What does it take for a craft to perform a flight simulating weightlessness without having to fly a steep parabola/ellipse?

Question

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.

Answer 1

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)." 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.

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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 $h$:

And here for much larger ones

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:

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.

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

Answer 6

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.

Answer 7

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.