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Imagining the pilots have fainted in the cockpit and are pointing the aircraft upwards at approximately 90° . Not sure if it would get really cold, or really hot before entering "space". Or if a normal aircraft, like a Boeing777 or Boeing787 could make it past a certain altitude without certain pressurization issues.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – Federico Nov 27 '17 at 8:22
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The weightless experienced in earth orbit is a balance between centrifugal effect of rotation around the earth, and the gravitational force of the earth. Because that balance is seldom perfect, we call it microgravity.

Other ways microgravity can be obtained is by flying "parabolic flight paths" where the aircraft is accelerated towards the earth with similar acceleration as gravity, hence within the reference of the aircraft, the acceleration of the aircraft matches the acceleration of the earth.

Back to the height question, one would have to go well beyond the bulk of the atmosphere, and certainly not in a jet, which requires atmospheric oxygen to help power it. In a very general sense, for low earth orbit (LEO) to be reached, the speeds required are on the order of 17,000 mph.

So since the velocity of the object escaping from earth must reach a point where the object has a centrifugal effect equal to the attraction of the object by the gravitation of earth, higher velocities result in higher orbits. Kepler applied the inverse square relationship of gravity between two bodies, resulting in what now is known as a combination of Newton's Laws and Kepler's laws. These were intended to deal with celestial bodies, and does not have provision for the high energy requirement of fighting a dense atmosphere. Once an object is substantially outside the atmosphere, then the numbers work. The Satellite Tool Kit (commonly called STK) is a useful tool for determining orbits, velocities and interrelationships between objects.

A 787 would not go fast enough to balance the gravitational forces of the earth, and because it could not escape the atmospheric drag of the atmosphere, it would have a very high power requirement to reach high speeds. So there is no point that microgravity would be achieved in sustained flight. Therefore, the conditions in this question cannot be met.

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    $\begingroup$ I'm sorry, but whilst your premise is right, your undestanding of the physics is just wrong. Centrifugal forces is categorically not a real thing and things in orbit are not weightless. In fact, on the ISS, gravity is still about 90% that on earth, and as such, their weight is near identical. What is actually happening in orbit is simply that due to your speed and the curvature of the earth the surface is falling away from you at the same speed as gravity pulls you down. There is an imbalanced resultant force here which is causing your constant accellaration (velocity is a vector quantity) $\endgroup$ – Persistence Nov 24 '17 at 14:54
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    $\begingroup$ @JamesHughes While this answer does have some technicalities ill-defined, your comment is also not trivially true. To quote Wikipedia: "There is also a rival tradition [...] which sees weight as [...] a measure of the magnitude of the reaction force exerted on a body. Typically, in measuring an object's weight, the object is placed on scales at rest with respect to the earth, but the definition can be extended to other states of motion. Thus, in a state of free fall, the weight would be zero. In this second sense of weight, terrestrial objects can be weightless". $\endgroup$ – Sanchises Nov 24 '17 at 15:16
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    $\begingroup$ @JamesHughes this is not Physics SE, and I intentionally answered in the context of Aviation SE, and addressed the response to the poster. Perhaps I used the old school centrifugal force term, but it is still taught today in high school. Perhaps a more correct term would have been centrifugal effect which more accurately describes the inertial property of an object in orbit. $\endgroup$ – mongo Nov 24 '17 at 15:51
  • $\begingroup$ Centrifugal effect is measurable, in that a 200# person on the poles, will weigh 199 # on the equator due to the rotation around the earth with a velocity of about 1000mph (~435m/s). $\endgroup$ – mongo Nov 24 '17 at 16:00
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    $\begingroup$ There is such a thing as "centrifugal force" in non-inertial reference frames in classical mechanics. While centrifugal force is often called a "fictitious" force in that context, the term "centrifugal force" is still used; and for that matter, gravity itself is a "fictitious" force in general relativity: physics.stackexchange.com/questions/33875/… -- Add the obligatory link to xkcd.com/123 -- Having said all that, I think "effect" fits better than "force" did in the first sentence of this answer. $\endgroup$ – David K Nov 24 '17 at 19:08
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Let's assume that the thrust of the Boeing 777 is approximately 20% of the weight, and that with no lift in vertical flight the average drag during the zoom climb is 5% of the weight of the aircraft.

Let's say we start the 90 degree climb at a cruise speed of 250 m/s (900 km/h), and the deceleration due to gravity corrected for thrust and drag will be $ 85 \% \times 9.81 = 8.35$ m/s2.

After $\frac{V}{a} = \frac{250}{0.85 \cdot 9.81} \approx 30$ seconds the aircraft would have climbed approximately 3.75 km (12 300 feet) and lost all its speed. If we would start this zoom climb at 35 000 feet (a common cruise level for the B777), the apogee would be at approximately 47 000 feet (approx. 14.3 km) which is well below 100 km, the height of the arbitrary line between atmosphere and space. Since it is above the maximum certified altitude of 43 100 ft, you may expect some pressurisation problems.

It turns out that the B777 is not suitable for space flight.

After having reached the apogee, the aircraft will start to fall back to earth. If the thrust is cut before the fall starts, then weightlessness will be at the moment the speed is 0 (zero-drag). At that precise moment the acceleration will be only due to gravity since there are no external forces applied to the aircraft; for a moment it will be weightless.

The outside temperature is approximately the same as at 35 000 level and the pull of gravity is not changed significantly either.

A far more efficient way of achieving weightlessness in a B777 is using a parabolic flight path.

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    $\begingroup$ "It turns out that the B777 is not suitable for space flight" What a pity:( $\endgroup$ – el.pescado Nov 24 '17 at 14:38
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Weightlessness doesn't come from height, it comes from lateral speed. ISS is not very far away, it just flies really fast.

The apparently humorous statement from Hithchiker's guide to Galaxy is not actually a joke at all when you replace "flying" with "orbiting": "There is an art to flying, or rather a knack. The knack lies in learning how to throw yourself at the ground and miss. ... Clearly, it is this second part, the missing, that presents the difficulties."

This is pretty much what satellite in orbit does: it flies so fast sideways that before it can hit the ground, it have already missed the Earth.

So, your Boeing 777 doesn't actually have to fly very high to achieve weightlessness. It "merely" has to fly very, very fast. For ground level, that's approximately 7910m/s, or a little over Mach 23. Sadly, even ballistic missiles can barely do half of that.

SpaceShipOne does exactly what you propose: it flies nearly vertically very high (over 100km), but even there it doesn't achieve weightlessness. As high as it seems, still 7850m/s is required to stay there (the difference in orbital speed isn't big, because 100km isn't much when you're adding it to Earth's 6371km radius). SSO's lateral speed is close to zero instead, so when the rocket motor burns out, it falls back to Earth like a rock. The people inside experience weightlessness for a while then, but it's no different than what you experience when jumping down.

XKCD link, as requested

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The biggest flaw in your question is your understanding of weightlessness.

It is often misconstrued as a state when in fact it is actually the lack of a state.

Weightless simply means you can not feel gravity, it does not mean gravity is not there.

When you stand on the ground you feel heavy because gravity is pulling you towards the centre of the earth, but, obviously, the ground is not going to move out of the way, so pushes back with an equal force. The weight you feel is that compression between the two opposing forces and the drag of your bones and organs being held up by your body tissue.

If however you are falling in a vacuum, there is no opposing force, and you feel no weight. Similarly, if you are in an elevator, and the cable breaks, you will fall at the same rate as the elevator and will feel weightless. Or at least, you will for a short while.

If you jump off a cliff, ignoring air resistance, you will be weightless till you hit the ground. If you run really fast before you jump, you will fall and hit the ground further away from the base of the cliff. If the cliff is high enough and you run really, really, fast, you will fall slower than the curvature of the earth, and will never hit it. You will then be in orbit.

None of that has anything to do with reaching a specific height. (Well other than the crashing into the side of Mt Everest part in the last example.)

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First of all, what do we mean by vertical? If we mean 90 degrees straight up staying directly over the runway, you are still rotating with the Earth at a rate of one circle every 24 hours. So the answer to your question in that case would be geosynchronous altitude.

This assumes that your plane is space-capable, has plenty of energetic fuel, and is taking off from the equator. If you are not starting at 0 degrees latitude, celestial mechanics would tend to force you to enter a great circle route, crossing the equator twice with each "orbit". This means part of your fuel would have to be spent maintaining your latitude; more so the higher you go.

If you don't mind "going off the vertical" by the time you get up there, you would again achieve weightlessness by the time your initial rotational speed (from the fact that the ground was moving to begin with) is fast enough to go into orbit.

However, if by vertical you mean exactly straight up while the Earth turns under you (sidereally vertical), or if you take off from either pole, then you never break free of Earth's gravity -- you could be as far away as Arcturus and there would still be an infinitesimal amount of gravity pulling you back to Earth.

But for practical purposes, you are "free" from Earth's gravity if the contribution from Earth's mass is trivial compared to other nearby bodies.

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    $\begingroup$ While this is a correct explanation of orbiting body, it doesn't really apply to the context assumed in the question. You need to take into consideration how a commercial airliner flies, with a maximum altitude of about 11 km, else this doesn't really answer the question. $\endgroup$ – mins Nov 24 '17 at 16:06
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    $\begingroup$ I was assuming the question was about what would happen if the aircraft could fly indefinitely upward -- sort of like how much wood could a woodchuck chuck if a woodchuck could chuck wood. If not, it wouldn't be much of a question. Keep in mind an actual commercial aircraft cannot fly straight up -- with the exception that if you turn towards the zenith, you can proceed so for a short time, losing momentum swiftly. In that context, my answer is the best one of all. $\endgroup$ – Jennifer Nov 25 '17 at 5:34
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    $\begingroup$ Also, the question was at what altitude would you achieve weightlessness -- not about what you would have to do to simulate a zero-g environment. Too much XYing! $\endgroup$ – Jennifer Nov 25 '17 at 5:36
  • $\begingroup$ I stand corrected. Looking again, the question is indeed not related to aviation, except for pressurization. The geosynchronous orbit height is the answer. +1 $\endgroup$ – mins Nov 25 '17 at 9:39
  • $\begingroup$ @mins, geosync is not a special orbit with respect to microgravity effects. Rather it is merely the point that the orbital rotation matches the rotation of the earth. If the earth changed mass, or spun at a different rate, then the altitude of a geosync orbit would change. Jennifer is correct in that the microgravity effect comes into play at the time that the inertial centrifugal effect matches the pull of planetary gravity, because at that point there is equilibrum. That equilibrium will exist for any object, regardless of mass (assuming non-planetary masses which changes things a little). $\endgroup$ – mongo Nov 26 '17 at 0:22
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Weightlessness as observed in the near space (ISS, etc) is not a result of the Earth being far enough, the Earth is still close enough and would pull with the gravity. The spacecraft is in weightlessness because of the path (trajectory) it follows (including speed changes also, not just co-ordinates).

Hence all you need to get the weightlessness is to fly a certain pre-computed path that an artillery shell would follow if launched with the same speed and direction as the aircraft is currently flying. Looks like Airbus 300 can do this. The company site also claims zero gravity. Unfortunately, the aircraft is too low to complete the whole orbit around the Earth because the planet is too big and gets into the way.

enter image description here

As seen from the diagram, the path might be within the capabilities of the aircraft.

(from Wikipedia Commons)

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  • $\begingroup$ the "zero-g" situation nominally exists when there is either an opposing "force" to the "force" of gravity, OR, when there is no force against the "force" of gravity. Shooting parabolic paths, provides for a short term acceleration of the aircraft to match the acceleration of gravity, so the aircraft has no opposition to gravity. Orbit is different, in that there is a centrifugal effect (some call a force) which opposes gravity, causing a local balance of those two effects, and creating a place of zero-g. A plane in a dive will not approach the velocity needed for inertial balance. $\endgroup$ – mongo Nov 26 '17 at 0:29
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Because a plane cannot fly straight up, it will stall and fall until the pilot can resolve the emergency. During that time, all passengers and baggage will appear to be weightless because they are falling at the same speed as the plane.

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  • $\begingroup$ "Because a plane cannot fly straight up" mh, depends which plane you talking about. $\endgroup$ – Federico Nov 27 '17 at 8:45
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As explained in another answers, weightlessness in orbit doesn't appear because of high altitude. But answering the actual question "what if" an aircraft could climb vertically without limits - the answer is - few tens of thousands km or miles. The gravity of Earth extends to infinity and you can never truly escape it, but it decreases quickly beyond certain distance, so the passengers could feel weightless even without orbiting.

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