Gravity on a flat Earth.

                                      Hello people on this beautiful round blue dot, or blue pancake 

I want to ask a question.
Forgive me, English is not my native tongue, so i guess it reads kinda crooked.

Gravity, the acceleration due to curvature of space-time.

(g=G* the mass of the Earth divided by the radius of the Earth squared)
At the surface( sea level) g=9.8ms2.
I read the flat earth wiki page and came across this line, regarding Universal Acceleration.

"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)."

https://wiki.google.com/Universal_Acceleration

My question,

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

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. Determining R acceleration due to gravity without determining the mechanism isn't a big ask.

"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?

"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?

You do need a little more information but it is not impossible. Drop it off a 100 m building marked at 10  m increments. Check your time at each increment. A higher cliff would be even better. A steel arrow like shaft with a weighted end might be better than a ball.

"You do need a little more information but it is not impossible. Drop it off a 100 m building marked at 10  m increments. Check your time at each increment. "

You are talking about the acceleration a of the object in F = m*a.
(Newton laws of motion)

I need F=m*g
(Newton law of gravity)

How do we determine g.

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

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.

"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.

"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..

"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.

In the first second the object dropped 4.9m. (Average velocity of 4.9m/s) In the second second it dropped 14.7m. (Average velocity of 14.7m/s) In the third it dropped 24.5m. (Average velocity of 24.5m/s) 24.5-14.7=9.8 14.7-4.9=9.8

Thus you can conclude that the average velocity is increasing by 9.8m/s/s.

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..

I love it when Kami comes in and crushes dreams lol.

Hello Kami

Nice job doing calculus with g=9.8ms2 


Also we know.......g=9.8ms2


s= int v dt  (An alternate unit is the g per second.)

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

a=2s/t^2"



I ask you again how do you determine g=9.8ms2
(g=G*massEarth/r2)

Quote from: BalGehakt on October 30, 2016, 01:35:54 PM

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

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.

But Flat Earthers do not seem prepared to accept the word of "globulists", so where did this and other values come from?

Could I dare suggest that these Flat Earthers just borrow values and reject values "when it suits them"?
A few cases in point:

• The value of "g":
  Quote from: the Wiki

  UNIVERSAL ACCELERATION
  In the Universal Acceleration model, all the celestial bodies including the earth are being accelerated in one uniform direction at roughly 9.81 m/s^2.

  No justification or indication of source. It just matches the "typical" value of "g" from experments done by Globe supporters, including Newton.
  
  
• The "Shadow Object's" orbital inclination:
  Quote from: the Wiki

  The Lunar Eclipse
  A Lunar Eclipse occurs about twice a year when a satellite of the sun passes between the sun and moon.
  This satellite is called the Shadow Object. Its orbital plane is tilted at an angle of about 5°10' to the sun's orbital plane
  

  There is never any justification for this 5°10' inclination. The Globe model has no "shadow object", so the Flat Earth Wiki just borrows the 5°10' inclination of the moon's orbit.
  
  
• The size of the "known earth":
  Quote from: the Wiki

  The Ice Wall
  The figure of 24,900 miles is the diameter of the known world; the area which the light from the sun affects.

  No justification, but clearly just taken over from the circumference of the Globe.
  

I venture to say that one of the few actual "measurements" done by Flat Earthers is Rowbotham's "True Distance of the Sun" measurement. (See the link for details of how it was done.)

CHAPTER V. THE TRUE DISTANCE OF THE SUN.
. . . . . . . . . . . . . . . .
so that it is perfectly safe to affirm that the under edge of the sun is considerably less than 700 statute miles above the earth.
From Zetetic Astronomy, by Samuel Birley Rowbotham, CHAPTER V. THE TRUE DISTANCE OF THE SUN.

Note that Rowbotham claims "that it is perfectly safe to affirm that the under edge of the sun is considerably less than 700 statute miles above the earth" so differs markedly from the current "a bit over 3,000 miles.

But his calculations are based on very inaccurate sun elevation angles. This will come as no surprise if you look at his measurement methods.
If these same calculations are repeated with accurate sun elevation angles (from a site such as http://www.sunearthtools.com/)
Yet, Rowbotham carries this 700 miles through his writings, using values related to it for Moon, Planets and Stat's distances.

"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?

Your question was how do FEers get to 9.8m/s^2 without using the mass and radius of the Earth. Neither is necessary to measure the rate a falling object accelerates, regardless of the shape of the Earth.

Quote from: Boots on October 30, 2016, 02:40:30 PM

Quote from: BalGehakt on October 30, 2016, 01:35:54 PM

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

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.

But Flat Earthers do not seem prepared to accept the word of "globulists", so where did this and other values come from?

Could I dare suggest that these Flat Earthers just borrow values and reject values "when it suits them"?

Well they certainly do that. But an acceleration rate can be calculated by observing that an object falls 4.9m in the first second and 14.7m in the second second which comes to a total of 19.6m. In the third second it drops 24.5m bringing the total distance dropped to 44.1m. Assuming a constant acceleration it is very simple math to deduce that the velocity is increasing at a rate of 9.8m/s/s. Like this. Or am I missing something. As far as I can tell, the only two things assumed by this method are the observations holding true and a constant rate of acceleration.

Admittedly, it would be difficult to arrive at that level of precision just by observing and timing falling objects. Technically it would be possible though. Since they are using that level of precision, and as you say, they have not offered "any indication or source," it is not likely that they actually arrived at that number through the observation method. I concede that they probably just borrowed it because this was one of the instances where it suited them to do so.

Well they certainly do that. But an acceleration rate can be calculated by observing that an object falls 4.9m in the first second and 14.7m in the second second which comes to a total of 19.6m. In the third second it drops 24.5m bringing the total distance dropped to 44.1m. Assuming a constant acceleration it is very simple math to deduce that the velocity is increasing at a rate of 9.8m/s/s. Like this. Or am I missing something. As far as I can tell, the only two things assumed by this method are the observations holding true and a constant rate of acceleration.

Admittedly, it would be difficult to arrive at that level of precision just by observing and timing falling objects. Technically it would be possible though. Since they are using that level of precision, it is not likely that they actually arrived at that number through this method. I concede that they probably just borrowed it because this was one of the instances where it suited them to do so.

I was not intending to criticise your post in any way, just to cast doubt that any flat earther made the measurement of 9.81 m/s² quoted in "the Wiki".

I was not intending to criticise your post in any way, just to cast doubt that any flat earther made the measurement of 9.81 m/s² quoted in "the Wiki".

No problem. I doubt it as well.

Quote from: rabinoz on October 30, 2016, 03:50:49 PM

Quote from: Boots on October 30, 2016, 02:40:30 PM

Quote from: BalGehakt on October 30, 2016, 01:35:54 PM

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

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.

But Flat Earthers do not seem prepared to accept the word of "globulists", so where did this and other values come from?

Could I dare suggest that these Flat Earthers just borrow values and reject values "when it suits them"?

Well they certainly do that. But an acceleration rate can be calculated by observing that an object falls 4.9m in the first second and 14.7m in the second second which comes to a total of 19.6m. In the third second it drops 24.5m bringing the total distance dropped to 44.1m. Assuming a constant acceleration it is very simple math to deduce that the velocity is increasing at a rate of 9.8m/s/s. Like this. Or am I missing something. As far as I can tell, the only two things assumed by this method are the observations holding true and a constant rate of acceleration.

Admittedly, it would be difficult to arrive at that level of precision just by observing and timing falling objects. Technically it would be possible though. Since they are using that level of precision, and as you say, they have not offered "any indication or source," it is not likely that they actually arrived at that number through the observation method. I concede that they probably just borrowed it because this was one of the instances where it suited them to do so.

Couldnt you just drop 10 things from different heights in controlled conditions, time them and measure the heights to come to 9.81 m/s2?

At least hypothetically?

Couldnt you just drop 10 things from different heights in controlled conditions, time them and measure the heights to come to 9.81 m/s2?

At least hypothetically?

I think so. For example if you dropped object from 5m, 15m, and 44m and timed them, you would pretty much have the same information and come to the same conclusion.

That's what I was thinking thanks for the clarification.

So, resolved right?

These threads debunking UA always make me chuckle.

That's what I was thinking thanks for the clarification.

So, resolved right?

These threads debunking UA always make me chuckle.

I think that the gist of this thread is not that we can't easily measure "g", but that "the Wiki" simply quotes the 9.81m/s² with no indication of where the figure came from. I gave a number of instances where it seems obvious that they just "borrowed" values from similar Globe measurements (for example the "shadow object" orbital inclination).

Maybe we should do the experiment Boots suggested to use as a citation for the wiki.

I just want to point out that g=9.8 m/s² was known way before the gravitational constant G was calculated to much accuracy. Or the mass or radius of the earth for that matter.

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. Measuring g is trivially easy. Kami's answer was correct, despite OP's objection.

In the natural world, nothing pulls others (except exceptions) ; (except exceptions) everything push others. Earth isn't pulling us, sky is pushing us.

In the natural world, nothing pulls others (except exceptions) ; (except exceptions) everything push others. Earth isn't pulling us, sky is pushing us.

Are those exceptions the four fundamental forces of physics?

Should I post pictures of me both pushing and pulling?

Got some of my last 4x4 trip actually, plenty of pushing and pulling.

                                             Good morning people of this beautiful blue dot.

"a) a CONSTANT acceleration on the ball"

-Like to see your calculus with g=G*MEarth/r2 in equation, giving the ball an gravitational acceleration.
Meaning with acceleration and not with constant speed due to gravity.
I'l give you guys some info on this topic so you know where kami and Meatball are talking about.
With constant "force"

The standard example of motion with a constant force is the effect of gravity
here on earth. This is a slight cheat, since of course the gravitational pull
should depend on how far we are from the center of the earth. But if we
do our experiments in a room (even a large room) it’s hard to change this
distance by more than a few meters, while the radius of the earth is measured
in thousands of kilometers, so the changes in distance are only one part in
a million. One can measure forces with enough accuracy to see such effects,
but for now let’s neglect them.

So, in the approximation that we don’t move too far, and hence the pull
of the earth’s gravity is constant, we write

F = −mg   You guys see the minus
gravity is negative.
Putting this together with F = ma, we have

m=d2x(t) /dt2 = −mg.

The extraordinary thing is that the mass m appears on both sides of the
equation, so we can cancel it, leaving

d2x(t)/dt2 = −g.

Just so that you know all the words, the mass which appears in F = ma
is called the inertial mass, since this is what determines the inertia of an
object. Inertia expresses the tendency of objects to keep moving in the
absence of forces, and corresponds intuitively to the effort that we have to
expend in stopping of deflecting the object. We also use inertia in everyday
English to mean something quite similar, although not only in reference to
mechanics. In contrast, the mass in F = −mg is called the gravitational
mass, for more obvious reasons. The statement that the masses cancel thus
 is the “equivalence of gravitational and inertial masses,” or simply the
“principle of equivalence.”
The essential content of the principle of equivalence is clear from Eq
You actually can’t tell the difference between a little extra acceleration
(on the left hand side of the equation) and slightly stronger gravity
(on the right). Einstein made the point in a thought experiment, imagining
himself trapped in an elevator. Unable to see outside, he argued that
he couldn’t tell the difference between falling freely in a gravitational field
and being accelerated (e.g. by rocket jets attached to the elevator). From
the Newtonian point of view, this equivalence is a coincidence. After all,
there are other forces such as electricity and magnetism which aren’t proportional
to mass, and thus one could have imagined that the gravitational
force wasn’t proportional to mass either. Indeed, you may remember that
when we go beyond the approximation of gravity as a constant force, if two
objects with masses m1 and m2 are a distance r apart, then the force that
one objects exerts on the other is given by

F = −Gm1m2/r2

OR IN OTHER WORDS THE CENTER FROM THE MASS IS NEEDED, THE RADIUS OF THE EARTH.

So i ask again for the fifth time, where does the flat earth society come up with this number 9.8ms2
The gravitational acceleration due to warped space ,
g=G * the Mass of the Earth divided by the radius of the Earth squared.

g=G*EarthMASS/r2

We need,
F=m*g not F=m*a

https://www.princeton.edu/~wbialek/intsci_web/dynamics1.1.pdf

We came up with 9.8m/s2 cause that's how fast things fall.

Who is this guy?

Edit, I read your pdf link, we are aware of Newtons laws, you realise it didn't support anything you said tho right?

"I'll come to the FES and show them, no one there will notice I don't have a working understanding of physics"

We did.

"We came up with 9.8m/s2 cause that's how fast things fall."

Dunning- kruger.

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. Determining R acceleration due to gravity without determining the mechanism isn't a big ask.

Boydster answered you on the first reply.

I just want to point out that g=9.8 m/s² was known way before the gravitational constant G was calculated to much accuracy. Or the mass or radius of the earth for that matter.

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. Measuring g is trivially easy. Kami's answer was correct, despite OP's objection.

Also this.

Edit.

"We came up with 9.8m/s2 cause that's how fast things fall."

Dunning- kruger.

I had to translate that.

Well, how fast do things fall mister shmarty phants?

"Well, how fast do things fall mister shmarty phants?"

-That depends on the objects's distance to the center of the Earth.
I'm trying to explain this with subtle detail