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Physics of falling objects

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Physics of falling objects

Old 14th Mar 2014, 09:45
  #101 (permalink)  
 
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Yes I was referring to the reduction in g due to the curved path at speed, the classic in a lift with no windows experiment.


You are wrong about deep oceanic tides. they are pretty much non existent in the middle of the oceans and build up around continental masses.

My point is that your earlier explanation

"The tidal acceleration moves water about two meters in six hours. The free fall acceleration moves water two meters in about 0.4 seconds. The force is related to the square of the ratio of those times. That's gives a measure of the low-down scale of tidal forces"

was codswallop. It is vastly more complex. I thought of a few more factors including coriolis effect, centripetal effects etc.
To characterise it as (and I am paraphrasing you here)
"tides go up and down with very little acceleration compared to water falling in free fall therefore tidal effects are negligible"

is just wrong.
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Old 14th Mar 2014, 09:59
  #102 (permalink)  
 
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Oggers:

"Deep ocean tides on Earth don't exist"
"I don't understand when factors are too small to matter"
"I can't estimate the size of physical quantities and effects"

Tourist:

"I can't use a library or a search engine"
"I have difficulty comprehending the statement 'By including unnecessary terms in the gravity, and ignoring the larger degree of ignorance about the density profile, and lift effects, it's not very helpful to the discussion about falling objects.'"

Thanks for the ongoing valuable and voluminous contributions, chaps. No doubt we'll be having more.
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Old 14th Mar 2014, 10:28
  #103 (permalink)  
 
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awblain

Superficially sounds knowledgeable, but when pressed seems strangely unable to maintain and defend a position.
Makes statements which are simplistic but when taken to task slides away in a soapy fashion.....
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Old 14th Mar 2014, 10:38
  #104 (permalink)  
 
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As has been pointed out, to understand tides you have to realise its a dynamic problem which must be framed appropriately as two revolving extensible bodies rotating about their common cm. The problem with the poor professor's paper is that it is quite simply unintelligent to include the range variation of gravity as its very small perturbing effect relative to a constant model is less than trivial when compared to the other forces acting and the inherent limitations of the model. That's what was meant by another poster when he said the beginning of the paper was wacky and not worth the read. He is entirely correct on that. I must have another look at it to see if he includes coriolis forces: I doubt he understands that either.

So can someone now explain this variable gravity theory of tides to me as opposed to the standard theory ?
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Old 14th Mar 2014, 10:48
  #105 (permalink)  
 
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Originally Posted by Mr Optimistic View Post
The problem with the poor professor's paper is that it is quite simply unintelligent to include the range variation of gravity as its very small perturbing effect relative to a constant model is less than trivial when compared to the other forces acting and the inherent limitations of the model. That's what was meant by another poster when he said the beginning of the paper was wacky and not worth the read. He is entirely correct on that. I must have another look at it to see if he includes coriolis forces: I doubt he understands that either.
Ehh, when I read the paper, my takeaway was that he was talking about the variation of gravity due to altitude, a la the free air gradient vs assuming gravitational acceleration to be constant at all heights along the trajectory of the falling airplane parts. I didn't see anything in the actual paper to suggest that he was considering adding terms for temporal variations.
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Old 14th Mar 2014, 10:50
  #106 (permalink)  
 
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In very rough terms, the earth and moon can be considered an object with a centre of mass somewhere between the centre of the earth and the moon.
This centre of mass will be near the earths core due to the relative sizes of the bodies.
This centre of mass will also be mobile as the moon revolves around the earth.
Thus, a person standing on the earths surface will be moving in and out relative to the centre of mass of the earth/moon pair and will experience (very tiny! 2micro m/s2) variations in gravity.
They would also be moving in and out relative to the L1 Lagrange point, ie varying their range from a point of zero g
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Old 14th Mar 2014, 10:53
  #107 (permalink)  
 
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Tourist,

In a previous post you said:

Originally Posted by Tourist View Post
You have to bring in an incredibly complex range of factors including sea bottom shape, sea viscosity, flow patterns through constrictions, compressibility to name just the first ones off the top of my head.
Now, I know just enough about tides to be dangerous. I'm aware that bathymetric effects are major factors, and the other stuff makes sense. I'm wondering about "compressibility". I'm trying to imagine how/where that's a factor and coming up short. (no, I'm not baiting you. I don't know and I'm curious)
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Old 14th Mar 2014, 10:55
  #108 (permalink)  
 
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It is not that close to the earth's core. Awblain is correct in everything he says as far as I can tell a comment I can't make about said professor.

Edit: you are right to be sceptical about compressibility!
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Old 14th Mar 2014, 11:02
  #109 (permalink)  
 
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You are right, it is tenuous!

My idle thought was that since water has some limited ability to compress, the (agreed it is tiny!) variation in g with the moons motion will cause the ocean to expand and contract by admittedly tiny amounts.
That's all!
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Old 14th Mar 2014, 11:07
  #110 (permalink)  
 
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For the genuinely curious a good explanation of the equilibrium theory of tides can be found in Newtonian Mechanics A P French p531 et seq.
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Old 14th Mar 2014, 12:13
  #111 (permalink)  
 
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Oggers:

"Deep ocean tides on Earth don't exist"
"I don't understand when factors are too small to matter"
"I can't estimate the size of physical quantities and effects"
awblain, having seen you perform on other threads it seems you are more comfortable with that kind of post than one which includes a coherent technical argument to support your point.

I think at this stage it is fair comment to say that - in my subjective opinion of course - you seem to lack the technical nous to make an argument that will carry your point in a technical forum such as this.

You have failed to explain anything at all on this thread, you have merely piled up a list of vague rhetorical responses stuffed with ambiguous and woolly jargon.

Take this one as a case in point:

The poor guy's written a decidedly mediocre conference proceeding that advances Mr Stokes' work from the 19th century not one jot.

If you can't see that, then you need more than a hundred-word comment to explain.
It is a purely subjective comment. But this is a technical forum. Why can't you explain what is wrong with his model? Because there is nothing wrong with it except in your subjective opinion it isn't much use. But you are not an accident investigator whilst he is. You are an anonymous lay-person, whilst he is a professional in the field who sees utility in the trajectory analysis he has developed.

Let's face it awblain, you aren't going to be using that or any other trajectory analysis to investigate in flight break-ups. I'm not either.

If you can't win by arguments strong - and you haven't provided anything resembling one - can you at least point me to your publishing history? For the third time of asking Because, all I have to go on with you, is about 6 pages of waffle that any schoolboy with a search engine could cobble together.
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Old 14th Mar 2014, 13:03
  #112 (permalink)  
 
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I do not understand what you are discussing.

This is what I know:

Newton:1 - If a body is not subject to any net external force it either remains at rest or continues in uniform motion.

Newton:2 - F = ma (Law of acceleration)

Newton:3 - When two particles interact, the force on one particle is equal to and opposite to the force on the other.

So, when dealing with real life stuff you must consider air resistance as a force that wants to slow things down per N:3. The force D working on the object is calculated by the drag equation:

D = 1/2(rho)CA(v^2)

C is the drag coefficient for that particular object. This is not constant but varies with its shape and size - and of course what kind of air flows around it (wind conditions etc).

The force T acting on a falling object in air would then be:

T = F+D

which is dependent on mass to some extent (greater size usually means greater mass).

As for the debate about varying gravitational forces, I do not understand what you are fuzzing about. The gravitational coefficient g is calculated like:

g = GM/R^2

The interesting part here is R^2 which means the radius squared. As the earth is elliptical and not a sphere, this value will vary, depending both on altitude and geographical position relative to the earth's core. Thus, the 'g' will vary across the earth. But it is decided for all practical engineering reasons that g=9,81 uniformly. (Actually sometimes engineers use g=10 just to build in extra margins of safety and make calculations easier.)

Tidal water again, is explained by the gravitational forces distributed between earth and moon. Newton summed it up in his Law of gravitation:

F = G(Mm)/r^2

The moon will therefore exert some gravitational force on that part of earth that is closest to it. Since land does not move that easily, water is the main reason we notice this mutual force between out space home and its moon. The water is "attracted" to the force and moves in the direction of the moon which causes tidal flows.
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Old 14th Mar 2014, 13:28
  #113 (permalink)  
 
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MrSnuggles

"The moon will therefore exert some gravitational force on that part of earth that is closest to it. Since land does not move that easily, water is the main reason we notice this mutual force between out space home and its moon. The water is "attracted" to the force and moves in the direction of the moon which causes tidal flows. "

Don't forget the increase in centripetal action on the far side of the earth moon pair due to the increase in radius of orbit around the centre of earth mass that acts to reduce gravity causing the other "lump" of water.
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Old 14th Mar 2014, 13:30
  #114 (permalink)  
 
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And the tidal bulge on the other side is caused by...?
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Old 14th Mar 2014, 14:03
  #115 (permalink)  
 
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Are you asking him or me?
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Old 14th Mar 2014, 14:09
  #116 (permalink)  
 
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Him....you got there first. A couple of complications owing to the cm of the system being within the earth (3000 miles from centre) and the earth not revolving owing to this motion ie not a dumbbell which makes it hard to visualise as the cm of the earth moves about the common cm.
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Old 14th Mar 2014, 14:38
  #117 (permalink)  
 
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Oggers, I'm sure you are as capable of using a search engine now as you were six hours ago when I answered your question.

You are an anonymous lay-person
So you say. Once again, your knowledge and intuition appear to be letting you down. I'm not sure how posting under a name is in some way anonymous?

You're right that I'm probably not going to fix wreckage positions anytime soon, but I could if I was asked to, as could pretty much anyone with even a first degree in physics.

I don't understand why you've been trying to pick a fight with me about physical reality for a couple of days.
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Old 14th Mar 2014, 15:00
  #118 (permalink)  
 
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But I and many others do understand....
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Old 14th Mar 2014, 15:01
  #119 (permalink)  
 
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A Squared, yes it was considering the variation with height that is the guys undoing. As for temporal variation I was trying to figure out if that was on anyone's mind.
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Old 14th Mar 2014, 15:05
  #120 (permalink)  
 
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Originally Posted by awblain View Post
Oggers, I'm sure you are as capable of using a search engine now as you were six hours ago when I answered your question.
Well without intending to take sides, I have to say that oggers' ad-hominem approach regarding publications is a spectacular failure.
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