Science a GoGo's Home Page Forums General Discussion General Science Discussion Forum Mansfield's Earth Formation Hypothesis: Evidence.

ImagingGeek,

I am declaring a red herring alert for your last post. Let’s start with your links that supposedly invalidate a change in surface gravitation.

The ‘Limits to the expansion of Earth, Moon, Mars and Mercury and to changes in the gravitational constant’ clearly purports to invalidate the Earth Expansion Theory, which I do not support. You stated:

“The first of those papers directly measured paleogravity at several sites on the earth. Had your little magical gravitational change happened they would have observed it. Instead they found that the force of gravity remained constant, throughout the earths history, at the sites they tested. Given that they tested site which were part of pangea, that's a pretty big hole in your hypothesis.”

The first of the papers did not measure paleogravity, it measured PALEOMAGNETISM. If you don’t know the difference, we’re in real trouble. Only measurements of paleogravity can deduce the size of the Earth in the past. That eliminates that link.

The second link is still not working. When I copy/paste the URL at the top of the screen, nothing happens. How about summarizing the website in a paragraph with direct quotes supporting your position. Or, better yet, supply a link that works.

I’ll repeat my statement:
“In my prior post, I proved that you were solving the wrong problem. You were solving the problem as though the Earth was not spinning.”

If you go back and reread your statements, even though you used the term COR, you were solving the problem as though the Earth were not rotating or rotating at a very slow rate, like the moon. In other words, you were ignoring the moment of inertia (i.e., the rotational mass) of the core(s)/Pangea.
This is why you came to the erroneous conclusion that equilibrium would be established when the tiny center of mass shift of the Earth due to the consolidation of Pangea would be offset by an equal center of mass shift due to the wobble exerting a reaction-force on the core(s).

You claim that I’m correcting you with lies???? Signs of desperation.

My example for a 54% change in surface gravity on Pangea (i.e., it would have been 54% of current “g”) was based on a shift of the center of mass of the Earth from the current center by a distance of one sixth of the diameter of the Earth. Considering we would be dealing with the shift of the inner core, outer core and the densest part of the mantle, I don’t find this to be unobtainable. And yes, my r^2/d^2 (remember ‘d’ is not diameter but distance from Pangea’s COM to the new COM of the Earth) was derived from Newton’s law.

Laze

Just correcting my statement:

"The first of the papers did not measure paleogravity, it measured PALEOMAGNETISM. If you don’t know the difference, we’re in real trouble. Only measurements of paleogravity can deduce the size of the Earth in the past. That eliminates that link."

Should read:
Only measurements of paleomagnetism can deduce the size of the Earth in the past.

Laze

And Laze confirms once again that he did not read the paper, and that he also has limited understanding of the implications of findings such as these.

The angle of the crystals they use to measure the strength of paleomagnetism is dependent on the local gravitational force. The local gravitational force is dependent (obviously) on the amount and distribution of mass below a test site. From this we can come up with three simple hypothesis:

1) If the mass of the earth increased/decreased over time, this would be apparent as universal changes in paleomagnetism, appearing as a larger/smaller radius, or
2) If the earth had periods of time with areas having significantly different levels of surface gravity, than we would see this as local differences in the paleomagitism measured at different points on the earth, or
3) If the earths surface gravity has been continious over time, this will be apparent in the form of a constant G, and thus constant radius.

As you can see in table 1 of the paper, no significant variations of paleomagnitism were seen in any of the >200 test sites, which are scattered all over the globe. Ergo, there was never any significant deviations in the local gravitational fields at those sample locations.

In fact, this paper puts a cap on the size of shifts that could have occurred, as larger shifts would have been identifiable through the statistical noise. This particular paper puts this at 4-7%, depending on the geological period under question (Table 2).


Works for me, try clicking this link.

Basically, this paper analyzed the paleoorbit of the moon, using tidal deposits as proxy data for the period of rotation and height of tides:

"Hence the thickness of successive laminae deposited by tidal currents can be a proxy tidal record, with paleotidal and paleorotational values being determined by analysis of measured records of lamina and cycle thickness. "

Tidal heights are determined solely by the ratio of lunar gravity to local gravity. The higher the earths gravitational pull at the measurement site, the smaller the tide.

Once again, was your model correct they would have seen changes outside of those caused by the moons regression, as their sample site was part of pangea. This is not the case (see table 1, Figs 10 & 15).


Really? Perhaps you can show me exactly where I made that assumption.

Oh wait, you cannot show me where I made that assumption, as I never did. This is simply yet another excuse by you to not address the faults in your math.


And yet, a basic understanding of the universal law of gravitation shows it to be impossible. I'd point out again that I solved using the law of gravity, to calcluate the delta (change) in Fg, and I was unable to replicate your result and you've been unable to show my math to be wrong.


And, as I pointed out several times before, your formula only gives the correct answer if you shift the enter mass of the earth, meaning it is completely useless for measuring the gravitational shift when only the core moves.

I'd also point out that your formula ignores all of the things you claim I ignore - rotation of the earth, distortion of the core, etc.

Bryan

ImagingGeek,

You wrote:
“The angle of the crystals they use to measure the strength of paleomagnetism is dependent on the local gravitational force. The local gravitational force is dependent (obviously) on the amount and distribution of mass below a test site.”

You’ll have to explain how the “angle of the crystals” has any relationship to local gravity.

You then listed three hypotheses:
“1) If the mass of the earth increased/decreased over time, this would be apparent as universal changes in paleomagnetism, appearing as a larger/smaller radius, or
2) If the earth had periods of time with areas having significantly different levels of surface gravity, than we would see this as local differences in the paleomagitism measured at different points on the earth, or
3) If the earths surface gravity has been continious(sic) over time, this will be apparent in the form of a constant G, and thus constant radius.”

Again, you have to explain what you believe to be the relationship between paleomagnetism and paleogravity. You also seem to be trying to disprove the Expanding Earth Theory with your references to changes in “radius” and “mass.” The theory we are evaluating requires no change in radius or mass.

You wrote:
“As you can see in table 1 of the paper, no significant variations of paleomagnitism were seen in any of the >200 test sites, which are scattered all over the globe. Ergo, there was never any significant deviations in the local gravitational fields at those sample locations.”

At the risk of sounding like a broken record,
Again, you have to explain what you believe to be the relationship between paleomagnetism and paleogravity.

Your second link works this time. However, I don’t believe your conclusion that variations in sedimentation thickness due to tidal variations can measure changes in surface gravity on the Earth.

You wrote:
“Tidal heights are determined solely by the ratio of lunar gravity to local gravity. The higher the earths gravitational pull at the measurement site, the smaller the tide.”

This is your opinion (i.e., what is local gravity?). Remember that the current theory posits a gravitational gradient; lowest “g” at Pangea’s COM and gradually getting higher toward both poles; and highest at the antipode. Since the moon would be facing significant areas of varying values of “g” on the Earth continuously as the Earth spinned, any attempt to come up with meaningful data on tidal variations would be fruitless.

In response to my example of a 54% change in “g” produced by a shift in the Earth’s COM by a distance of 1/6 th of the diameter of the Earth, you repeatedly write:

“And, as I pointed out several times before, your formula only gives the correct answer if you shift the enter(sic) mass of the earth, meaning it is completely useless for measuring the gravitational shift when only the core moves.”

Again, as I have repeated many times, not only both inner and outer core move but also the densest part of the mantle that surrounds the cores also moves further from Pangea. And yes, the “entire mass of the Earth” (i.e., the COM of the Earth) must shift.

Your previous calculation is totally incomprehensible:
Fg = G*m1*,2/r^2
0.56G = 1/r^2
r = 1.33631 earth radi


Finally, you wrote:
“I'd also point out that your formula ignores all of the things you claim I ignore - rotation of the earth, distortion of the core, etc.”

Red herring alert! My formula is very clear....and simple. The formula simply states that ratio of the local “g” at any point on Pangea to today's "g" is equal to r^2/d^2, where “r” is the radius of the Earth and “d” is the distance between any point on Pangea and the shifted COM of the Earth. It doesn’t have to explain how the outer core becomes distorted or anything else.


Laze

You could try reading the paper, or reading on the relationship between gravity and magnetism...

That said, it isn't rocket science. As paramagnetic crystals form in molten rock they orientate such that they align with the local magnetic field. Gravity bends magnetic fields, therefore the angle of these crystals can be used to determine the force of gravity bending those magnetic lines.


Already stated, and described in detail in the paper I provided.


Exactly. There are three possible options in terms of the paleogravity record - consistent change over the entirety of the earth (i.e. expanding earth), local changes (your hypothesis) or no changes (i.e. you and the expanding earth hypothesis are wrong).

The record shows the later - no change. Therefore neither your, nor the expanding earth, hypothesis are correct. I would be remiss to not consider all possibilities - which is why I included the expanding earth option.


What you believe is irrelevant, it is what you can prove that matters.

I provided scientific evidence in two separate forms which refutes your claim. Your only reply has been "I don't believe these papers are right".

You can either find evidence they are wrong, or the only logical thing to do is to concede that those papers refute your hypothesis.


No, that is basic physics. The height of a tide, relative to mean sea level, is determined by the force of gravity pulling down on the water (i.e. the gravity in the local region) and the gravitational force pulling the water upwards (determined by the distance between the moon and earth, in the case of lunar tides).

Keep in mind we use variations in the earths tidal forces on orbiting satellites to map existent gravitational anomalies (used to map the sea floor). Given that you're proposing changes in gravity approx seven orders of magnitude larger than the ones we observe today, the tidal record should have recorded those changes.

So if your hypothesis was correct, the lower gravity at pangea would have resulted in much larger tides during that time period (approx. double the height) - instead we see continually shrinking tides on a pangeal tidal flat.


But what moves is irrelevant; its how far it has to move that invalidates your hypothesis.

Lets assume that 50% the mass of the earth is shifted away (i.e. the 32% that constitutes the cores, plus a whopping portion of the denser mantle). To get your 54% reduction in gravity, ignoring the gravity of the lower-density backfill, would require a shift of:

Fg=0.5/r^2
0.54 = 0.5/r^2
r = sqrt(0.5/0.54) = sqrt(0.9259) = 0.96 earth radi

Or, in otherwords, to get your 54%G via a core/mantle shift, you would have to move the cores COM to a few km beneath the oceans crust - as in you'd have a tens-of-thousands of km high bulge opposite pangea.

If you take into account the "backfill", and in order to get your 54%G would require that your remove the core, and dense mantle, completely from the earth.

That's a far cry from the 1/6th earth radius your "formula" calculates.


But that is not the issue with your formula.

The only time your formula gives the correct answer is if you move the entirety of the earth relative to the observer - i.e. calculate the Fg on the observer standing on the earth, verses being several thousand km above the earths surface.

So it will not work for the situation you are trying to apply it to - where the earths distribution of mass changes, but the observer and earth remain stationary in regards to eachother.

Bryan

ImagingGeek,

You wrote:
“Gravity bends magnetic fields..........”

How about some proof? Maybe Wikipedia? Please don’t provide an obscure 100 page pdf for me to wade through.

You wrote:
“Already stated, and described in detail in the paper I provided.”

Again, provide a specific, concise reference. If you can’t copy and paste the relevant info then your links are worthless.

When repeating your assertions that tidal effects can measure past surface gravity, which is total nonsense, you conveniently omit my initial response, which was:

Remember that the current theory posits a gravitational gradient; lowest “g” at Pangea’s COM and gradually getting higher toward both poles; and highest at the antipode. Since the moon would be facing significant areas of varying values of “g” on the Earth continuously as the Earth spinned, any attempt to come up with meaningful data on tidal variations would be fruitless.

You then add your traditional double-speak:
“I provided scientific evidence in two separate forms which refutes your claim. Your only reply has been "I don't believe these papers are right".

If you can’t explain what’s in links that you provide in a paragraph or two, don’t expect me to read through 20 pages or more to try to ferret out what you claim. This is just a stalling and diversionary tactic.

You wrote:
“Given that you're proposing changes in gravity approx seven orders of magnitude larger than the ones we observe today, the tidal record should have recorded those changes.”

Seven order of magnitude larger than today??? Show me where I stated that!!!!!
Again, your distorting what has been written is unconscionable.



You keep repeating your nonsensical math without any explanation:


Fg=0.5/r^2
0.54 = 0.5/r^2
r = sqrt(0.5/0.54) = sqrt(0.9259) = 0.96 earth radi

The above is totally meaningless! If this is the best that you can do, we are in trouble. Either provide a complete set of equations or admit that you are fudging it.

Laze

I already provided the citation - its in the first of those papers you still haven't bothered to read yet. That gravity determines the angle of a planets gravitational field lines is a very well understood phenomena, and is thoroughly discussed and cited in the forst paper I provided earlier.


And your evidence it is nonsense is...nothing.

Reality is that the hight of tides is directly determined by surface gravity, and thus tidal deposits can, and are, used as a proxy data for gravitational strength. Once again, the methodology and relevant citations for this fact are all throughly covered in the second paper I provided.

Just because you deliberately ignore that paper, and the citations within, doesn't change that reality one iota.


Actually, I took that into account. In short, its nothing but wishful thinking on your part. In your situation tidal effects would be very obvious in the record - as tides approach the lower G area they would increase in height - dramatically if we assume your 54% value to be correct. While you may not get an exact measure of G, the decrease in G (as observed in the form of an increase in tidal height) would be eminently obvious.

Once again, your wishful thinking does not eliminate the reality of science. At the end of the day, tidal heights are nothing more than a product of the relative gravitational force of the earth beneath the tidal bulge and the moon:
http://en.wikipedia.org/wiki/Tide#Physics


That isn't double-speak; that is a statement of fact.

I have provided two papers, which measure the force of gravity at the earth surface, over geological time. Both of them refute your hypothesis, and both of them extensively cite the scientific literature in support of their methods and results. Your response to those papers has continually been "my opinion is they are wrong", without one iota of evidence that is the case.


LOL, your continued insistence to not read those papers is a stalling and diversionary tactic. After all, if you were serious about refuting these papers, it would only take you a half hour or so to read them.

But you and I both know that all of this is an excuse to not read something that challenges your belief system.

And, I'd point out, that I've explained what is in these papers extensivly - even quoted directly from them - in several of my posts. For example, #36322, 36349 and 36274.

Now, seeing as your premise is false - I've extensively described the papers, quoted from them, etc, what is your excuse now for not reading them?


You didn't; math does (although 7 is an exaggeration on my part). You are claiming a decrease of 0.46G. The size of the anomalies measured today are as small as 5uGal. Earth standard gravity is 980mGal.

So you claim: .46*980 = 450.8mGal change
We can detect: 0.005mGal chage
Fold difference: 90,160, AKA ~5 magnitudes of order difference.


Distorting = using math. Call the presses! LOL.


Meaningless? I've shown the math since day 1. And quite ironically, I've made the same assumptions that you did in yours - that we can treat the earths mass and radius as constants. Otherwise, it is newtons law of gravitation. That said, I did make a mistake, and have corrected it below - note that I fix my mistakes; maybe you can learn something from that...

Now, let me take your hand and walk you through this...

Fg = GM1M2/r^2, newtons law of gravitation.

On earth we can "convert" all of these into units of earthlyness, i.e. Fg = 1 earth gravity, GM1M2 = 1 earth mass*G, r = 1 earth radius.

So for the earth, Fg = GM1M2/r^2
Fg = 1/1^2
Fg = 1G

For any component of the earth we can then use fractions of these values to solve for the relative contribution of that component. I.E. for the unshifted core + mantle you are shifting (which I assumed weighted ~50% an earth mass), we get:

Fg(core) = 0.5/1^2 = 0.5G

For a 0.01 earth mass block of matter, located 1.2 radi away from the observer on the surface:
Fg = 0.01/1.2^2
Fg = 0.00694 earth G's

Now here is my small mistake. You are claiming a pangea G of 54%, I used this value for Fg. But my error is that we want to calculate the change in earths Fg as the core moves, not just the Fg of the core.

So before the shift the core+moving mantle provides 0.5G worth of gravity. It moves, reducing (by your claim) gravity by 0.46G, giving a final surface G of 0.54G at pangea.

Assuming the shift in G is due solely to the movement of the core (i.e. no "backflow" filling the space left behind and whatnot), the contribution of the core must be reduced to 0.5-0.46 = 0.04G. So we need to calculate the radius at which the core+mantle's gravitational attraction is reduced from 0.5 to 0.04G:

Fg(start) = 0.5 = 0.5/1^2 (r = 1)
Fg(end) = 0.04 = 0.5/r^2 (r = ?)

So:
0.04 = 0.5/r^2,
r = sqrt(0.5/0.04)
r = 3.54 earth radi

Since the core started at 1R, this is a shift of 2.54 radi

My mistake didn't help you much...LOL.

And don't forget, this shift will be reduced further by the backflow of mantle into the void left by the core, as this will move mass towards pangea, thus adding additional gravity.

Bryan

ImagingGeek,

Your two references that purport to disprove the theory being discussed are:
---------------------------------------------------------------------------------------------------------------
STRIKE 1
‘LIMITS TO THE EXPANSION OF EARTH, MOON, MARS AND MERCURY AND TO CHANGES IN THE GRAVITATIONAL CONSTANT’

This conclusion of this study is:

“New estimates of the palaeoradius of the Earth for the past 400 Myr from palaeomagnetic data limit possible expansion to less than 0.8%, sufficient to exclude any current theory of Earth expansion.”

This paper, as I have repeatedly stated, provides evidence against the Earth Expansion Hypothesis by establishing narrow limits on the size of the Earth’s radius for the last 400 my. There is nothing in this paper that precludes variations in surface gravity due to shifting core(s).
--------------------------------------------------------------------------------------------------------------------

STRIKE 2
‘GEOLOGICAL CONSTRAINTS ON THE PRECAMBRIAN HISTORY OF EARTH’S ROTATION AND THE MOON’S ORBIT’

The conclusion of this paper is:
“The thickness of successive laminae deposited by tidal currents can be a proxy tidal record, with paleotidal and paleorotational values being determined by analysis of measured records of lamina and cycle thickness.”

Not only is the time period (i.e., Neoproterozoic ~620ma) not near the time of the existence of Pangea, but more important:
There are zero, no, nada references to the words “gravity” or paleogravity” in the entire pdf yet you insist that this paper invalidates the current theory. Absolutely no relationship is made, except by you, between paleomagnetism and paleogravity.
RED HERRINGS ABOUND!
---------------------------------------------------------------


STRIKE 3
Your reference to wikipedia in attempting to use tidal data to refute the current theory:

http://en.wikipedia.org/wiki/Tide#Physics

You should go back and reread it, it states:
“Tides vary on timescales ranging from hours to years due to numerous influences. To make accurate records, tide gauges at fixed stations measure the water level over time.”

And,

“The times and amplitude of the tides at the coast are influenced by the alignment of the Sun and Moon, by the pattern of tides in the deep ocean (see figure 4) and by the shape of the coastline and near-shore bathymetry.”

Got any tide gauges from 100mya?


STRIKE 4 (YOU’RE ALREADY OUT but I’ll let you keep trying)

On your claim that I had proposed a 7 fold increase in “g”, you stated:
“...7 is an exaggeration on my part.” From a “g” 1 to a “g” of .54 is close to a ½ decrease, some 14 times lower than you claim I made.
Yes, you do exaggerate!


Your calculations of the shift in COM are still based on erroneous assumptions. I’ll give you the problem to solve using a real-world scenario:

Tonight, you step on your bath scale and weight yourself. You get a reading of 100 lbs (you’re wondering how I knew you were a lightweight). Tomorrow morning you step on that scale and it only registers 54 lbs. You scratch your head and realize that surface gravity has decreased. The only explanation you can muster is that the inner/outer cores must have shifted. All you have is a pencil and paper and Newton’s Universal Gravity Law. You assume that the Earth’s COM has shifted away from you. That’s all you have, so now you figure out the ratio of the distance to the new COM vs. the old COM based on the reduction in your weight.

Let me know what you get as the answer.

Laze

And once again laze shows us he didn't read the paper.

You see, papers have data, and from that data you can answer questions other than the one the paper specifically addresses. In this study, these individuals took paleogravity measures of G from hundreds of sites scattered all over the world. This data directly refutes your hypothesis.

It's simple - if you were correct they would have observed variations in their measured G across the globe. Instead, their data clearly shows a consistent 'G' across the globe, throughout geological history.

Ergo, the paper directly refutes your claims.


Nothing except:
Table 1, which provides the mean and SD of 'G' measured at the varying sites across the globe based on magnetic field line angles.

Table 2, which provides the mean and SD of 'G' measured based on groupings of the varying sites across the globe

And on page 320 (page 5 of the article PDF) the formulas used to calculate radius are measured - and, had you read the article, you'd have noticed that R is calculated based on the changes in surface gravity (dG/dT) measured from the paleomagnetic data.

In otherwords, the "strike" is yours - the radius's they calculated were determined from the measured force of gravity at the various sites they analyzed.


Not working those critical thinking skills much, are we?

I'd first direct you to figure 2 - notice the time scale? Notice the data regarding the very time period we are discussing? Its ironic that you claim to have read the paper, while missing that one of the major points of the paper is the validification and complementation of the methods used to produce figure 2...

And , as with the last paper, the specific conclusion this paper is making is not the same as the one we are asking. But none-the-less, the data within the paper also answers our question.

Now I realize critically analyzing data isn't your forte, but first take into consideration the relationship between gravity, tides and momentum. I realize you're looking desperately for excuses to ignore this paper - like your wikipedia fail below, but none-the-less, that physical relationship is well understood.

Now, the momentum of the earth/moon system is essentially conserved (an assumption shown to be correct on page 55 of the journal; page 19 of the pdf). Ergo, any changes in tidal height must assume a constant momentum. Therefore, the only factor that can lead to broad-scale changes in the tidal heights at a set local is the relative gravitational force between the moon and the earths gravity at the site in question.

Here we have a site, situated on what becomes pangea. And there is no evidence of gravitational changes at that site; only evidence of the moon moving outwards. There is, in figure 2, simular observations acquired from numerous other sites, covering other points in history (including the de facto pangea), also showing no evidence for changes in earthly gravity.

What do YOU conclude from that? The correct conclusion is, of course, that strike 2 also belongs to you...


I direct your attention to figure 5, showing the near-shore bathymetry. And I direct you to figure 10, showing the patter of tides in the regional ocean. And lastly, to table 1, providing information as to the sun-moon alignment, orbital periods, day-length, etc.

In other words, we have all the data that wikipedia correctly states we need to calculate tidal heights. So, given that we have that data, what do you think we can do with it?

LOL, that's a pretty big fail on your part - proving in one quote that you neither read/understood the paper, or what wikipedia was saying.

So once again, strike 3 is also yours.


And now Laze is relying on lying to make his point...always the last tactic of the pseudoscientist when faced with the physical impossibility of their claims.

Simply put, I never made the above statement. What I actually said was:
"Keep in mind we use variations in the earths tidal forces on orbiting satellites to map existent gravitational anomalies (used to map the sea floor). Given that you're proposing changes in gravity approx seven orders of magnitude larger than the ones we observe today, the tidal record should have recorded those changes."

Post #36349

Pretty clear what I said. And in post #36363 I clarified what was said, since you couldn't figure it out on your own:
"You are claiming a decrease of 0.46G. The size of the anomalies measured today are as small as 5uGal. Earth standard gravity is 980mGal.

So you claim: .46*980 = 450.8mGal change
We can detect: 0.005mGal chage
Fold difference: 90,160, AKA ~5 magnitudes of order difference."

I'm assuming that catching you in a bold-faced lie counts as a strike. I'd also like to assume that such on obvious demonstration of reading incomprehension is also worth a strike...


If that were truly the case, you'd be able to list those erroneous assumptions. Since you didn't we can only conclude that once again, you are lying through your teeth.


I've done this before, and once again you didn't provide enough data, so I'll assume the core + mantle that has moved is 50% the mass of the earth. For simplicity I'll assume the gravitational contribution of the rest of the earth remains constant (keep in mind this will underestimate the size of shift you will need - backflow of mantle would increase local G relative to what is calc'd here):

Fg(start) = 100lbs (I'm flattered, btw) = 45kg * G = 441N
Fg(end) = 54lbs = 25kg * G = 245N
Me = 5.98e24kg
Mcore = 0.5*5.98e24kg = 2.99e24kg
Fg(unshifted core) = 0.5G

Portion of our starting Fg determined by the core = 441N * 0.5G = 220.5N

What this has to be reduced to, to reduce Fg(end) to 245N:
delta(Fg) = dFg = 220.5N - (441N-245N) = 220.5N - 196N = 24.5N

So the core has to move to a location where it provides 24.5N of force.

24.5N = G*Me*M/r^2
24.5N = [6.674e-11 * 2.99e24kg * 100kg]/r^2
24.5N = 1.995e16/r^2
r = sqrt(1.995e16/24.5)
r = 2.853945354e+7m = 28539km

r(earth) = 6371km

therefore, the core must be shifted (28539-6371)/6371 = 3.48 earth radi.

Bryan

In other words, to get this shift in my weight, the contribution of the gravitational force of the core must be reduced from 220.5N to 24.5N

ImagingGeek,

I asked you to solve a simple problem, one that any high school student could solve. Yet, you claim:

"I've done this before, and once again you didn't provide enough data..."

There is enough data for you to solve the problem without making any other assumptions. Instead, you would rather make invalid assumptions to come up with an answer that you want. Go back and reread Laze's Assumptions.

Assuming a value of .5 of the mass of part of the Earth in your calculation is your biggest error. Without knowing what the effective mass of the densest part of the mantle is, relative to the upper, less dense mantle, your assumption of .5 is erroneously misleading. Even you, in an early post, suggested that the near-core mantle might be too viscous to allow core movement, yet you ignore this when estimating total mass movement.

First, solve the problem I gave. If you can't do this, you don't understand Newton's Gravity Law.

Laze

ImagingGeek and his sock-puppet Laze certainly did take this thread of on a tangent, didn't they?

Hey preearth,

Funny thing.........I was starting to think that you were ImagingGeek's stocking-puppet.

Haven't heard much from him but if you want to fill-in for him, welcome aboard. Or would that be re-aboard?

Laze

Hi Laze;

21 out of your 21 posts at scienceagogo.com are in this thread. Not a single post to any other thread.

You disappear for a few months, from Oct 1st till now, but are still able to reply within an hour of being "called".

This is all evidence that you are a sock-puppet, in fact, probably one of the people who run/control this website.

Hey preearth,

The only reason I knew you referenced me in your post is because I have a cadre of sock-puppets monitoring the web.
You are now officially on their Watch-List!

Laze

Pretty much every one of the Geek's comments in the 2nd post (#35758) of this thread, is wrong. After weeks of argument with the Geek, something akin to bashing one's head against a wall, just to prove his first claim wrong, his line 3, I decided I couldn't be bothered any more.

However, I recently decided that I should once again take up the task of pointing out his errors.

We move on to his lines 5 & 6.


This is another example of the Geeks misleading replies

In fact, there is no smiley that can express my derision at the unscientific nature of the Geek's reply.

POINT ONE; Even though the Geek ignores it, the velocity of the projectile is important. For example, a bullet travelling near the speed of light would split the apple into subatomic particles. The Geek shows you what a high velocity bullet does to an apple.

Here, is a video of a slightly slower bullet hitting an apple;

http://www.youtube.com/watch?v=zV4X_O_Rrrs

where most of the skin is unscathed (i.e., still recognizable as skin and, in this case, the unscathed skin has not even split into pieces).

Another example of this type of thing, is a bullet hitting a human skull. The bullet rips through the skull, leaving most of the skull (and skin, etc) around the entry wound unscathed,... Same basic idea.

POINT TWO; The Geek's claim that "Unscathed skin is nowhere to be found." is, of course, false.

Even in this instance of a high velocity bullet, the apple skin is still recognizable as apple skin. It has just been split into pieces, just as the continental crust was still recognizable as continental crust, and was just split in pieces (called continents, which then expanded apart).

Preearth, this is not a criticism, it is a genuine attempt to learn.
I find myself wondering to what extent a bullet hitting an apple, or a human skull, is analogous to a scenario in which two objects of similar mass and composition collide. The importance attached to this analogy makes me suspect that there is a greater similarity than I can detect.

What are you talking about?

Hardly any importance is attached to this analogy.

It just makes it easier to understand what is happening.

This is what I was talking about: " find myself wondering to what extent a bullet hitting an apple, or a human skull, is analogous to a scenario in which two objects of similar mass and composition collide".



It has figured quite largely in your threads. Is that not a sign that some importance is attached to it?



I find it difficult to accept that it can make anything easier to understand if you do not show the validity of the analogy.

agree

Perhaps, if you say why you agree, it might help with the general understanding. I don't seem to be getting my point across.