Would weightlessness (i.e. in thrill rides, planes, skydiving, etc.) be different on a Flat Earth?

Question

I was listening to a Scientific American podcast, dated March 27th, 2020, about what Flat Earthers believe and why they believe it. As I listened, Michael Marshall mentioned that FEers believe that the Earth is a plane which is accelerating upward at 9.8 m/s² (gravity 9.8 m/s² downward).

Scientifically, there is plenty of evidence (objects on the horizon or Eratosthenes' measurements) to disprove FEs' opinions, but if we are accelerating up, then wouldn't the experiences in a thrill ride be inverted or absent?

G-force is a mass specific force (force per unit mass). If you go on a thrill ride like the tower of terror and you drop, you experience a moment of weightlessness. If the Earth is accelerating upward, I shouldn't experience weightlessness. And the same for a tower ride which accelerates upward (you experience $g$-forces pulling you into your seat), but this would give you weightlessness (if 9.8 m/s² downward is true).

I am assuming that if the Earth were moving up according to the the FEs' opinion, then I would expect a sudden decrease in the upward force, because I am now accelerating at a different rate than the Earth. Wouldn't I feel heavier?

And I would think that all experiences in roller coasters, rockets, jet fighters, sky diving or even being in the "Vomit Comet" would be different.

Note - I've never been skydiving, but people tell me that you have the sensation of falling. I'm assuming that if the FEs' opinions were correct and the Earth accelerating upwards at 9.8 m/s², then I would be stationary while the Earth comes towards me. And with that, wouldn't parachutes not work? (edit) - Wouldn't skydivers need a jetpack rather than a parachute because they would need to match the upward acceleration of the Earth to land safely?

Answer 1

If we assume a flat disk sans gravity accelerating upwards at one $g$ (which is what the FEers propose), then the physics of thrill rides and parachutes and whatnot would actually be the same. If the disk is accelerating upwards, we should assume that its atmosphere is doing so as well, which means that even if you're stationary in some reference frame you still feel the sensation of air running past you faster and faster as you "fall".

The key aspect here is that this acceleration is, by definition in relativity, indistinguishable from gravitation. This is the result of the equivalence principle, which essentially equates inertial and gravitational mass. One of its consequences is that, in a box with no other information, you can't tell if you're on a rocket (or disk) accelerating at $9.81\,\mathrm{m}\,\mathrm{s}^{-2}$ or on Earth. In the case of a flat accelerating Earth, in the Earth's comoving reference frame (which is the one you're gonna want to work in) all objects appear to experience "gravitation" with "downward" acceleration at the exact same rate that in any inertial reference frame the disk is accelerating.

In reality you always experience weightlessness if you're not in direct contact with something that's putting a reaction force on your feet/back/etc. Falling from the sky, the only reason you feel anything but weightless is because of the air resisting your fall. If you were to go on a ride and briefly feel weightlessness, what that would mean is that the ride isn't accelerating you at all - this means that the Earth would be accelerating towards you at one $g$, which is equivalent to you accelerating towards the Earth at one $g$. In other words, weightlessness and related phenomena would still be the same.

Answer 2

Suppose for a moment that you are inside a closed, airtight box, with no view ports. According to the equivalence principle, there is no experiment that you could do to decide whether your box was resting on the surface of the Earth, or sitting on a platform that was accelerating upward at a constant $9.8\,\text{m}\,\text{s}^{-2}$. It would feel the same to you either way.

As for parachutes and airplanes and their like, So long as there were walls all around the edge of the disk of the flat Earth to contain the air (those walls would need to be at least $100\,\text{km}$ tall) flying and parachuting on the Flat Earth would feel no different from flying and parachuting on the actual, spherical Earth. The disk of the Earth would push the air upward with constant $9.8\,\text{m}\,\text{s}^{-2}$ just the same as it would push upward on the soles of your feet.

Now, Buckle up! 'cause here's where it gets weird.

According to the theory of relativity, the surface of the Earth actually is accelerating outward from the center in all directions at $9.8\,\text{m}\,\text{s}^{-2}$. That is to say, if you try to define an inertial coordinate system near the surface of the Earth, your coordinate system cannot be fixed to the Earth. An object in "free fall" would be stationary with respect to the "grid lines" of an inertial coordinate system, but we know that the same object, freely falling near the surface of the Earth, must be accelerating toward the surface at $9.8\,\text{m}\,\text{s}^{-2}$. The explanation in general relativity is that the "grid lines" themselves are continually being sucked inward, and relative to those grid lines, the surface of the Earth must be seen as accelerating outward at $9.8\,\text{m}\,\text{s}^{-2}$.

https://youtu.be/wrwgIjBUYVc

Answer 3

Would weightlessness (i.e. in thrill rides, planes, skydiving, etc.) be different on a Flat Earth? no

You might have heard "We can't actually feel speed. We can only feel acceleration and deceleration.", but while that is useful in the context of fx driving safety, it is not 100% correct/accurate. We can't strictly speaking feel acceleration/deceleration either. What we can feel is an external force acting/pressing upon us and causing acceleration/deceleration, but we cannot actually feel acceleration/deceleration by a uniform field (earths gravity field is practically uniform at the scale of a human).

Standing on earth with normal gravity (9.8 m/s²), you feel the normal force keeping you from accelerating further towards the center of the earth.

While standing on flat earth (without a gravity field but instead accelerating upwards at 9.8 m/s²) you would feel the force of earth pushing up towards you while your inertia is resisting the acceleration.

In both cases the force is the same, and the fluid in your inner ear will pool downwards giving you a clear sense of the direction down. It will feel identical.

Similarly in purely free fall without any reference you have no way to identify if you are in a gravity field or not, the fluid in your inner ear does not pool in any particular direction either way. So being in a gravity field accelerating you towards earth or being at constant speed with flat earth accelerating towards you will feel the exact same, weightless (i.e. the feeling of falling).

With skydiving, in addition to seeing the earth getting closer you also have the air rushing by you to add to your sense of direction, but whether you are accelerated through the air or the air is accelerated past you makes no difference to what you feel. In either case you will feel air resistance, so the fluid in your inner ear will pool and you will feel direction, but again it makes no difference if the fluid is pulled down or the rest of you is pushed up, it feels the same.

Answer 4

I've never been skydiving, but people tell me that you have the sensation of falling.

The "sensation of falling" that you feel when jumping off a plane while skydiving (or bungee jumping or in an amusement park drop tower ride) is actually the sensation of (full or partial) weightlessness.

Humans have evolved to live on the surface of Earth, where we always feel an acceleration of 9.81 m/s² in some direction, depending on the orientation of our body. That is, always except very briefly while jumping or falling. Our inner ear senses both the direction and the magnitude of this acceleration, and when the magnitude suddenly drops (unexpectedly, or for longer than a typical jump off the ground would last), it triggers a reflexive physiological response similar to a fight-or-flight or startle response that alerts us to the fact that we're (probably) falling from high up and urgently need to either grab something or brace for impact.

It's the exact same sensation that astronauts feel when the rocket engine cuts out and they enter the free-fall part of their flight, or when training in a "vomit comet" aircraft. The only difference is that those are pretty much the only situations where the sensation can last long enough for the brain to start acclimating to it. Otherwise the feeling of weightlessness can only last a few seconds at most: even skydivers accelerate in about 10 seconds or so to close to their terminal velocity, i.e. the speed at which atmospheric drag becomes approximately equal to gravity and stops them from accelerating further. At that point they no longer feel weightless, since their weight is now supported by onrushing air.

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So what if, instead of the Earth's surface, you instead lived in a giant spaceship (big enough for amusement park rides or even skydiving) equipped with magic rocket engines that continuously accelerated the ship at a constant 9.81 m/s²?

In that case, you'd still experience a constant 9.81 m/s² acceleration, just as on Earth, and would thus feel your normal weight. And if that acceleration ever temporarily stopped affecting you (e.g. because the engines turned off, or because you jumped off a platform), you'd still feel the "sensation of falling", just like on Earth.

Answer 5

While the Earth is spherical, its approximation by a flat surface accelerating upwards with magnitude $g$ is very good for all examples that you mention.

Whenever someone is in free fall, the surface accelerates to catch it, but before the impact there is no force acting on the body, so there is a weightless sensation.

All that problems of high school where it is said: consider the gravity acceleration constant equal to $9.8 \frac{m}{s^2}$, use implicitly that approximation.

Answer 6

The short answer is that there is not only no way to tell "gravity" from "acceleration" — it is actually the very same thing: Bent spacetime.

Things that would be different are:

• The effects of the (real) Earth's curvature leading to different velocities of Earth's surface at different latitudes. Especially space flight and wind and sea currents are strongly influenced by those differences which are perceived as the Coriolis force.
• The effects of the Earth's, Moon's and Sun's gravitational fields not being homogeneous leading to tides, different orbital mechanics and clock differences at different altitudes (but that won't convince your paranoid flat earther).

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As an excursion, it is interesting to really fundamentally dissect the semantics of gravitational (or any other) acceleration.

For an observer resting in the approximate inertial system of the Earth's surface, a body falling in a vacuum in Earth's gravitational field is perceived as undergoing acceleration, by the famous $9.81 m/{s^2}$. However, from the body's perspective, that body is not accelerated at all; it is following a spacetime geodesic, entirely imperturbed by all external influences. It is, if you want, in its natural state, as long as it falls freely. You, the observer, by contrast, are kept from following your geodesic by electromagnetic interaction with the Earth's surface, you are suspended in a gravitational field: It is you who is accelerated out of its geodesic. All geodesics are created equal; all accelerations out of them by one of the non-gravitational forces are created equal as well. There is no difference.

(This makes clear that the sentence above introducing the observer was utterly wrong: the Earth's surface is emphatically not an inertial system, not even approximately. It is an accelerated system.)

Answer 7

The Earth is locally flat: if I want to travel to a place that is within a few miles of my house, I can use a flat map to navigate there. The round Earth model is needed only when I work at a larger scale. For example, if I go the the seashore and look towards the North Island, Te Ika a Maui, the curvature of the Earth causes the horizon to get in the way.

I don't think there is any simple purely local experiment that you can perform, e.g. going on a thrill ride, which will exhibit the Earth's curvature, unless you are seriously rich, i.e. rich enough to charter a space rocket.

Answer 8

Of course there are differences.

Experiencing g on an upwards moving disk and experiencing g because moving upwards(=outwards) on a rotating, spherical surface, a globe, you are standing on and on which the gravitational pull of that mass counteracts that outward motion, is not logically equivalent.

The sensations are the same (and it is mathematically equivalent), but only as long as you are on the surface of either the disk or the globe. And you are not accelerating otherwise. Only free falling towards the surface of the disk feels the same as on the globe.

Also, it makes no sense that the disk should be moving upwards, if the disk is intended to have the same mass as the globe has. By moving upwards, the resulting gravity has the same direction (downward) as the gravitational force because of the mass has. Thus it would double the g force downwards. To make any sense, and assuming the disk has mass, this one has to fall/move downwards. This downward speed and the resulting accelerating force acting upwards together with the gravitational pull have to cancel each other out at 1g on the surface.

However, the following assumes the OP's stated scenario of a massless disk moving upwards in space and generating 1g at the surface.

The primary difference between both cases (disk vs. globe) is, that the disk only has one force to "simulate" gravity (moving upwards), while the globe has two (gravity and rotation/centrifugal acceleration).

Digging Holes

Imagine the disk being truly flat, of finite thickness and homogeneous mass distribution (but without weight). Now dig an imaginary hole downwards. Keep digging until you reach the other end. Because the disk is moving upwards you will experience the same 1g all the way down. Eventually you fall through the disk and end up being in weightlessness of space. The disk will continue its way without you.

Imagine the globe and dig a hole there. If you dig deeper you will experience a different g at increasing depths. You will feel lighter. At the center you experience total weightlessness and if you keep "digging down" you will experience increasing g again until it normalizes at the other side of the globe. You will continue staying on the globe.

You experience exactly one g only at the surface of the globe, because gravity from the mass of the planet and the rotational speed of the planet lift each other in a way that the gravitation towards the planet ends up 9.81 times higher than the centrifugal acceleration of the rotating planet also acting upon you.

Shooting Rockets into Orbit

If you imagine a rocket shot to orbit in a straight line upwards from the surface, this will also be different.

In the case of an upwards moving disk, there is no secondary force acting upon the rocket. As soon as you stop your thrusters, you will start "falling down" again (actually the disk is coming towards you), no matter on what orbit height you are.

In the case of the globe, this is different. At a certain orbit height you can deactivate your thrusters, because your centrifugal force (=velocity on that orbit) and the gravitation towards the planet have found another equilibrium to lift each other to almost 0. In this case, you are "free falling" around the globe all the time, without the need of external accelerations (=thrust).

Atmosphere

Atmosphere on a disk moving upwards in space will just be "compressed" and flow down at all sides. If the disk would be rotating, the atmosphere will be "spun off" to all sides as well.

On a globe this is different. The atmosphere is kept at the globe because every atom of it is subjected to the gravitational force towards the globe and its rotational force (centrifugal force) around its axis.

Acceleration

Imagine a Ferris wheel standing vertically, the seats are hanging free, always pointed down towards the ground. The wheel is spinning reasonable fast to let you feel force pushing you into the seat while spinning upwards. There is only one person (or a large mass) sitting in it which we focus here, there could be any number of persons.

On a flat disk moving upwards, you will experience the same sensation no matter where the wheel stands on the disk surface. You will feel more than 1g of weight when rising up and exactly 1g at the topmost point. Then you start free falling on the downward spin, feeling lighter, maybe weightless, depending on the spining velocity. At the lowest point you experience normal weight at 1g again and so forth.

On a globe this is different. If the Ferris wheel stands on the equator and is aligned with the globe's rotation, you are accelerated a little faster on the upward motion. If the wheel is aligned perpendicular to the globe's rotation the acceleration forces are similar to those on an upward moving disk (except there are small forces pressing you towards the side). On either pole those additional "catapulting" forces would not be experienced.