"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?
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. "Okay we will assume
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?
"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. "Okay we will assume
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?
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..
But Flat Earthers do not seem prepared to accept the word of "globulists", so where did this and other values come from?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.
UNIVERSAL ACCELERATIONNo justification or indication of source. It just matches the "typical" value of "g" from experments done by Globe supporters, including Newton.
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.
The Lunar EclipseThere 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.
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
The Ice WallNo justification, but clearly just taken over from the circumference of the Globe.
The figure of 24,900 miles is the diameter of the known world; the area which the light from the sun affects.
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.
. . . . . . . . . . . . . . . .
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. (http://www.sacred-texts.com/earth/za/za23.htm)
"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?
But Flat Earthers do not seem prepared to accept the word of "globulists", so where did this and other values come from?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.
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. (https://www.theflatearthsociety.org/forum/index.php?topic=68300.msg1833997#msg1833997) 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".
But Flat Earthers do not seem prepared to accept the word of "globulists", so where did this and other values come from?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.
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. (https://www.theflatearthsociety.org/forum/index.php?topic=68300.msg1833997#msg1833997) 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?
That's what I was thinking thanks for the clarification.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).
So, resolved right?
These threads debunking UA always make me chuckle.
In the natural world, nothing pulls others (except exceptions) ; (except exceptions) everything push others. Earth isn't pulling us, sky is pushing us.
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 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.
"We came up with 9.8m/s2 cause that's how fast things fall."
Dunning- kruger.
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?
"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. "Okay we will assume
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?
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 does the flat earth society determine this number 9.8ms2 without the mass of the Earth and radius of the Earth squared?
"Touchè.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 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.
"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.
"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."I suggest you read the link I posted.. it is done there.
-If that was true my friend ;D
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?
People of this beautiful blue dot,
Work calls, we talk later
Bye bye.
"Maybe a little less snideness next time."
I hate to be snide, but ignoring an equation given multiple times asks for a little snideness.. ;D
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.. ;D
Making a subtle point in English language is harder than i expected.
I meant, you were questioning why the UA theory uses 9.8ms2, the Equivalence Principle should help you understand.
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.
"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.
The mass of the earth was calculated using this gravitational formula, not vice versa! (There was no scale big enough, unfortunately...)
I'm no expert, but there is probably a way to derive Kepler's third law without universal gravitation.
If you observe then going in elliptical paths, you immediately known that the centripetal acceleration gets weaker with distance etc.
"If you want to play smart get your facts straight. The mass of the earth was calculated using this gravitational formula"Sorry to be throwing spanners around and getting them stuck in the works, but, look at the time-line:
Kepler's work (published between 1609 and 1619) improved the heliocentric theory of Nicolaus Copernicus, explaining how the planets' speeds varied, and using elliptical orbits rather than circular orbits with epicycles.And
Isaac Newton showed in 1687 that relationships like Kepler's would apply in the Solar System to a good approximation, as a consequence of his own laws of motion and law of universal gravitation.Both from Kepler's laws of planetary motion. (https://en.m.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion)
Kami
Maybe you could answer my question, what's wrong with my equation?
I made mistake on purpose there can you point it out please?
Hard putting them in a online calculater isn't it ;)
Kami,I agree with you there, whole-heartedly. But just because you believe in a round earth does not mean that you are free from error.
The flat earth society has no theory, it's a believe.
Everything points in the direction of a round beautiful blue dot.
It's not a manner of perception, Mathematics proves the earth is a sphere.
Even without knowing the Earth's mass, we know the Earth is a sphere, because of Eratosthenes.
Why are you doing this, are you agnostic wtf!
I leave it be.
Kami,The upper one is correct (i suppose), considering you assume a nearly circular orbit and the mass of the satellite to be neglible compared to earth's mass. In the lower one you forgot the "Mearth". And yes, this is exactly what one can find by simple googling without putting an effort into this.
"Mearth= 4pi2a3: T2G
4 times pi squared times the half-axis of the elliptical orbit cubed = the orbital period squared times the gravitational constant. ( oops i forgot something in the equation, i bet you can't see what it is ;) )"
V=4/3 *Pi r3 , and we know 1 m3 Earth rock weights approximately 5000 kg.It is just that the earth is much denser at the core than on the surface and that the density of earth rock is approximately 2.9t/m^3, not 5 (compare rabinoz' post). If you take some surface layer and propose that its density is roughly equal to earth's without knowing a single thing about the interior, then you might have an approximation that is correct by order of magnitude, but nothing more. To get a good estimate of earth's mass you need to analyze its gravitational field OR somehow measure its density (i dont know much about this but i think it could be possible to do this by analyzing seismic waves or something similar; this is just a guess though and those methods are quite recent, the mass of the earth has been known by then).
Put a big piece of rock (1m3) on a scale, with this number we can determine roughly the Earth's mass :)
We know the volume, we know the radius and we can weigh a m3 of rock !
Calculate with a m3 of rock and you will see you that you are close determining the Earth's mass without g and G.
Oh my, the irony is so strong it burns.
Dunning- kruger.
The density and mass were found more or less together in 1797-98 by Henry Carpenter - look him up in Wikipedia.Kami
Maybe you could answer my question, what's wrong with my equation?
I made mistake on purpose there can you point it out please?
Hard putting them in a online calculater isn't it ;)
Which equation do you mean?
Then, by all means, tell me how I can measure the density of the earth. I have no problems with Eratosthenes' work, but I am fairly certain that the (average) density of the earth was derived from its measured mass, not vice versa.
The density and mass were found more or less together in 1797-98 by Henry Carpenter - look him up in Wikipedia.That is interesting, although I can't seem to find him.. could you provide a link?
The Universal Gravitational Constant could then be calculated from that mass.
Kami,Did not see the second equation. The m there stands for the mass of the second object; as I said, if this is neglible (so M+m roughly equals M), then the upper one is correct.
Kami said,
"In the lower one you forgot the Mass of the earth"
My equations,
-Upper one,
Mearth= 4pi2a3: T2G
4 times pi squared times the half-axis of the elliptical orbit cubed = the orbital period squared times the gravitational constant. ( oops i forgot something in the equation, i bet you can't see what it is ;) )
Lower one i said,
T2= 4Pi2 divided by G(M+m)*a3
The Big M stands for the mass of the earth.Thus the correct equation..
The first one i gave is incorrect, can you see it?
You don't have to answer this.I really don't mind, this from my textbooks, i don't think you are able simple Google it :)
On topic,Let us not argue about words. Certainly the earth appears to be a globe and all evidence points to it, but you can not rule out that we all live in a matrix constructed by our alien overlords making us believe that the earth is round. Not saying this is true, but you can not know anything for sure in physics, there are just assumptions that were verified to such a high degree of certainty that we believe them to be true (for example we believe that wenn you jump, you fall down. This has happened every time. But it could be that when you jump next time, you will fly into space and die. Granted, the possibility is very, very small, but not entirely zero).
Round Earth is also just a believe? Explain please why you think that.
And,Your first question was answered correctly. Certainly, when you go into rocket science F=m*g will not suffice anymore, but for all surface-based observations (and even airplanes) this suffices. The inaccuracies due to varying g over the first 10 kilometers are 0.3%. That is really small compared to other possible errors like air resistance and minor things like a not exactly timed clock, a slightly incorrectly measured height, etc.
"You cant calculate the mass of the Earth exactly without g, only roughly with a m3 rock"
Let's use your analogy,
My first question was, how did the flat Earth community came up with this figure 9.8ms2?
Just by observing....and we know we can neglect the changing of g over a SHORT distance.
So s=v*t is sufficient, you say.
Wrong!
Roughly 9.8ms2, because you can neglect short distances like 100 meters above the surface.
But with rocket science you cant neglect even minimal change in g.
I want to measure a falling object from 2000 m to the ground.
Do you need to know the distance to the earth's center or not, roughly isn't enough.
No you can't, only roughly without the distance to the center of the Earth squared......, again if you want to be sure, you need to know the object's distance to the Earth squared!
g=9.8ms2 at the Earth's surface is like saying the Earth is round.
With simple geometry you can calculate the Earth circumference, this is math and proof the Earth is a beautiful round ball.
Why is this just belief the Earth is round, kami?
Sorry for the rough text.
Don't have a lot of time ....
But Don't you want to know why it falls :) ?I do! But to be fair, while the bending of spacetime does explain gravity, why do masses bend spacetime?
g= not uniform( the gravitational field around the Earth is not a constant and g depends on the distance of the object to the Earth's center, the Earth is not an infinite plain and the light of the Sun takes 8 minuten to reach our surface of the Earth, do you agree with these facts so far?Yes.
My main concern is this, you said "we never put the Earth on a scale".I am just saying that the statement "we derived g from G and M" is wrong, the scale was just a joke
This tells me you are trying to somehow push all scientific work done in history and modern science today is based on wrong assumptions.
I want to know which wrong assumptions in history.I am a huge fan of general relativity, which contradicts this assumption on large scales. Otherwise, I agree with you
Because if you don't agree with g=G*Mearth/r2 i think it's reasonable to ask why.
I have learned that g comes from the earth center the mass that warpes space-time, and everyone else in modern science as well.
-We had a discussion my brother and i, about the scientific work done by Ligo, and the discovery they made in 2015.It is awesome, though not as awesome as the media put it. Gravitational waves have been proven for several years now, they have just been directly detected the first time. Most newspapers seem to confuse this.
Physically detecting gravitational waves, generated by black holes.
What's your opinion on this discovery?
-The matrix analogy....I don't mind. I do not believe it either. I just wanted to state that nothing is 100% certain.
Hmmm, yes nice science fiction story, don't want to go deeper in this analogy if you don't mind.
Disputeone
"Yeah he's been told that numerous times.
I'm betting he's gonna go the same way as physics teacher did." And that makes you chuckle right :'(
No, i'm quitte tough.
And I practice my English.
And i like talking to people,btw i acknowledge my lack of mutual understanding, but i think this is not all my fault!
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
My question,
How does the flat earth society determine this number 9.8m/s2 without the mass of the Earth and radius of the Earth squared?
Galileo Galilei
One prominent example is the "ball and ramp experiment." In this experiment Galileo used an inclined plane and several steel balls of different weights. With this design, Galileo was able to slow down the falling motion and record, with reasonable accuracy, the times at which a steel ball passed certain markings on a beam.
1673: Huygens' Horologium Oscillatorium
In 1673, Christiaan Huygens published his theory of the pendulum.
What about looking at the topic and real question in the OP. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
My question,
How does the flat earth society determine this number 9.8m/s2 without the mass of the Earth and radius of the Earth squared?
Yes, "How does the flat earth society determine this number 9.8m/s2"?
My simple answer is that there is no evidence in "the Wiki" or anywhere else that "The Flat Earth Society" ever did.
I suspect that they just "borrowed" it from main stream science - like they have done elsewhere when it suits them.
They could measure "g" by dropping a weight
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.
UA assumes a uniform acceleration.
What equation shows this uniform acceleration?
Who is sky?
I don't even know what the debate is about anymore.
UA assumes a uniform acceleration.
What equation shows this uniform acceleration?
Who is sky?
Drop a few things and calculus, uniform acceleration. This has been covered already.
I have absolutely no idea who sky is.
Learn how to quote people.
Ok great cause this thread has been resolved and you lost this debate. (the earth is still most likely a sphere)
Hope you have done your homework.
What about looking at the topic and real question in the OP. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
My question,
How does the flat earth society determine this number 9.8m/s2 without the mass of the Earth and radius of the Earth squared?
Yes, "How does the flat earth society determine this number 9.8m/s2"?
My simple answer is that there is no evidence in "the Wiki" or anywhere else that "The Flat Earth Society" ever did.
I suspect that they just "borrowed" it from main stream science - like they have done elsewhere when it suits them.
They could measure "g" by dropping a weight - Galileo did not have the means of measuring time accurately so he used a "ball and ramp" as in:Quote from: Wikipedia, History of experimentsGalileo Galilei
One prominent example is the "ball and ramp experiment." In this experiment Galileo used an inclined plane and several steel balls of different weights. With this design, Galileo was able to slow down the falling motion and record, with reasonable accuracy, the times at which a steel ball passed certain markings on a beam.
Once the theory of pendulums was developed the timing of a pendulum could be used to determine "g".Quote from: Wikipedia, Pendulum1673: Huygens' Horologium Oscillatorium
In 1673, Christiaan Huygens published his theory of the pendulum.
Robert Hooke, Isaac Newton and others (eg Jean Richer in a trip to Cayenne, French Guiana in 1671) used pendulums as a method of comparing "g" at various latitudes and altitudes.
So, there are a number of ways they could measure "g", but have they?
We need a reply from Flat Earthers for that! We cannot answer it.
.
What did I miss here in this topic!
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.
.
What did I miss here in this topic!
Obviously this...
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.
.
What did I miss here in this topic!
Obviously this...
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.
.
What did I miss here in this topic!
Obviously this...
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'm done man sorry.