Fixed Earth Opinion: A Closer Look

[Alkatran]

I've been reading this website for the past half hour.

A lot of this guy's evidence just... well it doesn't make much sense. Take his statement that since the Earth isn't going around the sun, parralax will give faulty results. Well, gee, I guess somebody measured something from the same spot on the planet twice (across a half a year) and found their distance to be infinity? But I never heard about it.

 

[anti_crank]

Wow. I never thought I'd see the words "lost cause" burned into my mind like that - and I've been in TD!

 

[Gza]

His use of an exclamation mark after every other sentence also has a way of detracting from his already non-existant credibility.

The Earth isn't moving!
The Big Bang never happened!
The Bible was right all along!

 

[Alkatran]

His use of an exclamation mark after every other sentence also has a way of detracting from his already non-existant credibility.

The Earth isn't moving!
The Big Bang never happened!
The Bible was right all along!

My personal preference is the CONSTANT allusion to holy books as "proof".

I've seen a site made by someone who believes he is jesus. The problem is, he curses every third word and is clearly ... well I won't curse here.


*edit* Man who thinks he's jesus. You have to go through three title pages to get to the 'very well' layed out site.

 

[Locrian]

Yes, I'm familiar with this one. Been looking at it for a couple of weeks. I'm 99% sure some of our friends spamming TD forum recently got their information from here.

He does link to another interesting argument though. Another site he links to claims you can prove stars cannot be far away by showing that since the energy density will fall off from the star as per the distance squared, by the time it reached us from many light years away there would be too few photons entering the retina to be detected by the human eye. I'm sure its bunk, but since I know nothing about the human eye, I can't make an argument... yet.

I'll see if i can dig through his mess to find that link.

 

[Locrian]

Here it is:

http://www.midclyth.supanet.com/page6.htm

I must admit, I've only skimmed it.

 

[Adam]

That's beautiful. I love it. Thanks for the link. :D

 

[Locrian]

I just did some very rough calculations.

He states (i'll play on his field when it comes to the human eye) that you need minimum 5 photons to touch an area of the eye whose diameter is 1.5 micrometers. This constitutes an area of $$1.76x10^-^1^2m^2$$. Given an approximate energy of light withing visible range to be about $$5.5x10^-^1^9 J/photon$$, converting joules to ergs (cause that's what my solar power emmision is in) and dividing I get an energy density requirement of 15.625 ergs/m^2. (convenient!)

Doing a rudimentary search on the web someone gave the energy output of the sun to be $$3.86x10^3^3 ergs/sec$$. The energy density of spherically emitted light falls off as the surphace of a sphere, or $$4(pi) r^2$$.

Solving for r gave me over 105 million light years. I did this rather quickly, so I'm sure there are errors... but unless they are methodological, I think this shows that a star like our sun can be seen from rather far away.

So by my estimates, we could see a star the size of the sun from 105 million light years away. By his it is less than 2 light days. I wonder what he did to make his answer so tiny.

Edit: for some reason my latex looks fine in the preview, and not in the post :/... ah well, hopefully still legible.

 

[enigma]

By his it is less than 2 light days. I wonder what he did to make his answer so tiny.

Either inconsistant units

or

Deliberate falsification

would be my guess.

 

[Alkatran]

I just did some very rough calculations.

He states (i'll play on his field when it comes to the human eye) that you need minimum 5 photons to touch an area of the eye whose diameter is 1.5 micrometers. This constitutes an area of $$1.76x10^-^1^2m^2$$. Given an approximate energy of light withing visible range to be about $$5.5x10^-^1^9 J/photon$$, converting joules to ergs (cause that's what my solar power emmision is in) and dividing I get an energy density requirement of 15.625 ergs/m^2. (convenient!)

Doing a rudimentary search on the web someone gave the energy output of the sun to be $$3.86x10^3^3 ergs/sec$$. The energy density of spherically emitted light falls off as the surphace of a sphere, or $$4(pi) r^2$$.

Solving for r gave me over 105 million light years. I did this rather quickly, so I'm sure there are errors... but unless they are methodological, I think this shows that a star like our sun can be seen from rather far away.

So by my estimates, we could see a star the size of the sun from 105 million light years away. By his it is less than 2 light days. I wonder what he did to make his answer so tiny.

Edit: for some reason my latex looks fine in the preview, and not in the post :/... ah well, hopefully still legible.

For some reason he's talking about photons per volume instead of photons per surface area on the first page. I would wager that has something to do with it.

 

[Locrian]

Yea, and he brings up movie cinematography... though why I don't know. I think it's safe to assume for the argument that the energy output is reasonably constant, meaning the amount of energy reaching your eye per unit time is reasonably constant. What could he mean? I must not be reading carefully enough.

Ahh, I didn't include affects of the atmosphere. I'll see if I can fix that a bit later. I don't think it is going to change anything by a factor of a hundred million though.

 

[Locrian]

Woa, I'm wrong.

That wasn't 105 million light years, that was 105 million light seconds. How embarassing.

A more carefull examination of the process I used produced a visual range of less than half a light year. This is not including atmospheric effects. This means that we shouldn't be able to see any star the size of our sun if it is more than half a light year away.

This isn't necessarily a strange conclusion. I mean, I used the same assumptions he did, and am now getting a similar result. So the question is, what assumption is incorrect? I feel comfortable saying the energy density from the light falls off spherically. Could it be something involving the human eye?

Sorry for the error. I'm sneaking this in at work and didn't have a chance to take the time to deal with it as carefully as I should have.

I'm going to go over this again more carefully. Someone else help me and see what kind of result they get. Finally, we have a (seemingly) reasonable and mathematically based argument for a small universe, and I'm itching to fight with it.

 

[Alkatran]

Ok, here's a good one from fixedearth:

Somewhere between the two bodies the pull of one body has got to be overcome by the pull of the other one. At this point (which NASA says and von Braun confirms averages c. 216,000 miles from the Earth) there is a neutral gravity zone where the Earth’s gravitational pull becomes weaker than the Moon’s gravitational pull and, at this same point, the Moon’s gravitational pull becomes weaker than the Earth’s gravitational pull.

This circumstance is demanded by the gravity concept and all the supposed facts and laws that describe the Earth and Moon gravitational fields. OK, Vern? Now keep that in mind as we read what "science" says about Tides on Earth...

What "science" says is that the Moon’s gravitational pull reaches out thru the neutral gravity zone (where it couldn’t attract a balloon full of feathers, a dead horse, or the Astrodome) and goes on beyond 50,000, 100,000, 216,000 miles and sucks the big ole Earth toward it like stretching a rubber ball! I know it’s incredible; I know it’s insane, but that’s what the naturalistic (i.e., "scientific") explanation demands that the Moon must do. Read it for yourself from these examples (which are repeated in every book on the subject):

Someone needs to explain to him that the moon pulls more on the "front" oceans than the earth, and more on the Earth than the "back" oceans. His argument is that gravity must "STOP" as soon as it hits his "neutral zone". If that was true... the moon would fly off.

 

[Alkatran]

And more:

Branley then gets into some numbers which grow out of the myth he believes, numbers so contradictory as to be comical if the world weren’t mesmerized by this deception:

"The tide-raising force of the moon is very small indeed, compared

to the force of gravity. The tide-raising force of the moon is about

1/9,000,000 that of the earth’s gravity..."28

Wow. One nine millionth. Let’s see; that means the earth’s pull toward its own center on its own crust and oceans and seas is nine million times as great as the Moon’s pull on the same features, doesn’t it? Nine million to one. Whew! Amazing, isn’t it that nine million mule power could be pulling something one way and one mule power could overcome all that every hour of every day and stretch the Earth and cause zillions of cubic feet of water to shift around everywhere! Yes, just amazing, especially when that one mule power stopped out there at the 216,000 mile neutral gravity zone! Momma mia...

I have no idea of the accuracy of those numbers, but he doesn't understand the concept that the ocean is being crushed into itself. Changing the force downwards by even a small amount is obviously going to lower the pressure and... raise the water! :surprise:

 

[Alkatran]

It just keeps coming! Here he explains how familiar he is with Kepler's laws.

Even if one can momentarily lock one’s brain into conceiving of a delicate balance between say the Earth and the Moon’s gravitational forces, a balance that would be achieved by the most precise, exact and unvarying distance between the two bodies, then that same brain is boggled when it is confronted with the fact that no such stable distance exists between these two bodies (or any other two!). Indeed, the undeniable reality is that the moon regularly varies its distance from the Earth by over 31,000 miles! When it comes closer and closer it gets in the stronger and stronger pull of Earth’s gravity. How can it then resist that pull and start going against that attraction? Contrariwise, as it goes out to the apogee and is moment by moment breaking loose from Earth’s gravitational pull at tremendous speed, how can it stop the outward movement and start back??

Gravity doesn’t explain this. Gravity can’t explain it. Gravity doesn’t explain the tides. Gravity can’t explain them. The same is true of the Earth’s supposed annual orbit around the sun. The simple fact is that we are closer by three million miles to the sun at certain times than we are at other times.

 

[K.J.Healey]

Maybe I am doing my calculations wrong but I get:

Assuming the sun at 105,000 LY :

Intensity is P / (4*Pi*R^2)
105,000 LY = 9.93355e20 meters
Power of the sun is 4e26 Watts

So power of the sun from 105,000 LY(cornea) = 32154.34 W/m^2 Intensity
Thats intensity at CORNEA.
Power at cornea would then be Icornea*PupilArea
Pupil Diameter is 2mm
So P_cornea = 0.1015e11 W
Then the Retina intensity is the P_cornea over the sopt area, which is about 150 square microns.
P/SpotArea = Retina Intensity = 0.1437e18 W/m^2 which is WELL above what we can see. My eye sensitivity measurement may be wrong, but we can see that thanks to the lense in our eyes it increases the intensity by 8 orders of magnitude. I think this was not in the calculations he did. The stars are relatively low intensity, which makes it hard to see reflections. I think I am right.

 

[Locrian]

I don't think you squared your radius, or something. Using your numbers I get 3.2x10^-11 W/m^2

In any case, I can see without calculating that 4e26/(9.9e20)^2 is going to be way, way less than one.

 

[K.J.Healey]

Lets see :

Assuming the sun at 105,000 LY :

Intensity is P / (4*Pi*R^2)
105,000 LY = 9.93355e20 meters
Power of the sun is 4e26 Watts

> I_pupil := (4*10^26)/(4*3.14159*(9.93355*10^20)^2);

I_pupil = .3225830396 10^(-16)


Pupil Diameter : 2mm = 2*10^-3 m
Pupil Area :

> Pupil_area := (3.14159265)*((2*10^(-3))/2)^2;

Pupil_area = .3141592650 10^(-5)

I_pupil*Pupil_area = Power_Pupil

> Power_Pupil:=I_pupil*Pupil_area;

Power_Pupil := .1013424506 10^(-21)

SpotArea = Pi*(150microns)^2

> SpotArea:=3.1415926*(150.*10^(-6))^2;

SpotArea := .7068583350 10^(-7)

I_Retina = Power_Pupil / SpotArea

> I_Retina = Power_Pupil/SpotArea;

I_Retina = .1433702421 10^(-14)


Thats in Watts/ m^2

Yea, i did bad calculations. Weird.

 

[The Binary Monster]

The Earth isn't moving!

I almost hate to say this, but... don't laugh at the guy. Well, not for that argument, at least.

Nobody here can prove that the Earth is moving. Or at least not that any particular point of it is moving.

If you can, I'll be interested to hear from you. I believe that relativity says that "all observers, regardless of their state of motion, may proclaim that they are stationary and 'the rest of the world is moving by them,' so long as they include a suitable gravitation field in the description of their own surroundings." (quoted from Brian Greene). So yes, if you can prove any particular point on Earth is moving, congratulations. You'll have just caused a lot of trouble in the world of physics by disproving (or at least contradicting) relativity.

 

[Alkatran]

I almost hate to say this, but... don't laugh at the guy. Well, not for that argument, at least.

Nobody here can prove that the Earth is moving. Or at least not that any particular point of it is moving.

If you can, I'll be interested to hear from you. I believe that relativity says that "all observers, regardless of their state of motion, may proclaim that they are stationary and 'the rest of the world is moving by them,' so long as they include a suitable gravitation field in the description of their own surroundings." (quoted from Brian Greene). So yes, if you can prove any particular point on Earth is moving, congratulations. You'll have just caused a lot of trouble in the world of physics by disproving (or at least contradicting) relativity.

You can't detect your own speed, but you can detect acceleration. This how you prove wether or not the Earth is moving. (It must be, since it is accelerating!)

 

[The Binary Monster]

You can't detect your own speed, but you can detect acceleration. This how you prove wether or not the Earth is moving. (It must be, since it is accelerating!)

Technically that's not true... It may not be accelerating, it may be subject to a gravitation field which makes it appear to be accelerating... and no-one can prove that it isn't.

 

[Gza]

I almost hate to say this, but... don't laugh at the guy. Well, not for that argument, at least.

Nobody here can prove that the Earth is moving. Or at least not that any particular point of it is moving.

If you can, I'll be interested to hear from you. I believe that relativity says that "all observers, regardless of their state of motion, may proclaim that they are stationary and 'the rest of the world is moving by them,' so long as they include a suitable gravitation field in the description of their own surroundings." (quoted from Brian Greene). So yes, if you can prove any particular point on Earth is moving, congratulations. You'll have just caused a lot of trouble in the world of physics by disproving (or at least contradicting) relativity.

You seem to be a bit confused on the concepts of circular motion and acceleration(we're not even talking relativity yet ) Since every point on the Earth is undergoing circular motion, there is an acceleration directed inward toward the center (the centripetal direction.) If you were to hold a pendulum steady on the surface of the Earth, you would find a deviation in the angle it makes with the ground(actually so small, you wouldn't notice it with your eyes alone.) This deviation is caused by virtue of the pendulum being in circular motion about the axis of the Earth, and thus experiencing a centripetal acceleration. So I guess I've brought the world of physics to its knees .


EDIT: Just saw your post above

Technically that's not true... It may not be accelerating, it may be subject to a gravitation field which makes it appear to be accelerating... and no-one can prove that it isn't.

Yes but the acceleration due to gravity had already been accounted for when calculating the deviation of the pendulum. The pendulum would point straight down if gravity was the only force acting (not to say that the centripetal acceleration of the pendulum is caused by some force, its just a circumstance of being in an accelerated reference frame, in this case the Earth undergoing circular motion.)

 

[The Binary Monster]

You have indeed! ...or at least just enlightened me, and corrected me a little. Whoops.

Since every point on the Earth is undergoing circular motion, there is an acceleration directed inward toward the center (the centripetal direction.)

Would this not also be able to be explained away with a suitably complex gravitational field? I was alway brought to understand that any observer could always claim to be stationary if a suitable gravitational field was taken into account.

 

[Gza]

Would this not also be able to be explained away with a suitably complex gravitational field? I was alway brought to understand that any observer could always claim to be stationary if a suitable gravitational field was taken into account.

Although you are correct in application of the equivalence principal, as you mentioned, trying to explain the centripetal acceleration as actually an acceleration due to gravity would not be trivial. The g field would be a function of longitude, and would be needlessly complex. It would be much easier to simply "separate out" the accelerations.

 

[enigma]

I almost hate to say this, but... don't laugh at the guy. Well, not for that argument, at least.

If the Earth is stationary, then the universe is rotating about us once every 23hr56min.

If the universe is rotating once every 23hr56min, then distant objects would have to be going many, many times the speed of light to return to the same spot that fast.

QED

 

[The Binary Monster]

Okay, okay, okay. I lose. That'll teach me to try and be big, clever, and make things up. I stand corrected for most if not all of my posts so far... Thank you for pointing out the flaws, even so. I feel like I learn through these stupid mistakes. :)

 

[physicsuser]

Why don't we put a camera up in space so that it can show us the solar system or at least part of it?

 

[Alkatran]

Why don't we put a camera up in space so that it can show us the solar system or at least part of it?

How would that prove anything either way? They would just argue something about the camera being rotated...

I think the best argument is: the moon is exerting a pull on the earth. This means the Earth's speed is always changing. That means the Earth's speed is never always 0, in any frame. If it isn't always 0, it isn't stationary.

Unless they want to reject gravity?

 

[The Binary Monster]

Or they want to make something up about equal and opposite forces to the moon's pull.

 

[Alkatran]

Or they want to make something up about equal and opposite forces to the moon's pull.

That's the reason it can't be true.