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Ranku

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In summary: This happens because the contact force between your body and the gravitational field is exactly canceled by the inertial force.

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Ranku

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Ibix

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A.T.

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Inertial forces appear if you choose to use a non-inertial frame of reference for your analysis. Depending on what frame you chose, they might or might not cancel the Newtonian gravitational force. In General Relativity, gravity itself is an inertial force.Ranku said:

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Ranku

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I guess another way of asking the question would be, is weightlessness in circular freefall for the exact same reason as in linear freefall - i.e., cancellation between gravitational and inertial forces?Ibix said:

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A.T.

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This is not the reason in either case:Ranku said:I guess another way of asking the question would be, is weightlessness in circular freefall for the exact same reason as in linear freefall - i.e., cancellation between gravitational and inertial forces?

- Weightlessness is a frame independent physical fact.

- Cancellation of gravitational by inertial forces is a frame dependent feature of your choice of analysis.

Weightlessness occurs when no forces are applied or if they are applied uniformly to a body, so that no internal stresses arise.

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Ibix

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I don't think weightlessness is due to cancellation of gravitational and inertial forces in either case. In Newtonian physics those forces don't, in general, cancel for a free-falling body. They only cancel in one particular frame.Ranku said:I guess another way of asking the question would be, is weightlessness in circular freefall for the exact same reason as in linear freefall - i.e., cancellation between gravitational and inertial forces?

The simplest explanation is that all parts of your body and your spaceship are experiencing the exact same acceleration due to gravity. This is different from standing on the Earth, where the gravitational force is opposed by a contact force that comes through the soles of your feet.

A better explanation is the general relativistic one, that gravity is not a force. So you feel weightless in free fall because the sensation of weight is actually just that contact force, which is not present in free fall.

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Here you are mixing Newtonian with general relativistic descriptions.A.T. said:Inertial forces appear if you choose to use a non-inertial frame of reference for your analysis. Depending on what frame you chose, they might or might not cancel the Newtonian gravitational force. In General Relativity, gravity itself is an inertial force.

In general relativity gravity is an interaction as all the fundamental interactions are and you can reinterpret the corresponding (gauge) field, as the pseudometric tensor, ##g_{\mu \nu}##, of a Lorentzian spacetime manifold. A true gravitational field is only locally, strictly speaking only in one point of the spacetime manifold, equivalent to inertial "forces", because you can at any spacetime point define a local inertial frame of reference. With a true gravitational field present, which is due to some energy-momentum-stress of matter, there's also non-vanishing curvature of the spacetime manifold and thus a true gravitational field cannot be entirely described as a pure inertial force in a non-inertial (global) reference frame in Minkowski space.

That's also true in the much more simple case of Newtonian gravity. If you have a true gravitational field you can define a free-falling frame of reference, and wrt. this reference frame within a small enough neighbourhood you have an inertial frame. This is a good approximation across scales within which the gravitational field can be assumed to be homogeneous. On larger scales within this locally inertial frame of reference there are tidal forces which cannot be transformed away by any accelerated frame of reference.

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A.T.

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I did't mix them, I merely mentioned both. I also don't see you addressing weightlessness at all.vanhees71 said:Here you are mixing Newtonian with general relativistic descriptions.

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jbriggs444

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I would say yes. The situations are the same.Ranku said:I guess another way of asking the question would be, is weightlessness in circular freefall for the exact same reason as in linear freefall - i.e., cancellation between gravitational and inertial forces?

In both cases we have adopted a non-inertial frame where the inertial force from the chosen frame is equal and opposite to the gravitational force.

In both cases, the frame we have adopted is a free falling frame. So it is inevitable that the inertial force will cancel the gravitational force exactly - we chose the frame to make it so.

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It's easy to show within Newtonian gravity that in a free-falling reference frame in a region of an extension small enough that the gravitational field can be considered homogeneous you have an inertial frame ("weightlessness") for an observer at rest relative to this reference frame. However, "weightlessness" is an approximation only. In fact you'll have some residual tidal forces acting on any point mass at rest relative to this free falling frame. An example is the International Space Station: To 0th order the astronauts don't feel any gravitational forces, in reality there are however some tidal forces. That's why one talks more precisely about "micro-gravity experimentments" rather then "experiments in weightlessness".

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A.T.

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As an example: If you analyze yourself sitting on a chair from a free falling frame, the inertial and gravitational forces on you will cancel as well. That doesn't make you weightless.Ibix said:I don't think weightlessness is due to cancellation of gravitational and inertial forces in either case.

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Ranku

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So is the inertial force the same as the centrifugal force arising in the object due to circular motion? Is the cancellation therefore occurring between gravitational force and centrifugal force?jbriggs444 said:I would say yes. The situations are the same.

In both cases we have adopted a non-inertial frame where the inertial force from the chosen frame is equal and opposite to the gravitational force.

In both cases, the frame we have adopted is a free falling frame. So it is inevitable that the inertial force will cancel the gravitational force exactly - we chose the frame to make it so.

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In classical physics the answer is no. Circular and linear freefall both involve a singje gravitational force acting on the body. There is no cancellation of forces.Ranku said:

If you adopt an accelerating reference frame, then rhe real force is canceled by an inertial force. But, this is the case whether or not the real force is gravitational.

General Relativity does not apply.

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What I tried to say in several posts is that this cancellation can never be complete if you have a true gravitational field. The cancellation is complete in a homogeneous gravitational field, i.e., for real gravitational fields the cancellation is only partial and in a free-falling inertial system you always have tidal forces.A.T. said:As an example: If you analyze yourself sitting on a chair from a free falling frame, the inertial and gravitational forces on you will cancel as well. That doesn't make you weightless.

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There is no centrifugal force. The gravitational force is the real centripetal force and causes real, centripetal acceleration.Ranku said:So is the inertial force the same as the centrifugal force arising in the object due to circular motion? Is the cancellation therefore occurring between gravitational force and centrifugal force?

There is no cancellation of forces in circular motion.

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A.T.

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The centrifugal force arises from your choice of a reference frame, not from the motion of an object. The same goes for the linear inertial force.Ranku said:So is the inertial force the same as the centrifugal force arising in the object due to circular motion?

You can choose a reference frame where that's the case, regardless if the the object is in free fall, or not.Ranku said:Is the cancellation therefore occurring between gravitational force and centrifugal force?

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No! That's the specific feature of the gravitational interaction. If there are other forces involved in the reference frame moving freely in the corresponding fields there's no such cancellation of the inertial and these forces. Only in the case of the gravitational interaction is the source of the field (the gravitational mass) strictly proportional (in our usual system of units equal) to the inertial mass and thus all point particles feel the same acceleration when moving in a gravitational field. For, e.g., the electromagnetic interaction that's not the case. Here the acceleration is proportional to the charge-over-mass ratio, i.e., it's not a universal constant valid for all bodies.PeroK said:In classical physics the answer is no. Circular and linear freefall both involve a singje gravitational force acting on the body. There is no cancellation of forces.

If you adopt an accelerating reference frame, then rhe real force is canceled by an inertial force. But, this is the case whether or not the real force is gravitational.

General Relativity does not apply.

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In classical physics an object in freefall is by definition not weightless. Weight being by definition the (gravitational) force acting on the body.Ranku said:

As I understand it, weightlessness in classical physics is an illusion - an object still has weight although in a circular object it cannot be measured locally.

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Nevertheless an object in freefall has a single gravitational force acting on it. That that force acts on all bodies is irrelevant.vanhees71 said:No! That's the specific feature of the gravitational interaction. If there are other forces involved in the reference frame moving freely in the corresponding fields there's no such cancellation of the inertial and these forces. Only in the case of the gravitational interaction is the source of the field (the gravitational mass) strictly proportional (in our usual system of units equal) to the inertial mass and thus all point particles feel the same acceleration when moving in a gravitational field. For, e.g., the electromagnetic interaction that's not the case. Here the acceleration is proportional to the charge-over-mass ratio, i.e., it's not a universal constant valid for all bodies.

In Newtonian physics there is only ##F = ma##. There is no equivalence principle.

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The basic idea is very simple. Take some gravitational field ##\vec{a}(\vec{x})## and consider a mass point in free fall in a neighborhood of a point ##\vec{x}_0##. Then to "0th order" you can take ##\vec{a}(\vec{x}) \simeq \vec{a}(\vec{x}_0)=\vec{g}=\text{const}##.

Now consider an inertial reference frame. The motion of our reference body (coordinates ##\vec{y})## using Cartesian coordinates wrt. this inertial frame is

$$m \ddot{\vec{y}}=m \vec{g} \; \Rightarrow \; \vec{y}=\vec{x}_0 + \vec{v}_0 t + \frac{1}{2} \vec{g} t^2.$$

Now take a new reference frame consisting of a Cartesian basis (three perpendicular rods) fastened at the particle. The motion in this system is described by coordinates

$$\vec{x}'=\vec{x}-\vec{y}(t).$$

From this you get the equation of motion of any other body

$$m' \ddot{\vec{x}}'=m' (\ddot{\vec{x}}-\ddot{\vec{y}})=m'(\vec{g}-\vec{g})=\vec{0}.$$

This exact cancellation of course only holds if ##\vec{g}=\vec{a}(\vec{x}_0)## is strictly constant, i.e., for small neigborhoods around the reference point ##\vec{x}_0## of our reference body.

If you consider a more complicated case of a body in free fall along an orbit around a body like the Earth, it's more complicated. There the accelerated reference frame is in addition along a curved trajectory. To realize an approximate inertial reference frame you have to make sure to have (locally) non-rotating reference frames. You can realize this by fastening gyros to define the accelerated Cartesian basis fastened to the freely falling reference point. For such an accelerated reference frame you have only the tidal forces left from the "cancellation between the gravitational and the inertial forces".

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Ranku

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A.T. said:The centrifugal force arises from your choice of a reference frame, not from the motion of an object. The same goes for the linear inertial force.

To clarify, is centrifugal force in circular freefall the equivalent or analogue of inertial force in linear freefall?

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Seen from the inertial frame, there's no inertial/centrifugal force of course, but the gravitational force is the centripetal force of the circular motion.

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A.T.

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There no such thing as "the centrifugal force in circular freefall". There is a centrifugal force in certain reference frames, which can be used to analyze any scenario.Ranku said:...centrifugal force in circular freefall ...

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jbriggs444

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The force arising from the adoption of a rotating frame (centrifugal force) is identical in nature to the force arising from the adoption of an accelerating frame. Yes.Ranku said:To clarify, is centrifugal force in circular freefall the equivalent or analogue of inertial force in linear freefall?

Centrifugal force varies in direction and magnitude depending on position while the force arising from linear acceleration does not. But the essential nature of both forces is the same. They are both inertial forces.

I share other poster's discomfort with your emphasis on "free fall".

Circular freefall occurs when an object is falling in a circular path due to the influence of a centripetal force, while linear freefall occurs when an object is falling in a straight line due to the influence of gravity.

In circular freefall, the centripetal force is responsible for keeping the object in a circular path as it falls. This force is directed towards the center of the circular path and is equal to the object's mass multiplied by its centripetal acceleration.

The speed of an object in freefall is affected by the object's mass and the amount of air resistance it experiences. Objects with larger masses will fall faster than objects with smaller masses, while objects with more air resistance will fall slower.

Yes, an object can experience both circular and linear freefall at the same time. This is known as projectile motion, where an object is launched at an angle and experiences both horizontal and vertical motion simultaneously.

The acceleration due to gravity in freefall is approximately 9.8 m/s^2. This means that for every second an object falls, its speed increases by 9.8 meters per second.

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