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##### Maths
« on: May 31, 2014, 04:00:33 PM »
Now... I have posted a few things under ask a flat earth theorist anything, and I'm pretty sure I won't get an answer. So I thought I'd just go ahead and post the last 2 under the debate board. I am trying to avoid anything that could be viewed as flaming. Which is quite fun. I am simply doing a bit of maths, and looking at what it tells me.

So... The answer I got was the earth is accelerating at 9.8ms^-2, producing the same effect as gravity.
*Facepalm.
This is quite an interesting point however, as this implies certain things.
1. There is no gravitational force holding the moon in place.... So what, spooky force? Okay great explanation
2. Every single thing that does not orbit the earth is... A lie? Binary star systems? You know... The ones you can look at? The moons of every other planet?
Okay A*
3. So the earth accelerates at 9.8ms^2 and there is no gravity. The earth is approximately 4.54 billion years old (Following radioactive decay dating). I know you think the earth is something like 4000 years old though... So I'll go with that. Yeah this is.... I mean feel free to check my calculations but..... If the age of the earth is 4000 years old, that is about 1.2623 x 10^11 seconds. Now, you say the earth is accelerating at 9.8ms^-2. That means, following v = u + at, with a being 9.8ms^-2, t being 1.2623 x 10^11 and assuming u = 0 (Start at zero velocity), that the earth has a velocity of approximately 1.23705 x 10^12 ms^-1. The speed of light is about 2.998 x 10^8ms^-1.

Simplified
Following what this society "teaches" with the earth accelerating at 9.8ms^-2 and the age of the earth being 4000 years.
The earth is travelling at a velocity of at least 1,237,050,000,000 metres every second. (2,767,202,040,085.9 mph).
A little comparison, the speed of light is about 299,800,000 metres every second. (670,633,500.4 mph).

So this society is claiming all experimental evidence of the speed of light is incorrect. One of the cornerstones of modern physics is completely and utterly wrong. It's easy to claim certain things that the average person could accept, but the predictions based on that hypothesis need to be validated by experimental evidence. And quite simply.... This simple calculation based on what you're telling me is just mind boggling. The idea that the earth is travelling at that velocity is just.... How, how can nobody have realised what no gravity means? When you were thinking of a way to argue against gravity did it ever occur to you that acceleration means a change in velocity? And then this change in velocity adds up to a greater than speed of light value, utterly ridiculing the whole idea. If not... Then you are claiming that everybody else is wrong. Good luck with that

2:

Oh and another point, because I just remembered it.

On a flat earth accelerating at 9.8ms^-2, the acceleration is constant across the whole world. Which just so happens to be incorrect, the acceleration of free fall varies substantially from the equator to the poles, completely contradicting your hypothesis. The acceleration due to free fall can go from 9.76 to 9.83 ms^-2.

And because I was so amazed by my previous calculation, I decided to find the age of the earth, according to what you tell me (excluding the age you state).
So, with the earth travelling at a tenth of the speed of light, 29,980,000ms^-1, a reasonable velocity that should have little relativistic effects. Again, using v= u + at, where v = 29,980,000, a = 9.8 and u is zero, t would equal 3059183.7 seconds. In years, that is 0.09694 years. Now, I don't know how old you might be, but I'm pretty sure I've lived for longer than 35.4 days. However according to your fairly conflicting ideas, the earth is 35.4 days old. Thanks for the information, this is news worthy indeed.

P.S
In the likely eventuality that they run out of arguments and ban me, just remember that if you look into what they're actually saying, you can find some interesting statements, such as the age of the earth being little over a month

-Alex
« Last Edit: May 31, 2014, 04:02:41 PM by Pleaseexplain »

#### Gulliver

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##### Re: Maths
« Reply #1 on: May 31, 2014, 04:17:01 PM »
...
-Alex
You really should take a moment and read about how velocities add when Special Relativity is involved. Here's a good starting point: http://math.ucr.edu/home/baez/physics/Relativity/SR/velocity.html

However, the UA is still flawed. See the Flat Earth General board's topic on FE Models for a good way to demonstrate FET's UA's flaws.

Oh, and welcome. (Just to be clear, I'm a REer.)
Don't rely on FEers for history or physics.
[Hampton] never did [go to prison] and was never found guilty of libel.
The ISS doesn't accelerate.

• 11
##### Re: Maths
« Reply #2 on: May 31, 2014, 04:33:20 PM »
Hi, I've tried to avoid special relativity for now, as that's the next level for me. I do understand how nothing can reach the speed of light however, and so to say that the earth has a constant acceleration has really amused me. The whole UA thing is completely destroyed by different values for the acceleration of free fall, which just so happen to perfectly fit in with the laws of gravitation.

#### Pete Svarrior

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##### Re: Maths
« Reply #3 on: May 31, 2014, 04:35:11 PM »
Welcome! I can't help but notice that your first question is addressed in our FAQ (and a link to a more in-depth explanation is provided there).

As for the second one, the common response provided is that the heavens have a slight and uneven gravitational pull, causing the discrepancies in local perceived g.

Oh, and for the records, since these questions are exact copies of the ones you asked in "Ask a FE theorist anything", I removed them from that thread.
« Last Edit: May 31, 2014, 05:34:12 PM by pizaaplanet »

P.S.  All of us illiterate folks understood this the first time.

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##### Re: Maths
« Reply #4 on: May 31, 2014, 04:43:01 PM »
Ah well, university will solve that one. I'm sorry but the explanation for the UA is dark energy? The heavens have a slight and uneven gravitational pull? The answers in your FAQ are... Interesting. The explanation of seasons and day or night are absolutely golden. But sticking with UA, you're saying the "heavens" have a gravitational pull. So you're saying gravity is true. And therefore a flat earth contradicts everything we could know about gravity. What

#### Pete Svarrior

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##### Re: Maths
« Reply #5 on: May 31, 2014, 04:49:33 PM »
I'm going to skip any and all pointless snide remarks you've made. Try to avoid them in the future - they don't encourage people to respond.

I'm sorry but the explanation for the UA is dark energy?
Correct.

The heavens have a slight and uneven gravitational pull?
Indeed.

But sticking with UA, you're saying the "heavens" have a gravitational pull. So you're saying gravity is true.
Yes and no. Let's hope you don't jump into any rash conclu-

And therefore a flat earth contradicts everything we could know about gravity. What
...sions. Right, you did jump to rash conclusions. Just because gravitation (not to be confused with gravity) is exerted by some bodies does not mean all bodies necessarily have to exert it.

P.S.  All of us illiterate folks understood this the first time.

• 11
##### Re: Maths
« Reply #6 on: May 31, 2014, 04:58:18 PM »
Okay Gulliver, I had a look at some of the other threads you post in and.... I feel for you.

Anyway, so not all bodies exert gravity? So an object with mass does not always exert a gravitational force upon another mass. I don't know much about GR or SR but I do understand Newtons laws of gravitation. Quite simply, everything with mass experiences a gravitational force. Is this saying the earth does not have mass, but then the stars with mass still affect the earth?

I think this can be summarised with
1. The earth has zero mass and is hence unaffected by gravity (Which would pull the earth into a sphere). With UA providing the acceleration of free fall.
2. The acceleration of free fall is affected by the gravitational force of stars.

How are both true

#### Pete Svarrior

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##### Re: Maths
« Reply #7 on: May 31, 2014, 05:02:08 PM »
Quite simply, everything with mass experiences a gravitational force.
I have in the past asked people to show to me that bananas exert a gravitational force (I even suggested the Cavendish experiment when people started getting confused). Unfortunately, no one even attempted it. It is claimed that all bodies exert a gravitational force, and we quite simply disagree.

I think this can be summarised with
1. The earth has zero mass and is hence unaffected by gravity (Which would pull the earth into a sphere). With UA providing the acceleration of free fall.
Assumptions, assumptions, assumptions.

2. The acceleration of free fall is affected by the gravitational force of stars.
Almost. I'm not sure why you'd restrict yourself to just the stars. We're talking about the heavens.

How are both true
They're not.

P.S.  All of us illiterate folks understood this the first time.

#### Pete Svarrior

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##### Re: Maths
« Reply #8 on: May 31, 2014, 05:07:29 PM »
Also, I find your accusation of us banning people we disagree with quite inflammatory. We have public ban log, you know? There have only been 2 bans issued in the past half-year, both for excessive derailment and harassment. You posting a couple paragraphs on how you don't understand Round Earth physics and how you forgot to read the FAQ won't warrant a ban. The fact that you're talking to Gulliver (who thought the ISS accelerates towards the Earth at 9.8m/s2) should be the best proof of this.
« Last Edit: May 31, 2014, 05:12:42 PM by pizaaplanet »

P.S.  All of us illiterate folks understood this the first time.

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##### Re: Maths
« Reply #9 on: May 31, 2014, 05:14:17 PM »
Quite simply, everything with mass experiences a gravitational force.
I have in the past asked people to show to me that bananas exert a gravitational force (I even suggested the Cavendish experiment when people started getting confused). Unfortunately, no one even attempted it. It is claimed that all bodies exert a gravitational force, and we quite simply disagree.

g = -GM/r^2
It's really as simple as a single equation. The inverse square law tells us that the gravitational force due to stars are completely negligible. Indeed the objects in the heavens as you describe it, have no impact on the calculated value of g.

It's interesting how you separate FE physics from RE physics. One of the principal rules of physics is that the laws of physics are the same regardless of location, yet on earth there's an exception?

#### Pete Svarrior

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##### Re: Maths
« Reply #10 on: May 31, 2014, 05:17:35 PM »
g = -GM/r^2
It's really as simple as a single equation. The inverse square law tells us that the gravitational force due to stars are completely negligible. Indeed the objects in the heavens as you describe it, have no impact on the calculated value of g.

It's interesting how you separate FE physics from RE physics. One of the principal rules of physics is that the laws of physics are the same regardless of location, yet on earth there's an exception?
Yes, some parts of physics are fundamentally different between models. You'd think that would be obvious, given that FET introduces a whole new concept of dark energy accelerating the Universe.

Please specify M and r for the heavens. Your equation is rather useless without those.

Now, you say that physics applies equally regardless of location. Unfortunately, this is not true for the mainstream theories of gravitation: https://en.wikipedia.org/wiki/Gravitation#Anomalies_and_discrepancies

P.S.  All of us illiterate folks understood this the first time.

• 11
##### Re: Maths
« Reply #11 on: May 31, 2014, 05:22:50 PM »
And just to clarify as it seems that centripetal acceleration is poorly understood. The calculated value of g for the ISS using g = -GM/r^2.

r = radius from the centre of the earth to the ISS. The radius of the earth is about 6,371,000 m and the ISS orbits at about 370,000 m hence r = 6,741,000
M = mass of the earth, approx 5.97219 × 10^24 kg
G = 6.67 x 10^-11

Hence g (Of the ISS) = 8.77ms^2 or Nkg^-1

#### Pete Svarrior

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##### Re: Maths
« Reply #12 on: May 31, 2014, 05:24:08 PM »
And just to clarify as it seems that centripetal acceleration is poorly understood. The calculated value of g for the ISS using g = -GM/r^2.

r = radius from the centre of the earth to the ISS. The radius of the earth is about 6,371,000 m and the ISS orbits at about 370,000 m hence r = 6,741,000
M = mass of the earth, approx 5.97219 × 10^24 kg
G = 6.67 x 10^-11

Hence g (Of the ISS) = 8.77ms^2 or Nkg^-1
Yes, I've already explained this to Gulliver in the appropriate thread (although your r is slightly too low, causing an inflated result, but you're still closer than Gulliver's 9.8m/s2, so all is forgiven). If you have anything to add to that thread, please post it there, not here.
« Last Edit: May 31, 2014, 05:26:23 PM by pizaaplanet »

P.S.  All of us illiterate folks understood this the first time.

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##### Re: Maths
« Reply #13 on: May 31, 2014, 05:27:46 PM »
You seem to think that the ISS is not accelerating towards the earth? Yes the value for g is less than 9.8ms^-2 for the ISS. That's precisely because it is orbiting at a height of 370km. g may be slightly lower than that on the surface of earth, but it is still an acceleration towards the centre of the earth.

#### Pete Svarrior

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##### Re: Maths
« Reply #14 on: May 31, 2014, 05:28:49 PM »
You seem to think that the ISS is not accelerating towards the earth?
No. Have you tried actually reading the thread I linked you to? And I already told you you got your r wrong, but I suppose that didn't come through, so I'll try again:

It's not 370km, you numpty. It's about 420km.

https://en.wikipedia.org/wiki/ISS

P.S.  All of us illiterate folks understood this the first time.

• 11
##### Re: Maths
« Reply #15 on: May 31, 2014, 05:39:16 PM »
I'm going to skip any and all pointless snide remarks you've made. Try to avoid them in the future - they don't encourage people to respond.

Well I'm glad you accept the ISS orbits the earth in a circular path, and as it has a centripetal acceleration of about 8.7ms^2, that would point towards a non UA. Now back to another point.

You claim that the variations seen in measured values of g are due to the "heavens". Can you explain where in g=-GM/r^2 the mass of stars, planets, moons, comets and general mass varies depending on the specific location where the measurements are taken?

#### Pete Svarrior

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##### Re: Maths
« Reply #16 on: May 31, 2014, 05:40:32 PM »
Well I'm glad you accept the ISS orbits the earth in a circular path, and as it has a centripetal acceleration of about 8.7ms^2, that would point towards a non UA. Now back to another point.
Yes, the Round Earth model does not have UA. I'm glad we can agree on that. It's a bit worrisome that you try to extrapolate from that to the Flat Earth model, but oh well.

Of course, it's worth noting that under the RE model, the ISS's path is not circular at all.

You claim that the variations seen in measured values of g are due to the "heavens". Can you explain where in g=-GM/r^2 the mass of stars, planets, moons, comets and general mass varies depending on the specific location where the measurements are taken?
Sorry, I've asked you to do this already, since you're the one who claims this equation would prove or disprove anything in FET. Also, it doesn't have to be mass. It could be the distance, which varies with location pretty much by default. Please don't arbitrarily pick one variable out of two, especially if you're going to pick the wrong one.
« Last Edit: May 31, 2014, 05:44:17 PM by pizaaplanet »

P.S.  All of us illiterate folks understood this the first time.

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##### Re: Maths
« Reply #17 on: May 31, 2014, 05:47:49 PM »
Okay let's try and stick with one thing please. No orbit is 100% circular with perigee and apogee attributed to orbits.

g= -GM/r^2

The radius of the earth is the one true variable in measurements. G is a constant and M is a constant.
At the north pole we see a greater value of g, indicating a reduced r. At the equator we see a lesser value of g, indicating an increased r. There are numerous explanations for this, including the moon, the oceans, mountains and hills etc. But I have never heard anybody claim the heavens can influence r. I am curious, please explain

#### Pete Svarrior

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##### Re: Maths
« Reply #18 on: May 31, 2014, 06:04:13 PM »
But I have never heard anybody claim the heavens can influence r. I am curious, please explain
No one claimed the heavens can influence the distance between two objects, that's likely why you never heard such a claim being made.

P.S.  All of us illiterate folks understood this the first time.

#### Rama Set

• 9914
• Round and round...
##### Re: Maths
« Reply #19 on: May 31, 2014, 06:16:37 PM »
Pizaa Planet: When you say Dark Energy, do you mean that there is obviously some energy causing UA, but you do not know what it is or how to detect it?

In regards to your banana story, I am not sure what you are getting at. The Cavendish experiment is performed by university students regularly, with consistent results. Do bananas hold a particular interest for you?
Th*rk is the worst person on this website.