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Alright, I'll try to ask a better question on what I mean so that we figure out how one becomes weightless in a craft without having to fly steep parabolae. Other than flying parabolae or nose-down towards the Earth, you don't have to achieve first cosmic speed (orbital velocity) in order to become weightless either. The X-15, the SpaceShipOne and the SpaceShipTwo all made suborbital flights at considerably lower speeds (and lower altitudes than where more than one stable orbit would be possible). The pilots of these spaceplanes are reported to have become weightless in flights in which they reached space (both according to the U.S. definition only (50 mi, 80.47 km) or the FAI definition too (62.14 mi, 100 km). Of course these are kind of/sort of parabolae too, but that would mean that pilots of the mentioned craft would also be weightless when they don't reach outer space.

  • On 30 March 1961 Joseph Walker reached 169,600 ft in the X-15 plane (thus he became the first man to reach the mesosphere and such high altitude) achieving a speed of 2760 mph and Mach 3.95.
  • Two weeks earlier, Robert White in the X-15 achieved a speed of 2905 mph and Mach 4.43 and an altitude of 77,450 ft.
  • On 21 April 1961 White achieved a speed of 3,074 mph and Mach 4.62 and an altitude of 105,000 ft.
  • In SpaceShipOne flight 14P Mike Melvill reached an altitude of 211,400 ft (64.3 km) and Mach 2.5.
  • In SpaceShipTwo flight PF03 two SS2 pilots reached an altitude of 170,800 ft (52 km) and Mach 2.47.

So while none of these above flights reached outer space according to any of the two definitions, I wonder whether they were fast and high enough for the pilots to become weightless in their planes. All of the mentioned pilots reached space in later flights and became weightless but I don't know of reports that they would feel weightless in the mentioned flights where they didn't reach space as yet.

From all of this I'm concluding the following: you need a certain speed (that doesn't have to be orbital speed) in order to achieve weightlessness in a craft. In order to achieve it, you should go to a certain altitude where the atmospheric pressure gets low enough. My question is: what speed / altitude / exterior air pressure are necessary in order to get weightless without having to fly a steep parabola and I'd appreciate if you'd tell me who first achieved weightlessness by the mentioned means (perhaps someone of those I mentioned?). Thank you.

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    $\begingroup$ Why is someone proposing to close the question? $\endgroup$ – user35272 Apr 11 at 8:48
  • $\begingroup$ (I will click now "leave open"). I believe the reason of the close vote is that it is more about athmospheric flights than about space exploration. Furthermore, the answer is trivial (about 7.8km/s is the answer, which is practically unreachable in 1atm air). But I believe that it is not enough far to make your question off-topic, and I think this will be the result of the review, too. $\endgroup$ – peterh - Reinstate Monica Apr 11 at 21:11
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    $\begingroup$ First, to be technically rigorous, it is not a "parabolic" flight path that is associated with a "weightless" environment, it is a segment of an elliptical flight path (note that said common "parabolic" reference assumes a "flat" Earth). Second, the "sub-orbital" craft mentioned in the OP (X-15, etc) achieved "weightlessness" by attaining an orbital path - albeit one which featured a perigee below the surface of the Earth (and, therefore, capable of not even one "stable" orbit). Third, it is easy to achieve "weightlessness" with zero horizontal velocity - jumped off of a curb lately ;-)? $\endgroup$ – Digger Apr 12 at 15:56
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    $\begingroup$ @user30007 think of the parabolic flightpath as a circular orbit where the lowest point in the orbit is below the surface of the planet. E.G. you'll still be in an "orbit" but the Earth stops you (violently) from completing that orbit. If you jump from the surface, at the height of your tiny jump, when you are travelling neither up, nor down, you are weightless. The top of your "orbit" is the height of your jump. The bottom of your "orbit" is somewhere near the core of the Earth. On your way back down, instead of plummeting to the core, you contact the ground, which stops your "orbit". $\endgroup$ – Magic Octopus Urn Apr 12 at 19:43
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    $\begingroup$ Also, even if your approach was wrong, don't delete the question if people have taken time to answer it. We've all been wrong before and we're all here to learn :). Maybe make an edit stating what you've learned from all the conversations here! Nobody is mocking you at all. I've rolled back the edit and hope you keep the question, feel free to cancel my rollback. $\endgroup$ – Magic Octopus Urn Apr 12 at 19:46
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How fast does a craft have to fly relative to the Earth's surface in order to be weightless within?

About 15,000 knots.

Astronauts in the ISS feel weightless with respect to their craft because they are "flying" at the orbital velocity for their altitude, which as everyone knows is "exactly 17500 MPH".

STS110-353-012 (8-19 April 2002) --- Astronaut Ellen Ochoa, STS-110 mission specialist, poses by the speed limit signs in the Unity node on the International Space Station (ISS)

STS110-353-012 (8-19 April 2002) --- Astronaut Ellen Ochoa, STS-110 mission specialist, poses by the speed limit signs in the Unity node on the International Space Station (ISS)." from here.

More photos in this answer to When was the ISS's “SPEED LIMIT 17500 MPH” sign originally posted?

Circular orbital speed at a distance $a$ from the Geocenter is approximate $\sqrt{GM/a}$ where the standard gravitational parameter of the Earth is 3.986E+14 m3/s2 and we can take $a$ to be 6378137 meters plus the altitude of the craft. There will be a small, roughly part-per-thousand correction due to Earth's oblateness (J2) but we can ignore that.

So your craft would have to "fly" at 7669 and 7844 m/sec at altitudes of 400 and 100 km (at the ISS's altitude and at the Karman line) and 7887 m/s at 100,000 feet, which range from about 14900 to 15300 knots.

To get that relative to the surface of the Earth is a small correction of less than 10% because the Earth's rotational velocity at the equator is about 900 knots and and direction of the orbit is unspecified. By "relative to the Earth's surface I assume you mean in an Earth centered Earth fixed coordinate system.

So while none of these above flights reached outer space according to any of the two definitions, I wonder whether they were fast and high enough for the pilots to become weightless in their planes.

No. It's not possible to fly those planes at 15,000 knots in air, aerodynamic heating would incinerate them. It might be possible for a rocket plane to achieve this kind of speed in the future (see the several good answers to What would a “Kármán plane” look like, a bird, or a plane?) and we do know real planes can briefly rise to the altitude of the Karman line (How did the X-15 control attitude above the Kármán line?) but currently there are no craft that can reach 15,000 knots or mach 22.5 and still be called airplanes.

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    $\begingroup$ You are speaking of orbital velocity. That's not what I meant but that's my fault, my input was wrong. You say these pilots didn't achieve that speed which is obvious since it weren't orbital spacecraft. But might they have become weightless anyway? $\endgroup$ – user35272 Apr 11 at 14:10
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    $\begingroup$ @user30007 the only way someone will feel weightless inside a craft is if the craft is on an orbit around the Earth. When those zero-gee simulator flights happen, while we call their trajectories parabolic they are actually elliptical and they precisely follow an elliptical orbit around the earth, but one that will quickly intercept the Earth's surface unless they pull up. When we jump off the ground vertically, while airborne we are on a vertical, straight-line orbit with eccentricity=1 as well. Seems counterintuitive but it's true. $\endgroup$ – uhoh Apr 11 at 15:49
  • $\begingroup$ The pilots did fly something like orbits so they should've been weightless in their spaceplanes, right? $\endgroup$ – user35272 Apr 11 at 15:51
  • $\begingroup$ You may enjoy the video discussed in this answer. $\endgroup$ – uhoh Apr 11 at 15:52
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    $\begingroup$ I can find, in about any tall building, a "craft" that is capable of giving me a ride in "weightlessness." Go to the top floor, get in an elevator, and have someone cut the cable (no horizontal velocity required). You'll get a relatively short and thrilling ride in weightlessness - the only problem is the sudden stop at the end... $\endgroup$ – Digger Apr 12 at 15:51
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In order to achieve "weightlessness", you don't need to achieve a certain speed, you need to achieve a certain acceleration. Earth pulls down at approximately 9.8 m/s^2 which means that any object falling gets faster by 9.8 m/s for every second that it falls. For example, a ball that falls from a tower (disregarding air resistance) and takes three seconds to hit the ground, will be moving at about 29.4 m/s when it hits because it had three seconds to accelerate.

Now, we humans are always feeling Earth's gravity pull on us and the acceleration it "wants" to cause. For example, if I'm in an elevator and it accelerates upwards, I feel heavier. If the elevator accelerates upwards at 1 m/s^2 until it reaches its cruising velocity, I feel Earth's gravity (9.8 m/s^2) plus the additional acceleration from the elevator (1.0 m/s^2) so in total I am feeling ~11 m/s^2 or 1.1 "G" of "gravity".

This means, to feel "weightless", one needs to accelerate in the direction of Earth's center at 9.8 m/s^2 to "cancel out" the normal gravitational effect we feel. This is achieved on any parabolic trajectory, so that includes basically every thrown object (disregarding air again). Any object, be it ball, brick, or person follows a parabolic flight path when thrown or dropped in an area with gravity. For example, if I throw a ball, as soon as it leaves my hand, it is no longer prevented from accelerating downwards (due to gravity) and starts being in free fall. It is now "weightless".

The reason that planes simulating artificial gravity go so high is because constantly accelerating downwards requires a lot of vertical space. You can easily calculate, if I am getting 9.8 m/s faster every second that passes, speed quickly begins to increase. Yes, you could build an elevator that causes the occupants to be briefly weightless, but it would be a very short elevator ride in even the tallest building.

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  • $\begingroup$ On parabolic airplane flights one usually isn't longer than 30 sec weightless. The "Vomit Comets" can't fly weightless forever because they'd reach terminal velocity eventually. I said that I didn't mean falling nose-down towards the Earth at 90 degrees. At what altitude is the air pressure low enough that there's no terminal velocity anymore and where you can be weightless for more than 30 sec? If you reach a certain speed there, you should fly fast enough relative to the surface to cancel out the Earth's gravity too. $\endgroup$ – user35272 Apr 11 at 8:17
  • $\begingroup$ A special case of a parabolic flight path is possible in the Bremen drop tower. Using the catapult of the tower, the flight path is a straight vertical line, just up and down again. $\endgroup$ – Uwe Apr 11 at 8:20
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    $\begingroup$ @user30007 horizontal and vertical components of velocity aren't linked, only the vertical component plays a role in becoming weightless. If you're talking about minimum 'ground speed' to achieve weightlessness, that's zero. $\endgroup$ – Dragongeek Apr 11 at 8:29
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    $\begingroup$ @user30007 sorry, but no, you are just completely wrong from a physics point of view. There seems to be a fundamental communication problem somewhere here. $\endgroup$ – Dragongeek Apr 11 at 9:05
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    $\begingroup$ The Vomit Comit can’t fly weightless forever mainly because there’s a planet in the way. $\endgroup$ – Russell Borogove Apr 11 at 12:02
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It's not so much a matter of speed but one of altitude: where the atmospheric pressure is low enough that there's no air drag so one can longer be weightless without any air limitation. Basically the altitude where there's a low enough air drag so your parabola can be of any size and where you don't necessarily have to immediately fall onto the Earth.

This is not correct. Air drag is only one reason why the airplane parabolas have the limits they do. The main reason is that the vertical acceleration in the parabola is nearly fixed (even at very high altitudes, the gravitational acceleration is almost identical to acceleration at the surface).

So to fly a long parabola with that limitation, you only have two choices:

  • extend the parabola very high
  • make the parabola wide enough to not contact the earth (not really a parabola any longer).

The first is not possible with aircraft because they rely on air pressure to work. As you get higher, engines don't produce enough power and wings don't produce enough lift. There's almost no way to fly at 100k feet without rockets. And because of the vertical acceleration, doubling the altitude doesn't double the time.

The second isn't possible without the horizontal speed mentioned in the other answers. It allows the flight path to not intersect the earth's surface.

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My question is: what speed / altitude / exterior air pressure are necessary in order to get weightless without having to fly a steep parabola?

A speed of 2 m/s, an altitude of 1 m, and exterior air pressure of approximately 1 atmosphere will do the trick just fine.

How do you do that? Go find a creek, take a running start, and jump over it. This will cause you to feel weightless, albeit for an extremely brief amount of time.

You might be thinking, "That's not what I meant!" Actually, that is what you meant. The feeling of weightlessness is caused by being in freefall—that is, by not having any significant non-gravitational forces acting on you. There's absolutely no qualitative difference whatsoever between jumping over a creek and being aboard a suborbital spaceflight. In both cases, you feel weightless for exactly the same reason.

Okay, but I'm guessing that what you're really interested in is being weightless for a substantial amount of time, and with a shallow flight path.

Well, adding altitude will increase the duration of the weightlessness, but will cause your flight path to appear steeper. Adding speed will cause the flight path to appear shallower, and, if the speed is a significant fraction of orbital speed, it will also increase the duration of the weightlessness.

So, exactly what speed and altitude do you need in order to be weightless for a substantial amount of time, and for your flight path to be shallow? The answer depends entirely on what you consider to be "a substantial amount of time" and "a shallow flight path". It's impossible to give a more precise answer without having a more precise question.

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  • $\begingroup$ The Earth's gravity is so high that you don't feel quite weightless just by jumping, unless you jump on a strong-enough trampoline. A "shallow flight path" would be, let's say, 30 degrees or less. An amount of time let's say more than at Vomit Comet parabolic flights where weightlessness lasts up to 30 sec. So more than that I'd propose. $\endgroup$ – user35272 Apr 12 at 5:23
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    $\begingroup$ "The Earth's gravity is so high that you don't feel quite weightless just by jumping" – That's not so. If you have significant non-gravitational forces acting on you, then you feel weighty; if you don't, you feel weightless. Ironically, gravity has nothing to do with it. Note that astronauts aboard the ISS experience about 90% of the gravity that people on the surface experience. $\endgroup$ – Tanner Swett Apr 12 at 5:33
  • $\begingroup$ You aren't in the air long enough to fully feel weightless, same on Venus. On Mercury and Mars you might get weightless if you jump from an elevation of about 5 ft (1.5 m). On the Moon you probably indeed feel weightless by jumping from the ground (though if you wear the massive Apollo spacesuits you perhaps aren't in the "air" long enough either). ISS astronauts are on an altitude where there's ~0.9g but they experience microgravity because they're in free fall around the Earth. $\endgroup$ – user35272 Apr 12 at 6:23
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    $\begingroup$ @user30007: So, how long is long enough for you to "fully feel weightless"? The reason I'm asking you is that the answer to that question is subjective — objectively you are weightless as soon as you jump off the ground. The only question is how long it takes for you to feel it. $\endgroup$ – Ilmari Karonen Apr 12 at 6:47
  • $\begingroup$ @IlmariKaronen Well, if I'm jumping on a strong trampoline I really become what I call weightless. But it's not like I'd measure time. What about you, when would you consider starting to fully feel weightless? It's probably about the same, I don't think it's that subjective. $\endgroup$ – user35272 Apr 12 at 6:58
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This is just an addendum to the other answers whose aim is to give the actual formulae you need to work this sort of thing out.

Simplifying assumptions: I'll assume the Earth is spherical, its density depends only on radius and ignore its rotation. These are good first-order assumptions, but probably do not use them if you want to launch spacecraft to the Moon and you expect them to get there.

Working out acceleration

So, let's assume the Earth has radius $R \approx 6371\,\mathrm{km}$, and mass $M\approx 5.97\times 10^{24}\,\mathrm{kg}$. And we'll assume the thing which we want to be weightless has mass $m\ll M$ (so, this will be wrong for the Moon). $G$ is the universal gravitational constant, $G \approx 6.674\times 10^{-11}\,\mathrm{m^2 N kg^{-2}}$

So an object is weightless if it is falling freely in the Earth's gravitational field. So using Newton 2, Newton's law of gravitation and the shell theorem (also due to Newton!), if the object is a height $h$ above the surface, we get

$$ \begin{align} F &= \frac{GMm}{(R + h)^2} &&\quad\text{force on $m$, acting towards centre of Earth}\\ F &= ma &&\quad\text{Newton 2, on $m$}\\ a &= \frac{GM}{(R + h)^2} &&\quad\text{$a$ is directly down} \end{align} $$

So this tells you that in order to feel weightless, at a height $h$ above the surface, then you must accelerate towards the centre of the Earth with $a = GM/(R + h)^2$. We can look at how this varies with height. Here it is for fairly small values of $h$:

LEO

And here for much larger ones

HEO

So you can see that, for values of $h$ which are compatible with either being in the atmosphere of being in LEO, $a \approx 9.8\,\mathrm{ms^{-1}}$.

How you achieve that $a$ is up to you. Jumping in the air is an easy start.

Circular motion

One particularly interesting way of doing it is to try to move, very fast, in a circle around the Earth. This is interesting because it is, to a good approximation, what orbiting spacecraft do.

So, the acceleration of an object moving in with uniform angular velocity $\omega$ at a radius $r$ is $a = \omega^2 r$. Its linear speed is $v = \omega r$ & so $a = v^2/r$. So for this object to be weightless (aka to be in a circular orbit about the Earth) we have

$$ \begin{align} \frac{v^2}{r} &= \frac{GM}{r^2}\\ v^2 &= \frac{GM}{r} = \frac{GM}{R + h}\\ v &= \sqrt{\frac{GM}{R + h}} \end{align} $$

So this is how fast you need to move in a circular orbit at height $h$. Again, we can plot this:

Orbital velocity

Note that, at $h = 200\,\mathrm{km}$, $v \approx 7789\,\mathrm{ms^{-1}} \approx 17,420\,\mathrm{mph}$.

This is how fast you have to go, in a circle around the Earth, to be weightless.


If I have time later I'll add in the more general orbit cases.

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The question is what minimum altitude, speed and exterior air pressure/drag must be reached in order to become weightless without having to push the yoke for lowering a spaceplane's nose.

Reportedly humans start to perceive linear gravity "properly" at 0.007 g. So let's define considerable weightlessness as below 0.007 g (and above minus 0.007 g). 0.007 g is 0.069 m/s².

From all my atmospheric and spaceflights in Orbiter2016 I can conclude that you must get above 200,000 ft (61 km) in order to, when idling the engine, become weightless without having to push the yoke, but by keeping the spaceplane freely floating without controlling it. My spacecraft's accelerometer shows gravity at the x-, y- and z-axis. Above 200,000 ft gravity may fall below 0.069 m/s² (and remain above minus 0.069 m/s²) on all three axes if idling the engine. This does not necessarily occur immediately when you idle the engine but when the spaceplane reaches a certain cancelling-out-speed, de-/acceleration or angle. It also doesn't mean that you have to idle your plane's engine at or above 200,000 ft, you may idle it at lower altitudes and let the plane freely float on, surpassing 200,000 ft.

The speed above that altitude is usually around Mach 5-5.5 (5-5.5x the speed of sound) which at that altitude is a ground speed of around 1-1.1 mi/s or around 1.5-1.8 km/s. The atmospheric pressure above 200,000 ft falls below 0.003 psi.

Absolute weightlessness (so that all three axes show 0.000 m/s²) may be achieved above 54 mi (86.9 km) and your speed would be around Mach 6.5-7.0.

So as per the very realistic Orbiter2016 these seem to be the minimum requirements for achieving weightlessness without manually lowering a plane's nose. For comparison, the U.S.-defined space border is at 264,000 ft (50 mi, 80.47 km) and the FAI-defined space border at 100 km (330,000 ft, 62.14 mi).

From the five examples provided in the question, Mike Melvill in SS1 flight 14P may have become weightless.

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Just a vertical movement may do, a very special case of a parabolic flight. You only need the proper acceleration to get zero gravity for a short time. A drop tower with a vacuum tube like the Fallturm Bremen will do.

A drop from 110 m height delivers 4.74 seconds of weightlessness. Using a catapult from ground will double the time. For a free fall experiment without atmospheric drag the vacuum tube is evacuated.

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