Circular and linear freefall

[Ranku]

Does circular freefall of an object involve cancellation of gravitational and inertial forces, as it does in linear freefall?

 

[Ibix]

I'm not sure I understand the question. Are you asking if astronauts in an orbiting ship are weightless?

 

[A.T.]

Does circular freefall of an object involve cancellation of gravitational and inertial forces, as it does in linear freefall?

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]

I'm not sure I understand the question. Are you asking if astronauts in an orbiting ship are weightless?

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?

 

[A.T.]

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?

This is not the reason in either case:
- 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.

 

[Ibix]

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?

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.

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.

 

[vanhees71]

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.

Here you are mixing Newtonian with general relativistic descriptions.

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.

 

[A.T.]

Here you are mixing Newtonian with general relativistic descriptions.

I did't mix them, I merely mentioned both. I also don't see you addressing weightlessness at all.

 

[jbriggs444]

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?

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.

 

[vanhees71]

Obviously I didn't make clear enough what I wanted to say. Forget about GR for a moment and let's first concentrate on Newtonian gravity.

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".

 

[A.T.]

I don't think weightlessness is due to cancellation of gravitational and inertial forces in either case.

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.

 

[Ranku]

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.

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?

 

[PeroK]

Does circular freefall of an object involve cancellation of gravitational and inertial forces, as it does in linear freefall?

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.

 

[vanhees71]

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.

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.

 

[PeroK]

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 centrifugal force. The gravitational force is the real centripetal force and causes real, centripetal acceleration.

There is no cancellation of forces in circular motion.

 

[A.T.]

So is the inertial force the same as the centrifugal force arising in the object due to circular motion?

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.

Is the cancellation therefore occurring between gravitational force and centrifugal force?

You can choose a reference frame where that's the case, regardless if the the object is in free fall, or not.

 

[vanhees71]

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.

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]

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

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.

 

[PeroK]

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.

Nevertheless an object in freefall has a single gravitational force acting on it. That that force acts on all bodies is irrelevant.

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

 

[vanhees71]

No. In a free-falling reference frame you have the said approximate cancellation. You must only not forget that inertial forces only occur only in accelerating reference frames.

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".

 

[Ranku]

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?

 

[vanhees71]

In circular free fall the centrifugal force is the inertial force and compensates the gravititational force (up to the tidal forces) in the comoving (free falling) reference frame.

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.

 

[A.T.]

...centrifugal force in circular freefall ...

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.

 

[jbriggs444]

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

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.

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".