"Falling" towards the earth

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"Falling" towards the earth
« on: March 28, 2009, 10:49:46 PM »
I was browsing this forum when I thought of this flaw in the FE theory.  According to it, objects never "fall" on Earth; the earth just catches up to it, thus causing it to experience the sensation of falling through the atmosphere.  If this is so, then why do things like accelerometers detect motion when an object falls from the sky?  I'm sorry if this is in the FAQ or is totally retarded -- just wondering.

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Jack

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Re: "Falling" towards the earth
« Reply #1 on: March 28, 2009, 11:11:59 PM »
If this is so, then why do things like accelerometers detect motion when an object falls from the sky?
??? Free-fall isn't an acceleration...

Re: "Falling" towards the earth
« Reply #2 on: March 28, 2009, 11:16:39 PM »
Well, on RE free fall is acceleration, you know gravity.  Though i haven't tried it myself, I would assume an accelerometer would detect the 9.8 m/s^2 provided by a free-fall, or else the established laws of physics would be null and void in a very obvious way.

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Jack

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Re: "Falling" towards the earth
« Reply #3 on: March 28, 2009, 11:19:41 PM »
Free-fall is an inertial motion. An accelerometer would read 0 for a free-falling object.

Re: "Falling" towards the earth
« Reply #4 on: March 28, 2009, 11:23:37 PM »
I'm sorry, but I haven't even taken high school physics yet; please explain why free-fall (on RE) doesn't count as acceleration in simple terms.

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Jack

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Re: "Falling" towards the earth
« Reply #5 on: March 28, 2009, 11:27:52 PM »
Free-fall is basically a motion along geodesics in space-time. An object following geodesics is considered inertial, and any force applied to such object will deviate it from the geodesics (Einstein's version of Newton's first law).  Thus, free-fall is an inertial motion.

Re: "Falling" towards the earth
« Reply #6 on: March 28, 2009, 11:34:15 PM »
I guess you didn't understand the fact that I don't know anything about physics.  However, I still have logic and it is as follows:
-an object moving an increasing rate = acceleration
-objects in free fall = moving at an increasing rate
-objects in free fall = accelerating

I have no idea what inertial motion is, but I can tell you that if I jumped out of an airplane I would fall faster and faster until I hit terminal velocity.

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Jack

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Re: "Falling" towards the earth
« Reply #7 on: March 28, 2009, 11:37:40 PM »
Inertial motion = a motion without acceleration.

You don't fall faster. It's relativity physics. In your frame of reference, the floor is coming toward you at 9.8m/s2. It appears that you are falling faster, which in fact you aren't if you attach yourself with an accelerometer.

Re: "Falling" towards the earth
« Reply #8 on: March 28, 2009, 11:42:50 PM »
If I don't fall at an increasing speed out of an airplane, then why is landing on the ground any different from jumping off a wall?  If the speeds are identical, then the impact of my landing should be equal, yet I break my legs when I land after jumping from the airplane.  I'm sorry if I'm being difficult; I just want to understand.

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Jack

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Re: "Falling" towards the earth
« Reply #9 on: March 29, 2009, 12:05:40 AM »
If I don't fall at an increasing speed out of an airplane, then why is landing on the ground any different from jumping off a wall?
You don't increase in speed no matter where you fall. It's the floor that's accelerating, in RE or FE.

The free-falling elevator is a perfect example. Inside such elevator, you're floating and feeling weightless (or feeling no acceleration). You don't know if you're under the influence of "gravity" or in deep space without gravity. However, if you grab an accelerometer, you will see that it reads 0 as there's no force acting on you.

Alternatively, inside an accelerating elevator, you drop an apple and it falls to the floor. You don't know if the apple is accelerating toward the floor due to gravity or the floor accelerating toward the apple. Interesting.

Read more to get you started: http://www.einstein-online.info/en/spotlights/equivalence_principle/index.html

If the speeds are identical, then the impact of my landing should be equal, yet I break my legs when I land after jumping from the airplane.
That's because the ground is coming toward you at 9.8m/s2 while you undergo constant velocity. It will break your legs.

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RAFboiMF

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Re: "Falling" towards the earth
« Reply #10 on: March 29, 2009, 09:38:42 AM »
Free-fall isn't an acceleration...
WRONG!!! (1)

Free-fall is an inertial motion. An accelerometer would read 0 for a free-falling object.
WRONG!!! (2)

Inertial motion = a motion without acceleration.

You don't fall faster. It's relativity physics. In your frame of reference, the floor is coming toward you at 9.8m/s2. It appears that you are falling faster, which in fact you aren't if you attach yourself with an accelerometer.
WRONG!!! (3)

1. Free Fall is an un-aided acceleration of a mass due to gravity. This acceleration continues until terminal velocity is reached (terminal velocity is when the drag forces acting in the opposite direction to gravity equal gravity)

2. An accelerometer detects the acceleration due to gravity until terminal velocity is reached. Likewise an accelerometer aligned with the vertical axis measures a constant 9.81... m/s2 when stationary on the ground due to gravity.

3. Again, try a free fall skydive with an accelerometer (I have). You will see your acceleration decrease to zero as you reach terminal velocity.
(remember tho that the accelerometer you use must be calibrated to compensate for gravity when stationary)

As another note, aircraft use 3 accelerometers in an inertial navigation system, one aligned north, one aligned east and one aligned vertical.  The system knows when they are all in alignment by measuring the acceleration due to gravity on the vertical gyro.

Please Get your facts straight before posting so much false information.
« Last Edit: March 29, 2009, 09:42:50 AM by RAFboiMF »
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Re: "Falling" towards the earth
« Reply #11 on: March 29, 2009, 10:15:03 AM »
Actually jack has a point. Under relativity thing that are falling are not accelerating, they are moving in a straight line in the forth dimension. one something is on the ground it is being acted upon by the force of the ground pushing up on it and is accelerating in the fourth dimension. at least that is the way I understood it. I have been wrong before.
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Dr Matrix

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Re: "Falling" towards the earth
« Reply #12 on: March 29, 2009, 10:32:46 AM »
I have discussed this subject in detail before...

Check out this part of the gravity sticky thread.

Allow me to summarise:

- an accelerometer in your hand while you sit on the sofa reads +9.8ms-2.  This is the result of what is commonly called the normal contact force of your arse with the sofa.  It acts in an upwards direction, hence the '+'.

- an accelerometer in your hand as you stand in the doorway of a plane flying at constant altitude reads +9.8ms-2, again due to the contact force of you on the plane.  The plane maintains altitude by generating lift, essentially pushing on the air to generate an upwards force.

Now you jump out of the plane door...

- the accelerometer initially reads zero as you step out the door, then as you accelerate with respect to the air around you, you find that you measure a steadily increasing positive force resulting from air resistance.

- when this force equals +9.8ms-2, you stop accelerating with respect to the ground and continue at constant velocity, known as terminal velocity.

- as you hit the ground, the accelerometer records a rapid increase in positive acceleration, again as a result of the contact force with the ground, before returning to +9.8ms-2.

Note that at no point does the accelerometer measure a negative force in the form of 'gravity acting downwards'.  An accelerometer can never measure gravity directly, although it can measure it indirectly as in the contact force of an object stationary on the ground, or moving from one altitude to another.

Quote from: Arthur Schopenhauer
All truth passes through three stages. First, it is ridiculed. Second, it is violently opposed. Third, it is accepted as being self-evident.

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Jack

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Re: "Falling" towards the earth
« Reply #13 on: March 29, 2009, 01:17:20 PM »
1. Free Fall is an un-aided acceleration of a mass due to gravity. This acceleration continues until terminal velocity is reached (terminal velocity is when the drag forces acting in the opposite direction to gravity equal gravity)
Free-fall is an inertial motion along geodesics in space-time. Please review relativity.

2. An accelerometer detects the acceleration due to gravity until terminal velocity is reached. Likewise an accelerometer aligned with the vertical axis measures a constant 9.81... m/s2 when stationary on the ground due to gravity.
An accelerometer detects no acceleration on a free-falling observer, as there is no force acting on such observer at all. However, an accelerometer would read a value on an observer in contact with the surface, as there is an normal force exerted on him.

3. Again, try a free fall skydive with an accelerometer (I have). You will see your acceleration decrease to zero as you reach terminal velocity.
Nope, neglecting air resistance, it reads zero for me.

(remember tho that the accelerometer you use must be calibrated to compensate for gravity when stationary)

As another note, aircraft use 3 accelerometers in an inertial navigation system, one aligned north, one aligned east and one aligned vertical.  The system knows when they are all in alignment by measuring the acceleration due to gravity on the vertical gyro.
Get out of Newtonian physics, please. Modern physics already proved there is no such thing as "acceleration due to gravity".

Please Get your facts straight before posting so much false information.
Says the ignorant.

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Trekky0623

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Re: "Falling" towards the earth
« Reply #14 on: March 29, 2009, 01:24:19 PM »
An accelerometer reads 0 in free fall.  I might read air resistance, as in the air pushing you upwards or, in RE, the wind keeping you from falling at g, but in a vacuum, it will read 0.

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Dr Matrix

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Re: "Falling" towards the earth
« Reply #15 on: March 29, 2009, 02:11:06 PM »
I have discussed this subject in detail before...

Check out this part of the gravity sticky thread.

Allow me to summarise:

- an accelerometer in your hand while you sit on the sofa reads +9.8ms-2.  This is the result of what is commonly called the normal contact force of your arse with the sofa.  It acts in an upwards direction, hence the '+'.

- an accelerometer in your hand as you stand in the doorway of a plane flying at constant altitude reads +9.8ms-2, again due to the contact force of you on the plane.  The plane maintains altitude by generating lift, essentially pushing on the air to generate an upwards force.

Now you jump out of the plane door...

- the accelerometer initially reads zero as you step out the door, then as you accelerate with respect to the air around you, you find that you measure a steadily increasing positive force resulting from air resistance.

- when this force equals +9.8ms-2, you stop accelerating with respect to the ground and continue at constant velocity, known as terminal velocity.

- as you hit the ground, the accelerometer records a rapid increase in positive acceleration, again as a result of the contact force with the ground, before returning to +9.8ms-2.

Note that at no point does the accelerometer measure a negative force in the form of 'gravity acting downwards'.  An accelerometer can never measure gravity directly, although it can measure it indirectly as in the contact force of an object stationary on the ground, or moving from one altitude to another.


Quote from: Arthur Schopenhauer
All truth passes through three stages. First, it is ridiculed. Second, it is violently opposed. Third, it is accepted as being self-evident.

Re: "Falling" towards the earth
« Reply #16 on: March 29, 2009, 03:15:51 PM »
Just to clarify as far as the mechanics of freefall are concerned I have no argument with what Matrix and Jack says. However there still seems to be this opinion that modern physics somehow proves that gravity is a fictitious force. This is trivially a load of rubbish, modern physics provides a mechanism for how gravity can be a fictitious force.

F = ma => 2nd law of motion
F = mg => from NEWTON'S law of gravitation

Because a = g we cannot tell acceleration due to gravity apart from an inertial (fictitious) force. I prefer inertial because fictitious force makes it seem like it isnt there, put someone in a cylinder and spin it and I guarantee they will find the force very real. Anyway my point is that gravity is just as fictitious in Newtonian relativity, he just didn't present a mechanism. Anyway just wanted to reiterate that things are not this black and white. Inertial mass seems very likely to have its roots in the background dependent quantum field theory.

Re: "Falling" towards the earth
« Reply #17 on: March 29, 2009, 03:30:42 PM »
http://arxiv.org/abs/hep-th/0507235 - worth a read if anyone is interested

Re: "Falling" towards the earth
« Reply #18 on: March 29, 2009, 05:55:37 PM »
So if the earth is continually moving upwards why don't the clouds just fall the earths surface, surely they too can't me moving upwards to as they're only bodies of condensation..
« Last Edit: March 29, 2009, 05:57:47 PM by Mammon »

Re: "Falling" towards the earth
« Reply #19 on: March 29, 2009, 06:00:02 PM »
So if the earth is continually moving upwards why don't the clouds just fall the earths surface, surely they too can't me moving upwards to as they're only bodies of condensation..
the air can't fall of the edge of the earth because of dark energy and the could don't fall down is because there is something occupying the space on the ground(more air).
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Atom Man

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Re: "Falling" towards the earth
« Reply #20 on: March 31, 2009, 06:52:30 AM »
Sorry Matrix but how you describe acceleration due to "gravity" has no logical reasoning. For the sake of the argument lets say g = +9.8 m/s^2 both at ground level and in the aircraft. How come g changes from 9.8 m/s^2 to 0 m/s^2 as soon as you step out?

An initial acceleration of of 9.8 m/s^2 to 0 m/s^2 at terminal velocity just seems far more logical. If a = 0 m/s^2 then you have 0 N force and would just float with the clouds since you will not be applying a force to displace the air.

v = u + at if u = 0 m/s and a = 0 m/s^2 then v = 0 m/s
and
F = ma if a = 0 m/s^2 then F = 0 N
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Re: "Falling" towards the earth
« Reply #21 on: March 31, 2009, 08:40:25 AM »
Sorry Matrix but how you describe acceleration due to "gravity" has no logical reasoning. For the sake of the argument lets say g = +9.8 m/s^2 both at ground level and in the aircraft. How come g changes from 9.8 m/s^2 to 0 m/s^2 as soon as you step out?

An initial acceleration of of 9.8 m/s^2 to 0 m/s^2 at terminal velocity just seems far more logical. If a = 0 m/s^2 then you have 0 N force and would just float with the clouds since you will not be applying a force to displace the air.

v = u + at if u = 0 m/s and a = 0 m/s^2 then v = 0 m/s
and
F = ma if a = 0 m/s^2 then F = 0 N
Its using relativity. Lets say you are in a box with no windows. You take out an accelerometer. it says you are accelerating at 9.8 m/s^2. there is no way of knowing whether it is gravity or the box is accelerating to give you this reading. When you fall out of a plane and are in free fall lets say there is no air resistance. although it appears you are accelerating in the fourth dimension you are going in a constant speed.
You can't outrun death forever
But you can sure make the old bastard work for it.

Re: "Falling" towards the earth
« Reply #22 on: March 31, 2009, 11:09:37 AM »
Sorry Matrix but how you describe acceleration due to "gravity" has no logical reasoning. For the sake of the argument lets say g = +9.8 m/s^2 both at ground level and in the aircraft. How come g changes from 9.8 m/s^2 to 0 m/s^2 as soon as you step out?

An initial acceleration of of 9.8 m/s^2 to 0 m/s^2 at terminal velocity just seems far more logical. If a = 0 m/s^2 then you have 0 N force and would just float with the clouds since you will not be applying a force to displace the air.

v = u + at if u = 0 m/s and a = 0 m/s^2 then v = 0 m/s
and
F = ma if a = 0 m/s^2 then F = 0 N
Its using relativity. Lets say you are in a box with no windows. You take out an accelerometer. it says you are accelerating at 9.8 m/s^2. there is no way of knowing whether it is gravity or the box is accelerating to give you this reading. When you fall out of a plane and are in free fall lets say there is no air resistance. although it appears you are accelerating in the fourth dimension you are going in a constant speed.

No air resistance when you step out of a plane?

Also, should you decide to do that, and reach terminal velocity, you are not accelerating in the fourth dimension, as that would require us to be five-dimensional beings, and as far as I'm aware we only exist in four (layman here BTW, if we're in umpteen then no mockery required!).  You're accelerating in two, the velocity in the third (down) is constant, therefore it is also in the fourth, unless you're travelling faster through time than everyone else.  Please correct me if I'm wrong, I know there's something about doing so the faster you travel, but does that apply here?
"The Zetetic Astronomy has come into my hands ... if it be childish, it is clever; if it be mannish, it is unusually foolish."

A Budget of Paradoxes - A. de Morgan (pp 306-310)

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markjo

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Re: "Falling" towards the earth
« Reply #23 on: March 31, 2009, 12:27:08 PM »
No air resistance when you step out of a plane?

Most first year or so physics text books homework problems will tell you to disregard air resistance to make the math a whole lot easier while still getting the point across.  If you want to do the math required to calculate the resistance due to the constantly changing air pressure and aerodynamic drag of a falling body, then by all means, have at it.
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Re: "Falling" towards the earth
« Reply #24 on: March 31, 2009, 12:31:41 PM »
No air resistance when you step out of a plane?

Most first year or so physics text books homework problems will tell you to disregard air resistance to make the math a whole lot easier while still getting the point across.  If you want to do the math required to calculate the resistance due to the constantly changing air pressure and aerodynamic drag of a falling body, then by all means, have at it.

No thanks ;)
"The Zetetic Astronomy has come into my hands ... if it be childish, it is clever; if it be mannish, it is unusually foolish."

A Budget of Paradoxes - A. de Morgan (pp 306-310)

Re: "Falling" towards the earth
« Reply #25 on: March 31, 2009, 01:17:24 PM »
if you are in free fall you are not accelerating. In the fourth dimension you are moving in a straight line. it appears you are accelerating.
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Dr Matrix

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Re: "Falling" towards the earth
« Reply #26 on: March 31, 2009, 03:04:19 PM »
Sorry Matrix but how you describe acceleration due to "gravity" has no logical reasoning. For the sake of the argument lets say g = +9.8 m/s^2 both at ground level and in the aircraft. How come g changes from 9.8 m/s^2 to 0 m/s^2 as soon as you step out?

An initial acceleration of of 9.8 m/s^2 to 0 m/s^2 at terminal velocity just seems far more logical. If a = 0 m/s^2 then you have 0 N force and would just float with the clouds since you will not be applying a force to displace the air.

v = u + at if u = 0 m/s and a = 0 m/s^2 then v = 0 m/s
and
F = ma if a = 0 m/s^2 then F = 0 N

As the others have said, it doesn't matter what makes more sense to you, the accelerometer will say that once you are in free fall you experience no acceleration other than air resistance.  Please see the diagram I linked to for a few examples.
Quote from: Arthur Schopenhauer
All truth passes through three stages. First, it is ridiculed. Second, it is violently opposed. Third, it is accepted as being self-evident.

Re: "Falling" towards the earth
« Reply #27 on: March 31, 2009, 04:37:32 PM »
This is quite confusing because force is an observer dependent quantity. Take a trivial example, accelerating in a car. A person standing outside of an accelerating car will see a person inside who is accelerating forwards with the car. A person inside the car will fell a backwards acceleration forcing them back into the seat. If we now translate this to gravity, what you feel as weight is the force of the ground pushing against you. When you are in free fall you have no net force on you. However your velocity with respect to the ground is changing, though kinetic energy is also frame dependent :p . Although whether this is fundamentally true or only appears to be true due to the equivalence of inertial and gravitational mass is an open question. If it makes you feel better there are invariant quantities, unfortunately they aren't really every day quantities.

Re: "Falling" towards the earth
« Reply #28 on: April 03, 2009, 05:20:11 PM »
So if the earth is continually moving upwards why don't the clouds just fall the earths surface, surely they too can't me moving upwards to as they're only bodies of condensation..
the air can't fall of the edge of the earth because of dark energy and the could don't fall down is because there is something occupying the space on the ground(more air).

Why is it that this Dark Energy accelerates the Earth and the celestial bodies, yet has absolutely no effect on anything on the surface of the Earth itself?
Clothes are proof evolution never happened.

Re: "Falling" towards the earth
« Reply #29 on: April 03, 2009, 06:50:07 PM »
If the celestial bodies are as close as FE suggests then there is no need for dark energy as the metric that calls for dark energy is not valid anyway as I understand it.