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To experience "weightlessness" without actually traveling into space, and orbiting the earth, a parabolic flight is used. See the flight path of Mercury, as shown in this link: https://en.wikipedia.org/wiki/Project_Mercury#/media/File:Mr3-flight-timeline-simple-2.png

While I can understand feeling weightless in the downward path (from 5.14 to 7.48 min.) - where the force of gravity is acting on both the objects (the capsule and the astronauts) equally and simultaneously, I am unable to comprehend the weightlessness when the spacecraft is ascending (during 2.37 to 5.14) because that is the period when the capsule is being pushed upwards "against gravity" by the propulsion. I therefore feel that the astronauts would be experiencing acceleration against gravity, and hence not feel being weightless. However the flight path indicates this period i.e. some portion of the ascent also to be "weightless". I am confused because of this. Please help me overcome my confusion.

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  • $\begingroup$ Throw your ring of keys in the air. They will jingle while they are experiencing weightlessness. You'll note they are similarly weightless and jingling on the way up and on the way down. $\endgroup$
    – Wyck
    Jul 29, 2022 at 19:05

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When the spacecraft is "ascending" ... the capsule is being pushed upwards "against gravity" by the propulsion.

This is a critical misconception -- there is no propulsion after 2:37, it all occurs before that point.

I recommend going outside and throwing a ball high up into the air, far higher than you can reach.

The moment after the ball leaves your hand (analogous to the moment at 2:37), no upwards force is being applied to it, but it has a great amount of upwards speed. The difference now is that the ball is free to accelerate downwards at 10m/s due to gravity, unlike when it was being supported by the ground or your hand.

We often separate acceleration and deceleration into two different named things -- after all, if you keep holding the brake in a car after the car is stopped, it doesn't start moving backwards. But decleration isn't a different thing to acceleration: a deceleration of a spacecraft moving upwards is the spacecraft accelerating downwards. And it's been accelerating downwards, reducing its upwards speed, ever since the rockets stopped firing.

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    $\begingroup$ There is still propulsion in the parabola, else the passengers would actually feel “negative” gravity because of air drag. The pilots need to exactly cancel drag with thrust, as well as steering along the zero-G path. $\endgroup$
    – leftaroundabout
    Jul 25, 2022 at 20:22
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    $\begingroup$ @leftaroundabout you are referring to a parabolic flight in an airplane. But the link was to the Project Mercury suborbital launch profile, for the Mercury Redstone rocket. At 2:37 into the flight, it would be so high up that you wouldn't really feel much air drag anymore. $\endgroup$
    – Jordi Vermeulen
    Jul 27, 2022 at 10:32
  • $\begingroup$ @david Mckee: you said - This is a critical misconception -- there is no propulsion after 2:37, it all occurs before that point... Which is exactly what I also thought. The test is very logical. Thanks. $\endgroup$
    – Niranjan
    Aug 13, 2022 at 4:08
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I feel like the most intuitive way of explaining this is by realizing that the aircraft throws you up into the air and then actively follows your path.

You experience the throwing part as the overacceleration, e.g. 2g for several seconds in the case of Novespace's Air ZERO-G. Feels like your intestines are pulled to the ground. Then the airplane stops throwing you and just follows your parabolic flight, actively cancelling out air resistance (the transition from 2g to 0g is really weird). In that phase, if you were really sensitive and the g-jitter weren't that bad anyways, you would feel fictious forces due to the airplanes rotation. Then the airplane pulls up and you are pressed into the ground again.

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    $\begingroup$ They actually have a little rubber duck in the cockpit which they use to make sure that the airplane gets "thrown" the same way as the passengers. Otherwise, if the airplane reacts only a little different than the passengers do, the passengers would get slammed into the front or the rear. $\endgroup$
    – Jörg W Mittag
    Jul 25, 2022 at 20:03
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It’s all about acceleration.

It does not matter what speed you are going or in what direction, throw that out of the window. All that matters is your acceleration relative to the “container” that you are in.

If you drop a soda bottle you may notice that bubbles in it don’t rise while it is falling¹, that’s because they are accelerating at the exact same speed as the soda, and container and so they experience no gravity².

If you throw the bottle into the air, despite it moving upwards, it is still accelerating downwards, and so is the soda and bubbles. So it still experiences no gravity despite traveling upwards.

This is what happens in these flights, they stop applying a force to the container, and so gravity takes over, essentially making them in free fall.

¹ bubbles may still have motion, either from leftover inertia, drag on the bottle, or other forces that may cause small deviations.

² objects in a container in free fall are still under the influence of gravity, however they can’t actually tell that they are because everything around them is also falling, they have no non falling reference point.

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  • $\begingroup$ Note that the bubbles are less dense than the surrounding soda, so if anything, you'd expect them to move in the opposite direction to the acceleration. This is why if the bottle is just sitting on a table not moving, the bubbles rise instead of fall. Better comparison might be loose tea leaves in tea, or the little chopped veggies in vinaigrette dressing, etc., which are more dense than the surrounding liquid, hence they settle on the bottom when at rest. $\endgroup$
    – Darrel Hoffman
    Jul 25, 2022 at 19:03
  • $\begingroup$ @DarrelHoffman the reason I didn’t use that as an example is that once they are on the bottom they kinda just stay there and you can’t notice anything changing. In soda however, more bubbles are fairly consistently produced and moving upward, so it’s much easier to notice that something changed, although yes you could use that as an example and it would work. $\endgroup$
    – Topcode
    Jul 25, 2022 at 22:47
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Here's a demonstration I like: Take a clear plastic bottle and put some small items in it like pebbles or paperclips or maybe even little toy astronauts. Then put the lid back on and toss it in the air. You can see that the objects are basically floating around inside the bottle for the whole trip up and down. This is what is going on in parabolic flight and in orbit as well.

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  • $\begingroup$ Doesn't answer the question as to why it happens. $\endgroup$
    – Organic Marble
    Jul 25, 2022 at 22:38
  • $\begingroup$ Good analogy. This means the flight power is withdrawn at the point when time is 2.37. After that, its the "inertia or momentum" which keeps the flight moving up against gravity, and therefore, both are subjected to gravity. $\endgroup$
    – Niranjan
    Aug 13, 2022 at 4:04
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Good question. I liked the other answers but still wasn't quite getting it.

Then it came to me. The weightlessness can be summed up fairly simply as conservation of momentum.

Let's say the spacecraft's vertical speed is +20,000kph and thrust is cut. The craft will start decelerating at 1g due to gravity and some more due to drag. For an example, let's say the spacecraft's total deceleration with drag is 2g. So its velocity is 20,000kph - 2g*t (t=time, each second the spacecraft will decelerate further.) It is still going up very quickly until gravity and drag take away all that vertical speed.

Conservation of momentum says you too will start decelerating at 1g due to Earth's gravity. But you don't have the extra 1g of drag that the spacecraft does. So your velocity is 20,000kph - 1g*t. The spacecraft is slowing down a lot faster than you are, so unless you're wearing restraints you're going to slam right into the wall - and both you and the spacecraft are still going up very fast!

Since atmospheric drag is negligible way up high, then you and the spacecraft are decelerating at the same rate (-1g) and thus, in earth's gravity well of 1g, it feels like 0g (of acceleration) even though you still have forward (upward) velocity.

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    $\begingroup$ "20,000kph - 1g" is not a velocity. It's not even a speed. $\endgroup$
    – Organic Marble
    Jul 25, 2022 at 19:59
  • $\begingroup$ @OrganicMarble, agreed. Given the tenor of the question and my quest to keep things simple, I just didn't want to add the 't' to it and make it too 'mathy' ;), i.e. 20,000kph - 2g(t). I figured those who know, get it, and those who didn't know would get confused. But given your comment, I suppose I failed on both counts. I'll make an edit. $\endgroup$
    – Tracy Cramer
    Jul 25, 2022 at 21:48
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    $\begingroup$ What does "2g(t)" mean? Also note: The question wasn't about a spacecraft operating in the realm where "drag is negligible way up high." It's a question about an airplane. Like this one: youtube.com/watch?v=LWGJA9i18Co $\endgroup$
    – Solomon Slow
    Jul 25, 2022 at 23:27
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    $\begingroup$ @SolomonSlow: I think it's an acceleration multiplied by a time, like v = v_start + a * t to give a formula for the instantaneous velocity as a function of time, where a = 2g. Normally we wouldn't write it with parentheses around the time, which makes it look like g() itself is a function of t. The physics is correct, the notation is odd after the last edit (instead of nonsensical), but as you say, the situation being described is very different from parabolic zero-g flights by typical passenger jets at subsonic speeds. $\endgroup$
    – Peter Cordes
    Jul 26, 2022 at 0:26
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    $\begingroup$ @SolomonSlow, I believe the question is referring to the Mercury spaceflight and not an airplane (see the link and Wikipedia article associated with it.) "Spacecraft" and "astronauts" are mentioned in the question. $\endgroup$
    – Tracy Cramer
    Jul 26, 2022 at 19:22
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As the plane is ascending, imagine if it were to suddenly disappear. What would your free fall path look like? You would continue to ascend in the same angle but slowing down until you eventually reach a peak and then start to descend (this is the parabolic path). Now imagine hurtling in this same path, but you are surrounded by a large container which is also following that exact same free-fall path. While both you and the container are in free-fall, what you would experience inside the container is weightlessness relative to the container.

When the ascending plane follows the exact same path you would experience if the plane ceased to exist, then you will experience "weightlessness".

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It may be easier to intuitively understand weightlessness in “Vomit Comet” flights if you consider yourself in elliptical Earth orbit instead of on a parabolic flight through the Earth’s gravity field.

The weightless portion of the flight path is not really a parabola. If the flight path is mathematically modeled over an infinite Flat Earth with a uniform gravitational field then, yes, the path would be parabolic.

In reality, the flight path is the apogee of a highly eccentric elliptic orbit whose perigee is very close to the Earth’s center of mass. This flight plan includes a self-generated impact crater where it intersects the Earth’s surface. Obviously this impact event would be incompatible with desired mission objectives.

The parabolic flight path modeled by a Flat Earth is close enough to an ellipse for practical purposes and the calculus is easier. But the elliptic model may make orbital free-fall easier to visualize.

Once the ascending plane inserts you into the desired elliptical orbit, you are in free-fall exactly as if you were on the ISS. After insertion, the function of the plane is to shield you from aerodynamic forces for the duration of the elliptic orbit (free-fall) portion of the flight. And to rescue you from drilling an impact crater.

As with the ISS, the plane will rotate around you as the phase angle of the orbit progresses. As with the ISS, you are unlikely to notice the rotation in all the excitement.

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  • $\begingroup$ I feel the answer is excessively technical, and perhaps therefore confusing for me. $\endgroup$
    – Niranjan
    Aug 13, 2022 at 4:00

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