Weightlessness by a parabolic flight

Question

Do you actually achieve weightlessness during a parabolic flight? Because I believe I heard somewhere did you achieve 'near-weightlessness' and not 'weightlessness' (if this is true, why is this?) And, also, Is 'weightlessness' a correct term?

Answer 1

The answer depends on how the word weight is defined, which then dictates the meaning of the word weightless. Of course, there can be, and are, multiple definitions. As many (if not most) of the posters here are experts (whatever that may mean) in physics or astronomy, they of course have a greater intellectual library from which to draw. Some of us have specific audiences in mind when we answer questions here. I argue that audience must include introductory students who may, indeed certainly, look here for answers.

In introductory physics literature, weight is usually defined as

1) the force on an object due to gravitational interaction with Earth

or

2) the contact force on an object, as measured by a spring scale, due to gravitational interaction with Earth.

By definition 1 (which was the definition presented to me in my undergraduate and graduate courses at two universities), weight is the nickname for the gravitational force on an object due to Earth. No matter where the object is, it has a well defined weight. During a parabolic flight, the passengers experience essentially the same gravitational attraction to Earth that they experience standing on the ground. The attraction diminishes by only about 10% in low Earth orbit (e.g. aboard the ISS).

A potential consequence of definition 1 is that if the object on Mars, does it still have weight? It does, but that weight still involves interaction with Earth. There is also obviously a gravitational interaction with Mars, but if we adhere to definition 1 we can't name that force weight.

What most laymen and students mean by weightlessness is contactforcelessness, which is a bit of a made up word, but it accurately reflects the underlying physics. When the floor or seat that supports us is in free fall along with us, we don't sense its presence and thus we are tricked into thinking we are weightless. In actuality, what we sense is the absence of a supporting contact force on us. So, when the aircraft goes into free fall, the astronauts do too, and lose contact with the aircraft's floor. Understand that they are not floating. They are merely in free fall, but so is the aircraft. The gravitational force Earth exerts on them when on the ground still acts on them. They couldn't experience free fall if it didn't. The gravitational force Earth exerts on the aircraft when on the ground still acts on it too. It couldn't go into free fall if it didn't.

Some introductory textbooks define weight as the reading on a spring scale when the object is on the scale, and indeed this is a correct operational definition of weight, unless you expect numerical agreement with definition 1 all the time. The scale doesn't measure the object's gravitational weight if you're accelerating; the number depends on the object's state of motion. In general, the spring scale only measures the magnitude of the contact force pressing against the object, and it just so happens that if the object (and the scale) are not accelerating vertically, its weight is numerically equal to the contact force on you from the scale. This approach should not be used, especially in introductory courses because it's dependence on acceleration is confusing to non-experts.

A potential consequence of definition 2 is that without a contact force, one must then deal with questions about how and why things float inside spacecraft in low Earth orbit, and this issue is inevitable among even sophisticated students. Yes, the experts among us know that there's no floating involved and that the craft and occupants are merely free falling.

Ultimately, the concept of gravitational interaction underlies this issue, and no one has ever directly felt (in the traditional sense) such a force. We feel floors, seats, and other people pressing against us, but these are all contact forces, not gravitational forces. In certain circumstances, any of these forces may have the same magnitude as our weight, but not in general. Einstein would say this is perfectly logical, because gravitational forces don't actually exist in the first place. He would argue that gravitational attraction is caused by spacetime geometry, not some mysterious force. Note that force is a concept originating in Newtonian mechanics, but isn't necessary in all explanations of gravitational attraction and it thus of limited usefulness.

With respect to the sources using definition 1, NASA creates problems for students and interested laypersons by propagating the term weightless. For sources using definition 2, that particular complication doesn't exist, but another arises from the need to involve acceleration. Which one is better is not dictated by the most frequently used one, as that constitutes a fallacious appeal to the populous and perhaps also a fallacious appeal to authority. Which one is better should be dictated by which causes the less confusion and fewest misconceptions, a criterion easily measurable in the classroom.

Regardless of which definition of weight one uses, I hope we can agree that if the passengers and aircraft are all in a state of free fall, there is nevertheless a gravitational force due to Earth on them, and that this force exists no matter what, and that no contact force (aside from casually bumping into the aircraft's interior or another passenger) exists between aircraft and passenger.

Answer 2

Whether you achieve near or full weightlessness depends on the accuracy of the parabolic trajectory. Only if it is 100% accurate will the full effect of gravity be cancelled. In practice, nothing is ever perfect, so there will always be some residual weight. However, this weight can be positive or negative. If the arc is too shallow, it does not quite compensate for gravity and you will still feel a (positive) weight towards the earth. On the other hand, if the arc is too steep you will feel a force (negative weight) away from the earth. In other words you will drift towards the ceiling of the plane. So yes, in practice there is always some residual weight.

Note that the same thing even applies to satellites. They never experience a complete absence of weight; there are always minor residual effects. That is why people talk about a "micro-gravity environment".

I should also make clear that what (almost) disappears is the weight, not the gravity. Gravity is always present, no matter what you do or where you are. The only way to escape gravity is to be infinitely far away from any other masses - but even then the mass of your spacecraft will affect you gravitationally.

Answer 3

It depends on which definition of weight you use (see the wiki). Joe uses such a definition that there can be no weightlessness. I was taught at school (it was very long ago and very far from where I am now) that the weight of a body is a force with which the body acts on a support or a suspension. This definition is also mentioned in Wikipedia, and, according to this definition, weightlessness is possible in a parabolic flight. If the plane's motion is close enough to the "schedule" of free parabolic "fall", you will be able to stay away from the plane's interior surface during the entire parabolic flight, so your weight (according the definition above) will be zero.