Gravity on a flat Earth.

Touchè.

However I still feel you misunderstand the premise.

OP has it backward. We calculated big G, and other properties of the earth based on g=9.8 m/s2, not the other way around.

How does the flat earth society determine this number 9.8ms2 without the mass of the Earth and radius of the Earth squared?

So to clarify, the FES came up with 9.8ms2 by measuring how fast objects fell.

Example

Quote from: BalGehakt on October 30, 2016, 02:03:40 PM

"I think you can arrive at that figure by dropping a ball from a known height with a sufficiently accurate timer and doing a little math. "

Ok let's  play ball...say the ball weighs 1 kilogram and the ball is dropped from 10 meters.
And your accurate timer gives 2.5 sec.

What is g with these numbers?

Okay we will assume
a) a constant acceleration on the ball
b) the acceleration that works on the ball is due to gravity (meaning a=g)
c) there is a neglible amount of air resistance

We know:
v= int a dt
as we assume a to be constant
v=a*t

Also we know
s= int v dt

We conclude
s= 1/2*a*t^2
solving for a:

a=2s/t^2

Input s= 10m, t=2.5s
a= 3.2 m/s^2

Assuming a=g => g=3.2 m/s^2

not that hard..

How swift is your sword?

"Touchè.

However I still feel you misunderstand the premise.

Quote
OP has it backward. We calculated big G, and other properties of the earth based on g=9.8 m/s2, not the other way around.

Quote from: OP
How does the flat earth society determine this number 9.8ms2 without the mass of the Earth and radius of the Earth squared?

So to clarify, the FES came up with 9.8ms2 by measuring how fast objects fell.

Example

Quote from: Kami on October 30, 2016, 03:11:37 PM
Quote from: BalGehakt on October 30, 2016, 02:03:40 PM
"I think you can arrive at that figure by dropping a ball from a known height with a sufficiently accurate timer and doing a little math. "

Ok let's  play ball...say the ball weighs 1 kilogram and the ball is dropped from 10 meters.
And your accurate timer gives 2.5 sec.

What is g with these numbers?
Okay we will assume
a) a constant acceleration on the ball
b) the acceleration that works on the ball is due to gravity (meaning a=g)
c) there is a neglible amount of air resistance

We know:
v= int a dt
as we assume a to be constant
v=a*t

Also we know
s= int v dt

We conclude
s= 1/2*a*t^2
solving for a:

a=2s/t^2

Input s= 10m, t=2.5s
a= 3.2 m/s^2

Assuming a=g => g=3.2 m/s^2

not that hard..

How swift is your sword?"





-I explained this in my previous posts.
It's a calculation with a CONSTANT acceleration not with the gravitational acceleration, which depends on the distance of the object to the earth's center.

Maybe you should read my previous post again.

                                     
How does the flat earth society determine this number 9.8ms2 without the mass of the Earth and radius of the Earth squared?

The answer to your question is this. Nobody currently posting in this thread knows for 100% certain. We think they just borrowed it from RE theory. However, the consensus is that it could be determined to some level of accuracy by timing falling objects. And you are right in that this method doesn't account for the distance from the earths center. Neither does it account for air resistance. If this experiment is done at or near the surface of the earth with a heavy, streamlined object the effect from these two factors should be minimal. BTW I have done this experiment myself so I know it works.

Agreed your original question has been answered Bal, perhaps you would like to start another thread with another argument.

you know all the posters arguing with you on this are RE right?

Why spoil it.

"Touchè.

. . . . . . . . . . . . . .

-I explained this in my previous posts.
It's a calculation with a CONSTANT acceleration not with the gravitational acceleration, which depends on the distance of the object to the earth's center.

Maybe you should read my previous post again.

Come, come you are using equations and sums and maths! Sometimes I think that the flat earth is an EFZ (an Equation Free Zone).

I am being very cynical here, because I could name three who use equations prolifically, and for one of those, prolific is a gross understatement.

But, since the Newtonian Gravitation is generally denied by the Flat Earther's, they could hardly calculate "g", and I have seen no evidence in "the Wiki" or anywhere else where they have measured it.

"Drop an object off a high building. Observe that in the first second it drops 4.9m after the second second it has dropped 19.6m after the third second 44.1m etc. Do the math and come to the conclusion that when an object is dropped the earth will accelerate towards it (according to UA) at a rate of 9.8m/s2."

again,
No you can't calculate to that number (9.8ms2) with only newton's law of motion.(f=m*a)

You need  newton's gravitational law (f=m*g)
g=The gravitational constant G times the mass of the Earth divided by the radius of the Earth squared.
g=G*massEarth/r2

That's the point i'm trying to make here.

If you can get to that number with only using the law of motion let's see it.

Come on now don't defend that, he dug his own hole.

Nobody answered my previous post, so maybe take a step back and not use lectures, but instead wikipedia physics.

Maybe this will make things clear.

Defining the gravitational field strength, wikipedia-style Whoohoo !(not religious woowoo).

The gravitational field strength of a planet multiplied by an objects mass gives us the weight of that object.
We will now consider gravitational fields that are not uniform and how to calculate the value of g for any given mass.

"Any two point masses attract each other with a force that is proportional to each of their masses and inversely proportional to the square of the distance between them."

Finding the field strength of a mass.

F=-GMm/r2 (we know F=mg)

So,

mg=-GMm/r2 ( m cancels m )

So,

g=GM/r2

g=gravitational acceleration.
M=the mass of the Earth.
r2=is the distance to the Earth's center squared.
G= gravitational constant. (6.67408 × 10-11 m3 kg-1 s-2) small number ey

You can use this to find the gravitational field strength of a mass at a particular point, r.

Note that the gravitational field strength of the Earth near its surface is numerically equal to the acceleration of free fall, 9.8 ms2.

Also note,

g=depending on the object's distance to the earth center squared.(At the surface the distance to the center of the earth is 6.371 km)
The mass of the Earth can be considered a constant of course. (5,972E24 kg)
G=very small "force", we know gravity is a weak "force", but a huge mass like the Earth makes up for the small G if you know what i mean.

Can anyone show me the calculation of the gravitational acceleration of an object at a point r, 2000 km above earth's surface.Note the gravitational field is not uniform.

https://en.wikibooks.org/wiki/A-level_Physics/Forces,_Fields_and_Energy/Gravitational_fields








 







 

It is true that the gravitational acceleration varies with height, but you can (to a high degree of accuracy) neglect this fact over the first few kilometers.. (1/6300km)^2 is roughly equal to (1/6302km)^2..

Due to Einstein, an observer can not tell the difference between gravitational acceleration and acceleration due to forces (like in a car, rocket etc.), so UA and gravity would have the same effect.

Also, if you believe in an infinite plane with finite thickness you can derive g from newton's law of gravitation and even calculate the thickness of the plane from that.

Both theories fail to explain the lower gravitational acceleration at greater heights or the noticeable differences in the acceleration at different places of the earth though.

What you are talking about is orbital mechanics.. since most flat earthers deny the existence of those you will not find answers to those quesions here. But these formula are not necessary when you are close to the surface of the earth.

EDIT: Found this in an older thread, this is a calculation of g on a flat plane using newton's formula.
https://www.docdroid.net/abhC8KU/infplane.pdf.html

"Also, if you believe in an infinite plane with finite thickness you can derive g from newton's law of gravitation and even calculate the thickness of the plane from that."


-If that was true my friend 

Gravity won't work on a flat earth, only if the surface was infinite far, not only infinitely thick. this can be explained with General relativity.
(g=G* the mass of the Earth divided by the radius of the Earth squared)
An apple falling in the Netherlands falls just as fast in Australia.
Gravity accelerates us and all the objects in our environment towards the center of the earth. If the earth is round, then the attraction is everywhere the same.On a flat earth the gravitational attraction varies depending on where you stand. In that case, an apple in Australia would fall obliquely toward the Earth.
Is there someone here living in Australia maybe we can ask him how the Apple falls from a tree?

People who lived on the outside would be like always living on a really steep hill.

So the flat earth is an infinite plain?

"Also, if you believe in an infinite plane with finite thickness you can derive g from newton's law of gravitation and even calculate the thickness of the plane from that."


-If that was true my friend 

Gravity won't work on a flat earth, only if the surface was infinite far, not infinitely thick. this can be explained with General relativity.
(g=G* the mass of the Earth divided by the radius of the Earth squared)
An apple falling in the Netherlands falls just as fast in Australia.
Gravity accelerates us and all the objects in our environment towards the center of the earth. If the earth is round, then the attraction is everywhere the same.On a flat earth the gravitational attraction varies depending on where you stand. In that case, an apple in Australia would fall obliquely toward the Earth.
Is there someone here living in Australia maybe we can ask him how the Apple falls from a tree?

I suggest you read the link I posted.. it is done there.

The crucial point to this is that the plane has to be infinite; then the gravitational acceleration points downwards at all points because of symmetry.

EDIT: Of course I do not say that the earth is an infinite plane, I just say that, in theory, gravity would work on the surface similar to what we experience in our everyday life (it fails at higher altitudes though..)

"I suggest you read the link I posted.. it is done there."

-Dodging question, always the same.
I suggest reading my post again.

People of this beautiful blue dot,

Work calls, we talk later
Bye bye.

"Agreed your original question has been answered Bal, perhaps you would like to start another thread with another argument.

you know all the posters arguing with you on this are RE right?

Why spoil it."

-What?
Please kid go do your homework.

He's actually right. I think everybody who argued with youbsobfar have been globers. Including me.

People of this beautiful blue dot,

Work calls, we talk later
Bye bye.

this model also makes a full moon impossible to happen during the day time

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I think an infinite sphere could make gravity work, if you manipulated the density.

Maybe i need to be a little more snide....

We know the gravitational acceleration changes with the object's distance to the center of the Earth.
You can't calculate a falling object in a changing gravitational field,(-GmM/r^2 = Ma_r)without knowing the distance to the Earth's center and the mass of the Earth squared.

Let me ask you guys differently.
You say you don't need to know the Earth's mass and radius to determine g at a certain point in space ( constant gravity), but that's ridiculous, first of al,
We know gravity is not constant over distance,
and secondly,
You calculated this number to be 9.8ms2 by observation of a falling object near the earth surface by timing and measuring it, indeed possible with CONSTANT GRAVITY but you just don't want to know what this number tells you!! Thats the point here, i overlooked sorry

Well,
It comes from the radius and the mass of the Earth.
Lets get to the point and proving the Earth is a sphere, if you want to know the radius of the Earth you can use your observation that a object falls with a= 9.8ms2 and calculate the radius of the Earth with a=g.
If you calculate with g=9.8 ms2 and we know the mass of the Earth and we know G we can calculate the earth's radius.

So,
Indeed you can determine g at a certain point with observation and fixed gravity g= 9.8 ms2, but it also tells you the distance to the earth's center from that point.
It gives us the radius of the earth, your observation matches the radius of the Earth!
Because at the surface 9.8 ms2 matches exactly the Earth radius of 6.371 km

The mass of the Earth, and the radius of the earth  squared in relation with the the acceleration of gravity proofs the Earth is in fact a sphere!

Thank you.

Maybe a little less snideness next time.

"Maybe a little less snideness next time."

I hate to be snide, but ignoring an equation given multiple times asks for a little snideness..
Making a subtle point in English language is harder than i expected.

"Maybe a little less snideness next time."

I hate to be snide, but ignoring an equation given multiple times asks for a little snideness..
Making a subtle point in English language is harder than i expected.

Have you considered that maybe you're not understanding what's already been said to you in this thread by other round earthers?  Maybe instead of being snide, just try for mutual understanding. I realize it is difficult to be RE on the FE forums and resist the temptation to make fun of us, but I have faith in you.

Also, maybe this could help you a little https://en.wikipedia.org/wiki/Equivalence_principle

"Maybe a little less snideness next time."

I hate to be snide, but ignoring an equation given multiple times asks for a little snideness..
Making a subtle point in English language is harder than i expected.

Your original question was a very simple one and I believe it was adequately answered. No one disagrees with the point you are making now.

"no one disagrees"

Fine by me.
The Flat earth wiki page needs to be updated.

" I realize it is difficult to be RE on the FE forums and resist the temptation to make fun of us, but I have faith in you. "

-I'm sorry my intentions are not to be an asshole.

"Also, maybe this could help you a little https://en.wikipedia.org/wiki/Equivalence_principle"


-I dont want to be snide, but may i recommend this instead?
Lecture 1 of Leonard Susskind's Modern Physics concentrating on General Relativity. Recorded September 22, 2008 at Stanford University.
Equivalence
is discussed in one of the four lectures, just do not remember in which one.

I meant, you were questioning why the UA theory uses 9.8ms2, the Equivalence Principle should help you understand.

"I meant, you were questioning why the UA theory uses 9.8ms2, the Equivalence Principle should help you understand."

-Ik know what UA suggests, that's why i recommend Leonard Susskind's lectures.

Flat earth wiki;
The Basics
According to Flat Earth Theory, gravity does not exist. Instead, there is a force that produces identical effects as observed from the surface of the earth. This force is known as "Universal Acceleration" (abbreviated as UA).

Objects on the earth's surface have weight because all sufficiently massive celestial bodies are accelerating upward at the rate of 9.8 m/s^2. The mass of the earth is thought to shield the objects atop it from the direct force of UA. Alternatively, it is possible that the force of UA can actually pass through objects, but its effect on smaller bodies is negligible (similar to gravity in RET cosmology, which only has a noticeable affect on very large objects).

However, not all Flat Earth models dismiss the theory of gravity. The Davis Model proposes that the earth is an infinite plane exerting a finite gravitational pull (g), which is consistent with Gauss's Law.

- Yes a infinite plain, so the earth is infinite flat( sorry i meant an infinite plain)?
And we discussed earlier that g = not constant, but depends on your position to the center of bigM
Ok, better watch the lectures from prof Susskind he has something to say about infinite plains.

I meant, you were questioning why the UA theory uses 9.8ms2, the Equivalence Principle should help you understand.

I'm not sure he knows what he's questioning anymore.

BalGehakt may have thought that he could just barge in here and convince everyone of his point of view. Sorry, mr. meatball, but that's not how it works around here.