What is the highest possible altitude of a sub-orbital flight?

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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.

Answer 6

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.