude at all.

In real life, of course, things are a lot more messy. Even if your probe was below escape velocity, a gravity assist from, say, the moon could accelerate it above the line. Even the gravity from other planets and the sun can affect your probe's trajectory, either accelerating it, slowing it down, or turning it off course. SF's answer mentioned the Hill sphere of Earth (about 1.5 million km), which I think is about the closest you're going to get to an actual answer in the real world.

HiddenWindshield
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  • The gravity assist from the Moon will not increase the speed to the escape velocity? – Joe Jobs Aug 08 '20 at 10:41
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    @ Joe Jobs that depends entirely on where the moon is at the time of the flight. You could design your straight up launch to put your vehicle with zero velocity just below the moon's orbit, and if you timed / positioned it just right (in front of the approaching moon) the moon would slingshot the vehicle to above earth escape velocity. If the moon is far away though, the vehicle will just fall back to the earth. Same if you launch straight up from the north pole - the moon is then guaranteed to have no appreciable effect at all. – Level River St Aug 08 '20 at 13:43
  • @HiddenWindshield - to answer the question, I was asking about the real world situation, where getting far enough exposes the rocket to the gravity forces of other bodies too. So much exposed that it won't fall back to Earth – Joe Jobs Aug 09 '20 at 17:13
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If you disregard the gravity of other objects, then there is no upper limit.

If you toss a ball up with one hand so that it arches over and you catch it with your other, you can think of it as being in an elliptical orbit that happens to intersect the earth's surface. If all the mass of the earth were compressed into a point at the center, and if we could ignore relativistic effects, then the ball would actually complete a very long, skinny elliptical orbit. But, because in real life this orbit intersects the earth, it's not much of an orbit, so we call it a suborbital path.

So, that's a very low suborbital path. But earth's gravity, in the simple Newtonian approximation, goes to infinity. And it sounds to me like you are interested in the really simplified case where we ignore the rest of the solar system. In that case, if you start out below escape velocity, you will come back. The closer to escape velocity you start out at, the farther you go before you start to fall back. And there's no limit to how far that could be. If you leave right at escape velocity, you keep slowing down forever but never reach a stop and start falling back.

If you want to go into a full orbit that doesn't intersect the surface, you have to start by launching up and then turn and accelerate sideways. If you have no acceleration after you leave the surface, then you are already on an orbit that intersects the surface, so you can't help but hit it on the way back too.

Mark Foskey
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This whole question is about language lawyering, so you won't get definite answer.

All replies here seem to draw the border where "it's no longer Earth's orbit", so you get mentions of Hill's sphere and Moon influence and stuff.

I propose different view: sub-orbital flight should not be orbital.
In my opinion, the border between these 2 terms would be energy - if you got enough energy to do an orbit, it's only your problem that you wasted it flying up.

In other words, the question turns into "If rocket was moving at orbital speed but up, how high would it go?"

Difference in potential energy between 2 different distances $r_1$ and $r_2$ (measured from center of planet) is: $$\Delta U_{r_1\rightarrow r_2}=GMm\times\left(\frac{1}{r_1}-\frac{1}{r_2}\right)\\$$ And kinetic energy of orbital speed at $r_1$ is:
$$K_{r_1}=\frac{m}{2}\times v_{orb}^2 = \frac{m}{2}\times\frac{GM}{r_1}=GMm\times\frac{1}{2\times r_1}\\$$ Then we equate them:
$$GMm\times \left(\frac{1}{r_1}-\frac{1}{r_2}\right)=GMm\times \frac{1}{2\times r_1}\\[0.4in] \frac{1}{r_1}-\frac{1}{r_2}=\frac{1}{2\times r_1}\\[0.4in] \frac{1}{r_2}=\frac{1}{r_1}-\frac{1}{2\times r_1}\\[0.4in] \frac{1}{r_2}=\frac{1}{2\times r_1}\\[0.4in] r_2 = 2\times r_1\\$$ As you can see, if you use all your orbital energy on going up, you'll climb your initial distance from the body center.

So highest altitude for "sub-orbital flight" in this definition is about radius of Earth, i.e. ~6400km.

For comparison: geostationary orbit is ~36000km, the Moon is at ~30 Earth radii.

No Nonsense
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Noone AtAll
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Partial answer:

According to this answer to Is 678 km the new altitude record for a rocket shot “straight up” (vertical launch)? by a user who is likely to be Jonathan McDowell and @planet4589:

For a suborbital direct ascent trajectory, some early lunar probes (USSR's Luna-1 for example) would hold this record. Otherwise, early vertical research probes included the Blue Scout Junior, one of which reached 44,400 km on 1961 Dec 4 (mission O-2) - another may have reached 225,000 km on 1961 Aug 17 (mission O-1) although it wasn't tracked, so we're not sure if it really made it. A Chinese suborbital probe reached over 10,000 km and possibly 30,000 km on 2013 May 13.

I've added commas to the large numbers to make them more easily read.

Also this answer to How many hours long is Earth's longest possible sub-orbital flight? proposes using Earth's sphere of influence which it puts at about 0.92 million km from Earth, a somewhat more conservative estimate than the Hill sphere out at 1.5 million km.

Fred
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uhoh
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