Instigator / Pro
7
1702
rating
77
debates
70.13%
won
Topic
#2197

Resolved: If the Earth is a flat disc, any object would not fall off the edge by Newton's laws of motion

Status
Finished

The debate is finished. The distribution of the voting points and the winner are presented below.

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2
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After 1 vote and with 3 points ahead, the winner is...

fauxlaw
Parameters
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Last updated date
Type
Standard
Number of rounds
3
Time for argument
Two days
Max argument characters
12,000
Voting period
Two weeks
Point system
Multiple criterions
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Open
Contender / Con
4
1489
rating
19
debates
42.11%
won
Description

Resolved: If the Earth is a flat disc, any object would not fall off the edge by Newton's laws of motion [especially not by Newton's laws of motion, law #2: f = ma, or "gravity" as defined below]

Full Description:

Resolved: If the Earth is a flat disc, any object would not fall off the edge by Newton's laws of motion [specifically, by Newton's laws of motion, law #2: f = ma, or "gravity" as defined below] The force of gravity will not function that way with a celestial body that has the shape of a flat, limited plane, such as a disc. In any celestial body, Newton’s three laws of motion, [ref. 1, 2] apply:
1. Every object in a state of uniform motion will remain in that state of motion unless an external force acts on it. [a.k.a., the law of inertia]
2. Force equals mass times acceleration [f = ma]. Also considered the force of gravity
3. For every action there is an equal and opposite reaction.

Definitions: all definitions herein described are subject to the conditions of the resolution of a flat-disc-planetary body.

Earth: a planetary celestial body, proposed by Flat Earthers to be a limited flat plane; effectively, a disc.

Object: Any object, animate or inanimate, existing near the edge of Earth in a condition such as described in the definition of “Earth” above.

Fall: The action on an object set in motion by the effect of Newton’s first law of motion [a.k.a.: the law of inertia] when moved off the edge of the disc, Earth. The effect would theoretically be known as the force of “gravity.”

Gravity: The action on an object according to Newton’s second law of motion [a.k.a.: f = ma]

Debate protocol:
1. No waive of rounds. This action will be considered a round forfeit
2. Round 1, 2 Argument/rebuttal/defense
Round 3 No new argument, rebuttal/defense/conclusion

reference:
[1] https://ccrma.stanford.edu/~jos/pasp/Newton_s_Three_Laws_Motion.html

[2] https://www.britannica.com/science/gravity-physics/Newtons-law-of-gravity

Round 1
Pro
#1
Resolved: If the Earth is a flat disc, any object would not fall off the edge by the force of gravity
 
Thank you, Nikunj_sangahai, for accepting this debate. It is really just a bit of fun, but science is at the core.
 
I Argument: Newton’s three laws of motion are the basis for the resolution
 
I.a When “Newton discovered the relationship between the motion of the moon and the motion of a body falling freely on Earth… he established the modern quantitative science of gravitation,”[1]  which Newton proceeded to describe in the three laws of motion: 
 
I.b The first law invoked the law of inertia; that that bodies not acted upon by external force move in a straight line at a constant speed.[2] What, then, is the primary external force of one body acting upon another to alter that straight line? 
 
I.b.1 Newton observed that a body in free fall on Earth falls in a straight line, since an apple falls straight down from a tree to the Earth, because that apple is captured by the only force in the universe that can act upon it, at least in terms of its motion, unless another force, greater than that applied by the Earth, can act upon it to alter its course. And the apple falls at a constant rate, which formula for free fall is: 9.81m/s2  [where m = meters, s = second].[3]  This is true regardless of the mass of an object and absent air resistance.[4] Newton based his first law on the discovery by Galileo of the nature of objects in free fall, regardless of mass, and lacking air resistance, falling at the same rate of speed.
 
I.c Newton further realized that the Earth’s force of gravity acting on the Moon “pulled” the Moon into an orbit around the center of the earth, the point of origin of Earth’s, and virtually all celestial bodies’ gravitational force, expressing Newton’s second law, force equals mass times acceleration [f = ma].
 
I.d Finally, because the Moon, itself, a separate body in space, exerts the force of gravitation, it, applies that force to Earth. Being a weaker force than Earth’s, its effect is merely observed as tidal forces on the Earth’s bodies of water. This is the demonstration of Newton’s third law; an equal but opposite reaction. Given the relationship of the Earth and Moon, and the Moon’s weaker gravitational force, yet the force applied is equal, if not its effect. That is, the Moon does not and cannot pull the Earth into orbit about it.
 
I.e These three laws explain the reason why the resolution has merit: that if the Earth were a flat disk, a body at the edge of the disk would not fall off the edge in free fall by the force of gravity. Falling, in this instance, on a spherical body, would be a straight motion in the direction of that body’s center, and that is what is meant by the resolution’s statement, “fall off the edge” in the context of the resolution’s proposed flat disk; a “fall” in the direction perpendicular to the plane of the disk.  The nature of the fall of a dropped apple will be explained further on in this round 1 argument.
 
II Argument: Why Newton’s apple at the edge of a disk would not fall “down”
 
II.a Let’s suppose, for the sake of debate, the Earth, as according to Flat Earthers, was a limited plane, a disk for lack of a more complicated shape. After all, we observe the Moon, and other Planets of our Solar System, and, indeed, the Sun, itself, as disks in space, all presenting [usually] their full-circle disk to us. As we observe their motion in space [usually], they must be positioned, each upon its edge, as a coin set on edge on a table, and running in an invisible track around the Sun. We probably appear about the same to them. However, as a point in this discussion, the relative position of each disk, and specifically Earth’s disk, relative to a celestial North-South-East-West orientation, is irrelevant. It just complicates an already complicated definition of the size and shape of celestial bodies, which non-Flat Earthers consider to be uniformly spherical, with the exception of asteroids and comets, which the latter do not see as disks, either.
 
II.b This Earth disk has, just as the non-Flat Earth, a full circle horizon when perceived at a height sufficient to observe the complete, unbroken circumference of the horizon. At a height, because, even though “flat” for generic purposes of this debate, the Earth does have mountains and valleys.
 
II.c The flat Earth still has, as a feature of its construct, a force of gravity. This is so because virtually all celestial bodies of sufficient mass possess a force of gravity. All these bodies are in a state of perpetual motion, whether or not they are acted upon by bodies of greater mass, having, therefore, a greater gravitational force applied to other, smaller bodies, such as the Sun, the largest and most massive mass in our Solar System has gravitational influence over all other bodies permanently resident in the Solar System.
 
II.d However, as explained in argument I.c, any body “falling” by the force of gravity of a body greater than its own, will “fall” to the center of that greater body. If a series of towers of approximately 41m height [the probable height from which Galileo allegedly dropped objects to prove his free fall theory – and some scholars believe this was a “thought experiment” rather than an actual physical experiment conducted from the Tower of Pisa!][5] were constructed around the world [make the towers plumb for convenience!] from which an apple is dropped from each, their universal transit to the ground [on a spherical Earth] would all drop toward the center of the Earth, and not necessarily in a line parallel to the towers – although for practical purposes, the transit line of fall would meet both conditions.
 
II.d.1 However, on a flat disk Earth, because of the nature of free fall as described in I.c, above, an apple dropped from a hand while a person is standing at the edge of the disk, with the hand suspended over the edge, the apple would not fall “down” in a transit parallel with our standing orientation, but toward the center of the disk. This is the nature of Newton’s second law of motion relative to the function of gravity.[6]
 
III Argument: Flat Earth, no gravity? Or, no flat earth?
 
III.a Clearly, the answer to the first question above is, “no.” According to Newton’s laws of motion, the nature of gravity is that it creates spherical forms of celestial bodies because gravitational force is a single point drawing material to it by an equal force on material from all around it.[7] Equal force yields a sphere, not a disk.
 
III.a.1 Suppose for argument’s sake, however, that the Earth were flat, anyway. What, then, would be the nature of gravity? It would still function the same, with a central-point gravitational force located at the center of the diameter, and at the center of the thickness of the disk, presenting properties entirely different than on our spheroid model wherein an apple dropped from a tower, such as in argument II.d, above. Instead, as demonstrated in argument II.d.1, all apples dropped from anywhere on the surface of the disk would be drawn to the center of the disk. The gravitational force would have less force pulling on the apple if dropped at the edge, but the apple would still be drawn on a diagonal line of force toward the center of the disk; not straight down, i.e. perpendicular to the disk plane. The diagonal force is a phenomenon called negative gravitropism.[8],[9]
 
I’ll suspend the argument there, and toss the apple to Con.
 

Con
#2
There were a lot of scientific inaccuracies so I have will counter by stating exactly what will happen if the earth was flat. 

1.Tremendous loss of Mass: currently earth is known to be of ellipsoid shape, for argument sake lets consider it to be a perfect sphere, a sphere has tremendous volume . 
Volume=1,083,206,916,846 km^3 or  1.08* 10^12 km^3 is known to be the volume of the earth
Adjusting all the separate layers of the earth( crust,mantle, outer core, inner core, etc.) 
If we consider it to be a disk, for argument's sake lets take earth's surface area being equal to the surface area of the falt earth. 
for calculating the volume of the new proposed earth(flat earth) lets consider the thickness of disk to be 10km, since disks are arbitrarily known to be thin. 
Volume1 =(5.1* 10^8)*10 km^3
Volume1=5.1* 10^9 km^3. 
Compare Volume with Volume1 we notice that the proposed planet is 500 times less in Volume,and considering density of the crust to be 2550 kg/m^3. (average of upper and lower crust density) 
Weight1=1.3005*10^22 kg. 
Roughly about 500 times reduction in mass, now since we have established a tremendous amount of loss in mass we will begin to ponder about the implications. 

2.Gravitation: Gravitation does not work according to Newton's laws of motion(as described by PRO), it has its separate laws also invented by Newton, on the lines very similar to Coulumb's laws . Each object by virtue of it's mass/ weight exerts a force on a surrounding body. 
F( on either of the two bodies ) = G*(m1*m2) /R^2. 
where , m1= mass of body1
m2=mass of body2 
R=Distance between centre of mass of the bodies. 

Thus by this equation CON refutes points made by PRO
And the apple falls at a constant rate, which formula for free fall is: 9.81m/s2  [where m = meters, s = second].[3]  This is true regardless of the mass of an object and absent air resistance
The weight of an apple consider it as (m2) does effect the total force experienced by it from body1- consider it as earth, but the change in magnitude is so small, mostly it is not taught until a person takes up science after 10 th standard(Atleast in India, I dont know about USA). 

By the very same law, the new flat earth will have 500 times less gravity than the known earth, having the same speed of rotation consider it as a rotating disk. Hypothetically speaking a disk rotating the Sun, also revolving around its own axis. Point to note is that a body on the flat earth will experience500 times or  50,000% less gravity than current earth 


Now lets compare force experienced by mass of 70 kg , 
Force of gravitation= 1.4 Newtons (from the flat earth) 

CONCLUSION: 
The force is so small that other huge celestial bodies and their gravitational field
Sun= 0.4 Newtons , etc will effectively reduce it to 1N , with the average body of 70kg experiencing only 1N in force in gravitation rather than the expected normal 70*9.81 N= 686.7N gravitational force, anything can fall of the edge if earth is flat. Considering life is normal and people still live on the flat earth.  



Round 2
Pro
#3
I Rebuttal: Where did all these weeds come from?
 
I.a I fear my opponent wandered off the path into the weeds. His entre r1 argument consisted of: “1. Tremendous loss of Mass”  and“2. Gravitation.”
 
I.b “Tremendous Loss of Mass”  immediately draws a comparison with mathematic flourish between the current properties of Earth as a proper sphere [yes, technically not a precise sphere, but the debate is not about the current Earth in any event, so… what?].
 
I.b.1 It goes without saying that ifthe Earth were a disk, the resolution of the debate, the condition of which need not be so nearly detailed to arrive at the balance of the resolution, i.e. “any object  [I’ve applied Newton’s apple as an appropriate “object” in question] would not fall off the edge by Newton’s laws of motion”  can be addressed without so many weeds, such as comparisons of mass and volume of a spherical Earth, and a disk Earth. The spherical Earth, relative to this debate’s resolution, simply does not exist and need not become further weighted down than it is already by such calculations as F = GxM1xM2/r2  because that formula, erroneously claimed by Con to be the formula to apply in our gravitation resolution [Con stated in r1:“There were a lot of scientific inaccuracies so I will have to counter stating exactly what will happen if the earth was flat…”]  and proceeded to introduce the formula just noted in italics above. 
 
I.b.2 However, I rebut the weeds: the formula, F = GxM1xM2/r2   happens to be the formula used to calculate the gravitational forces applied by two celestial bodies, such as the Earth and the Moon, or the Sun, to whom we will later be introduced, to one another, and not the formula to apply to the gravitational force of the Earth, even in a disk shape [if that were possible], on an object, such as an apple, which has no gravitational force of its own to apply to anything, particularly relevant to this discussion, such as the flat Earth.
 
I.b.3 Since Con dismissed Newton’s second law of motion, which established f = ma,  in favor of Newton’s law of universal gravitation, F = GxM1xM2/r2,  which is shown above in I.b.2 to be irrelevant to the observation of an object, an apple, subjected to the gravitational force of Earth, flat or round, because we’re really not concerned with acceleration any more than we are of gravitational forces on two bodies when one of those bodies is as small as an apple, we should look to the nature of the “fall” of that apple in a proposed flat-Earth environment. Let’s defer that discussion to argument II, below.
 
I.b.3 I therefore claim Con’s entire first argument, “Tremendous loss of Mass”  to be irrelevant to the debate.
 
I.c “Gravitation,”Con’s second argument of r1, also calls on the formula, F = GxM1xM2/r2, to support his argument. I rebut Con’s argument on the very same principle as explained in I.b.2, above, since, again, the object dropped in question in the debate resolution, has no gravitational force to apply to our proposed disk Earth. Therefore, this second argument, as well, is entirely irrelevant.
 
I.d Con’s conclusion of his r1 becomes entangled in more weeds, saying, “…the Sun = 0.4 Newtons, etc…” and “…with the average body of 70kg experiencing only 1N in force…”
 
I.d.1 The Sun? A 70kg body? I challenge Con to demonstrate where and when these elements were invited to the debate, or, rather, where and when we were invited into these weeds? Pardon my correction, but we need not “consider whether or not life is normal and people still live on the flat earth.” These factors are not in the scope of the resolution, either, and shall not be further entertained.
 
I.d.2 The apple of my r1 is 175g, happens to be a Malus domestica  [golden delicious], from my own tree in my backyard. But, for purposes of this debate, you do not need to know these facts; they, too, are irrelevant, because the apple could be twice that mass, or half, and it would “fall” [see Definitions] on a spherical earth at the identical rate as my subject apple, because the proper formula to apply is still, and will always be 9.81m/s2,  as discovered by Galileo,[1] and upon which Newton created his second law of motion, f = ma.
 
II Argument: The Flat of the apple’s fall
 
II.a In his r1, argument 2, Con said, “…the new flat earth will have 500 times less gravity than the known earth, having the same speed of rotation consider it as a rotating disk.”  We can ignore the reduced gravity of 500x because we are not comparing two bodies in which both have gravitational force upon one another, as explained in my I.b.1 and I.c, above, and because one of those bodies is a mere 175g. So, we can get out of the weeds, get back on the path, and proceed. 
 
II.b The “new, flat Earth” is flat, and we do not care, either, whether it is 10m or 1,000 meters thick. This property, as well, is irrelevant, and Con’s cited law of universal gravitation can do nothing about that. However, when Con supposes that the flat Earth has “the same speed of rotation”  as, I presume he means our spherical Earth, he has just left Newton behind, and embarked on a new journey into unknown and non-cited weeds.
 
II.b.1 To shape a cosmic body into a disk (rather than a sphere), you've got to spin it very fast, says David Stevenson, a planetary scientist at Caltech in Pasadena, California. This would, unfortunately, destroy the planet by tearing it into tiny particles. In the 1850s, astronomer James Clerk Maxwell showed mathematically that a solid, disk-like shape isn't a stable configuration in the cosmos, in work he conducted regarding Saturn’s rings.”[2] 
 
II.b.2 That, alone, dispels the entire theory of flat-Earthers, but, for this debate, we are not concerned about proving or disproving a flat Earth. It is what it is, and for purposes of this debate, the proof of a flat Earth is not in the cards; it’s just the accepted resolution. I mention this because it is just other evidence that Con’s claim that my r1 was “…a lot of scientific inaccuracies” ought to consult a mirror.
 
II.b.3 Therefore, Con’s unsourced statement that our flat Earth will rotate at the same speed as our spheroid Earth, is false, but also irrelevant. It rotates; it matters not one whit at what rate it rotates.
 
II.c The referenced source for this explanation is this,[3] which contains some complicated calculations beyond my understanding, I’ll admit, but there is a handy illustration of a disk-Earth showing everything explained by the math, to wit:  Imagine a person standing at the edge of the disk, back to the center of the disk, and who drops an apple. Actually, according to the math, and the illustration, that person would be lying prone on the ground due to the unique gravitational forces that are featured, by the math, on a planetary disk: Gravitational force, at the center of the disk, is 90º perpendicular to the disk, and the lines of force bend progressively closer to the ground as one moves further from the center to the edge, until, at the edge, the gravitational force is parallel to the plane of the disk. Therefore, the apple, when dropped, will be pulled back toward the center of the disk as it falls. 
 
II.c.1 A video[4] shows this phenomenon, except that instead of a dropped apple, the object is a thrown ball. Observe the other strange properties of this disk gravity.
 
II.d Remember that Con said in his r1 that he would have to “…counter by stating exactly what will happen if the earth was flat.”  Only, Con did not then tell us what I have stated in argument II and all sub-paragraphs to this, did he? He gave us “masses of bodies, universal gravitational constant, and distance between centre of mass of the bodies…” And, “thus by  [the equation of Newton’s universal law of gravitation] CON refutes points made by Pro.”  What Con refuted is demonstrated to be a weeded field. I challenge Con to kill the weeds and get back on the path. After all, Con has not yet, as alleged, shown us “what will happen if the earth was flat,”  but I have, here, in argument II through II.c.1. Literally showed you by a video, no less. Who needs math but for back-up, which I have also shown. Besides, I didn’t serve popcorn for the video, but I did toss you an apple. Bon apétit.
 
 
 
 
 

Con
#4
PRO fails to define properties: PRO describes the flat earth in layman's terms, but the debate is scientific in nature, PRO from the beginning should have defined the dimenions ,weight, surface area, centre of mass whether or not the centroid and centre of mass coincide and other properties of the flat earth. 
Since PRO is the instigator , firstly PRO failed to provide exact details:
  • unknown density
  • unknown centre of mass
  • unknown centroid 
  • unknown mass 
  • unknown volume 
  • unknown surface area
  • unknown data about rotations 
  • did not define whether or not kepler's law are still valid for the given flat earth. (Kepler's laws outline our current understanding of how the celestial bodies in solar system revolve around the earth) 
  • did not define the causality of the force, F=ma holds true for an apple as used by PRO, but how is the force being generated if not for Newton's law of gravitation. 
  • Unknown status of rotation and revolution of the earth- PRO has just mentioned "a state of perpetual motion" how will CON rebut such intangible wording. 
There are simply no numbers to be found in both arguments published by PRO. Sorry Sir! Science is objective , with no numbers it is not resonable for anyone to talk about Science. Semantics and descriptions are of no use in Scientic jargon. All guidelines are pre-set, and are same to all scientific community, any violation renders any proposition, unscientific.

Breach of scietific guidelines: 
  • PRO skips talking about vectors, Newtons laws have a scalar as well as vector form. Scalar= having only magnitude no direction Vector= having both magnitude and direction. How it leads to a breach has been stated below. 
  • Since the word " off " cannot be defined without a use of directions , PRO should have stated all three laws will be used in vector format and specifically defined the guidelines or the field of plane to be able to ascertain those directions. He did not do it. 
  • He has not stated any quantifiable data, let alone usage of units rendering the debate intangible and prone to word- play. 
  • He has not stated the frame of reference from which the apple/body is to be observed. 
All these are severe consequences as it let CON ponder what exactly is PRO's resolution. CON has a reasonable understanding of the current laws of gravitation and kepler's laws for planets and newtonian mechanics , but all those fields  are being mixed without any proper outline.  

 Argument: Newton’s three laws of motion are the basis for the resolution
Rebuttal:  PRO's own sources undermine this stance , yes newton's law of motion talk about the motion of the body and in what specific direction the body will move when acted upon by a force, but to take in account those forces, one has to determine what is the exact magnitude and direction of those forces which are being applied, PRO simply states a= 9.81g/m^2 in R1 .Stating a=9.81 g/m^2 is  a blunder CON will substantiate on it. 

To be able to talk about flat earth, one must first be able to determine the current conditions on earth and then determine how different those conditions will be on the flat earth. When PRO failed to define those conditions CON assumed some basic universal laws and built his argument around it. 

And the apple falls at a constant rate, which formula for free fall is: 9.81m/s2  [where m = meters, s = second].[3]  This is true regardless of the mass of an object and absent air resistance. 
This is taught to 10-12 year old kids just being introduced to physics , in reality everything, every particle in the universe from stars to black holes to a pen is governed by Newton's law of gravitation. 

F=G*m1*m2/r^2 is UNIVERSAL , CANNOT BE ALTERED 
IF PRO had any desire to exclude a universal law of physics he should have stated in before the debate in the description. PRO's own source also states the same Encyclopaedia Britannica 
These three laws explain ........., would be a straight motion in the direction of that body’s center, and that is what is meant by the resolution’s statement, “fall off the edge” in the context of the resolution’s proposed flat disk; a “fall” in the direction perpendicular to the plane of the disk.  The nature of the fall of a dropped apple will be explained further on in this round 1 argument.
As stated previously scientific analysis does not allow usage of terms like straight line, body's centre. 
Body's centre of mass or centroid was is PRO talking about? 

II Argument: Why Newton’s apple at the edge of a disk would not fall “down”
Rebuttal:
However, as a point in this discussion, the relative position of each disk, and specifically Earth’s disk, relative to a celestial North-South-East-West orientation, is irrelevant.
By stating this PRO asked CON to exclude Kepler's laws from the debate and CON has done so not putting Kepler's laws in his R1 argument. 
the flat Earth still has, as a feature of its construct, a force of gravity.This is so because virtually all celestial bodies of sufficient mass possess a force of gravity.
Here PRO acknowledges there will be gravity, since gravity is only governed by F=G*m1*m2/r^2 . PRO is indirectly stating CON is allowed to use it. 
would not fall “down” in a transit parallel with our standing orientation, but toward the center of the disk
Again same problem: centroid or centre of mass? centroid and centre of mass coincide when the density of a body is uniform since earth's density is not uniform, PRO must state, if this "flat earth" has uniform density.
 It would still function the same, with a central-point gravitational force located at the center of the diameter, and at the center of the thickness of the disk, presenting properties entirely different than on our spheroid model wherein an apple dropped from a tower, such as in argument
Here PRO has defined a centroid, but fails to mention about density.  

Based on the disection of PRO's own R1 : he wants to exclude kepler laws but intends to keep laws of gravitation working. 
It goes without saying that ifthe Earth were a disk, the resolution of the debate, the condition of which need not be so nearly detailed to arrive at the balance of the resolution

No sir! I would beg to differ. Details are needed, because at any given moment a person on present earth is acted upon by: 
1. Centrifugal force because of earth's rotation
2.Gravitational pull of other simple objects like other persons, animals,etc.
3.Gravitational pull of other celestial body.[ PRO stated he wanted to exclude this] 
4.Gravitational pull of the Earth. 

In any scenario the 4 th force is pointed vectorically downward, and the massive magitude of 4th force along with it's vector direction keeps us on Earth. Otherwise every person or body would have been thrown into space because of centrifugal force. 

Centrifugal force is the reason why : a thing such as a gravitational slingshot exists and Space agencies use it to accelerate their space probes. If the flat earth is spinning along its centre / axis, every object will have a centrifugal force acting on it.

such as an apple, which has no gravitational force of its own to apply to anything, particularly relevant to this discussion, such as the flat Earth
PRO claims F=G*m1*m2/r^2, is only applicable to celestial bodies. Sir, laws of gravitation and Coloumb's laws(only if particle or body is charged) are applied in any galaxy of this universe to any particle no matter how small or large they might be. 
In nucleus of an atom, gravitational forces play the role of a stabalizing force apart from the strong forces already present in the nuclues of an atom. There are four forces which govern the stability of an atom and gravitational force is one of them so is electromagnetic force( Coloumb's forces).

Gravitational force of an apple on a human:- 
distance =10 cm or 0.1 m 
human=70kg 
apple=0.1kg 
F= (6.673*10^-11 N m^2 kg^-2 )*(70)(0.1)/(0.1)^2. 
F= 4.6711 *10^-8 N . 
Every object in the universe has a gravitational field and a gravitational pull. It is just that the gravitational constant is so low in magnitude that we dont observe this pulls in our day to day life. 

PRO again states that , gravitational force is irrelevant and we should just stick to nature of the fall? 
we should look to the nature of the “fall” of that apple in a proposed flat-Earth environment. Let’s defer that discussion to argument II, below.
It is impossible to ascertain nature of the fall without ascertaining the net magnitude and direction of the vector sum of all the forces acting on a body and the direction and way in which the force is being applied to the body, as well the mass of the body. That is the whole reason for the loss of mass and gravitation argument presented by CON, to determine the force acting on the body. 
The apple of my r1 is 175g, happens to be a Malus domestica  [golden delicious], from my own tree in my backyard. But, for purposes of this debate, you do not need to know these facts; they, too, are irrelevant, because the apple could be twice that mass, or half, and it would “fall” [see Definitions] on a spherical earth at the identical rate as my subject apple, because the proper formula to apply is still, and will always be 9.81m/s2
Sir, if you can prove this claim I will forfeit my degree in Mechanical engineering for which I have dedicated my entire life and rescind my name from my university. I am just sourcing wikipedia but you are welcome to attack me as many times as you like, Physics wont change it's laws for PRO to win his debate. 

At different points on Earth surface, the free fall acceleration ranges from 9.764 m/s2 to 9.834 m/s2[2] depending on altitude and latitude, with a conventional standard value of exactly 9.80665 m/s2 (approximately 32.17405 ft/s2).
derivation: 

F=G*m1*m2/r^2 = m2*a , where m2= body of a mass , cancelling m2 both sides 
F= 6.67*10^-11 * mass of the earth/ distance from the centre of mass^2(would depend on the latitude) = acceleration 
F=3.983324*10^14/distance from the centre of mass^2 (considering average=6371000m) = acceleration
acceleration= 9.813646 m/s^2 . 

9.81 is not some magic number hat scientists discovered, it depends on the weight of the earth, since weight of flat earth will be drastically less there will be no force to drag the body towards the centre of mass of the flat earth. 

I presume he means our spherical Earth, he has just left Newton behind, and embarked on a new journey into unknown and non-cited weeds.
 Sir, either you want the flat earth to exist or not, you can decide it, I am not arguing against it's existence , if I wanted to I would have, it is your flat earth , how am I suposed to know how it functions, but when you fail to set ground rules I have to assume. When I assumed the disk , its mass , the value of gravitational force was drastically low, I have applied a rudimentary yet universal formula. PRO cannot exclude universal formula's such as newtons law of gravitation and then speak as an expert on science citing a paper that explains how will the the magnitude of gravitational fields vary with respect to height. 
To obtain some numerical results, we fix one of the parameters of the problem—surface mass density σ—from the condition that the field strength near the surface (z = 0) at the center (r = 0) is equal to the acceleration g = 9.8 N/kg directed downwards
CON is quoting the same paper, they considered equal mass density and acceleration, why considering those are a problem has already been mentioned in the derivation above. To conclude the body will not have a force of gravitation strong enough to fall back on earth as stated in R1.
Round 3
Pro
#5
I Rebuttal: A new crop of weeds
 
I.a Con argues in his r2 that we must consider density, center of mass, mass, volume, surface area, rotations, Kepler, and a detailed understanding of a state of perpetual motion, scalers, vectors, … in order to understand a flat Earth. Do we really need all that? Consider this, instead: Newton’s laws of motion 1,2,3 are universal in their scope. That is, right now, with all their differences in materials, atmospheres, orbits, rotations, as well as their mass, volume, surface area, species named Kepler, etc, every planet in the Solar System, and, but for rogue planets in the universe, functions using the same force of gravity proposed by Newton, and, as I demonstrated in my r2, arguments I.b.1 through I.d.2. How those gravitational forces exert by values change, due to Con’s alphabet soup, but not the nature of the force. I also demonstrated that flat Earths, if they existed, would exert the same gravitational force explained by Newton, but would express by a different nature. The weeds are completely unnecessary to the debate, as is the fact that flat Earths do not and cannot exist [see my r2, arguments II.c through II.d]. But my resolution does not care if the resolved flat Earth exists, or not.
 
I.b So lost in the weeds is Con, he must ask, “What is Pro’s resolution?” It’s plain English: “If the Earth is a flat disc, any object would not fall off the edge by the force of gravity.”  I defined: earth, object, fall, and gravity in simple terms.  I referenced an article that demonstrated the math of the fall trajectory on a flat earth, and I referenced a video to see the effective nature of a flat Earth’s gravitational force. I am not the one trying to add an alphabet soup in a scientific lexicon, so I suggest Con stop trying to figure out what my debating strategy really is, because it really is simple: the nature of the gravitation force of a flat Earth is a different nature than of a spheroid Earth. That’s it. I would presume that means if Con has a BoP, it is that flats and spheroids have the same nature of gravitational force, and it matters not one whit about their relative shapes, masses, volumes, or inhabitants named Kepler. This was supposed to be fun; remember? Perhaps Con has forgotten how to be a 10-year-old kid. I’m likely twice, maybe three times Con’s age. I have not forgotten.
 
I.c Con argues that I acknowledge Con can use F = GxM1xM2/r2.  Con forgets my r2, I.b.2: “[That formula] happens to be the formula used to calculate the gravitational forces applied by two celestial bodies, such as the Earth and the Moon, or the Sun, to whom we will later be introduced, to one another, and not the formula to apply to the gravitational force of the Earth, even in a disk shape [if that were possible], on an object, such as an apple, which has no gravitational force of its own to apply to anything, particularly relevant to this discussion, such as the flat Earth.” What I meant was that other objects, smaller than planetoids, have no gravitational force of any relevance to larger, celestial bodies. That formula, therefore, is irrelevant to the debate. He may use it all he likes to take up space. Whatever gets his apple eaten.
 
I.d Con argues that the alphabet soup does matter, and offers still another list of soup ingredients: more weeds. One of the new weeds bears notice: Gravitational pull of other simple objects like other persons, animals, etc.”  I challenge Con to show me, by scholastic reference, that persons and animals, and all life on earth, combined, exert a measureable gravitational force sufficient to be the cause of moving the Earth any differently than the gravitational force of the Sun acting on Earth, let alone any of the other planets. The Moon exerts enough force to create tidal action, and not even Barack Obama can do that. Recall his reference that due to his administration, sea level would fall. Nice claim. Science doesn’t buy it; not even that called “Climate.”
 
Con argues the weight of a gnat [in no condition anywhere close to 70kg] can move the Earth. Okay, show me. Actually, Con did, to the tune of 70kg providing a force equal to fractions more than 4 x 10-8. I submit that may not be enough force to move the gnat, let alone the Earth. I argue it doesn’t matter. In my resolution, a gnat on the apple may eventually eat the apple, but, in the meantime, when dropped by hand at the edge of flat Earth, that gnat on the dropped apple needs firm jaws, because the apple is going bye-bye, behind the dropping hand, and will move, probably bouncing, toward the center of the disk, as demonstrated in my r2, source [4].[1] Gnats of the world, unite! You have the world to save from continuing its orbit about the Sun, with all the 70kg people on it.
 
I.e Con argued that my formula from r1, I.b.1, 9.81m/s2,was in error because he calculates it to a variance of 9.764 blah, blah to 9.834 blah, blah. More weeds. If my source[2] for 9.81 says that’s the general value to apply to the formula [it is not the average of Con’s calculations] am I really going to quibble that Con is wrong, as he quibbles I am? What is the bloody difference? In the scheme of a flat Earth v. spherical Earth proposal, does it really matter a tinker’s dam if the trajectory taken by the fall of an apple on a flat earth has a slightly altered trajectory depending on such weeds as at which coordinate in a 360º arc the apple dropper happens to be? Really??? Not all calculations need to be proven to an alphabet letters’ scope of decimal places, I don’t care what one’s Engineering degree says, or what one does with it. This debate does not require it, as demonstrated in my r1, argument III.a.1, and forward, which is why Con’s claim is in the weeds.
 
I.e.1 I conclude that Con has created a need that was never suggested, and made “sure” [only, he did not, as demonstrated further in this rebuttal section] his need, weeds, were arranged such that, by his calculations, which were never suggested, and, in fact, were warned against, result in a disk of specific alphabet soup parameters that will not provide sufficient gravitational force “…strong enough to fall back on earth as stated in R1.”  I presume Con means his r1, but, when specifics are a help to readers, Con lacks them.
 
I.e.2 But, let us recall that the weeds are of Con’s construct, not mine. My construct of the dropped apple falling behind the dropper and toward the center of the disk was proven by my reference to https://iopscience.iop.org/article/10.1088/1361-6404/ab0bba/meta,
which demonstrated, though it appears Con did not read it, mathematically, exactly the phenomenon on a flat Earth just described. Moreover, that scientific paper, titled, What the gravitation of a flat earth would look like and why thus Earth is not actually flat,  just happens to reference the video I referenced separately [see [1], below]. This positively refutes Con’s claim in his r2 conclusion. Engineering is not a perfect science. Not yet. Fortunately, even on sphere Earth, Engineering, being imperfect, has latitude: it’s called tolerance. Like 9.764 blah, blah to 9.834 blah, blah. 
 
I.e.3 The casual observer will note that 9.81 falls between Con’s two exhaustively calculated values. Not all tolerances in scientific engineering notation are always ±x;  of equivalent value on each side of the target specification. My Engineering education taught a simple principle: why have eight decimals of exacting specification tolerance when two will suffice. The result is unnecessarily precise, causing opportunity for measurement error simply because your measurement device is not capable of repeatable and reliable measurement at that exhaustive detail. Silly reason for a failure when the customer is delighted with the quality of your product at two-decimal tolerance. Who, after all, defines quality? Not the Engineer. Nor the gnat.
 
I.e.4 I therefore conclude I have proven my BoP: “If the Earth is a flat disc, any object would not fall off the edge by the force of gravity.”
 
I welcome the repetition of weeds, and the resulting dandelion soup. Then I’ll ask for your vote. Thank you for your attention.
 
 
 
 

Con
#6
I have never argued that a person can move the earth or any such thing merely demonstrated that that gravitational field is universal will exist whereever mass is present.  
Newton's law's or motion are actually not universal in scope as mentioned by my opponent, newtonian mechanics cannot be applied everywhere, fluid mechanics is a field where we dont use newtonian mechanics, it is mostly limited to solids. I will not argue about the relevance of my points , voters can judge for themselves whether the argument of a drastically less gravitational field is relevant or not. 

PRO fails to give credence to valid points put up by CON , that a gravitational field will depend on weight of the earth and the distance fromt the centre of mass of the earth to the body we are talking about, since both are uncertain in the flat earth as PRO did not shed light on the criteria, it cannot be said that PRO was able to prove his point. 

In my R2 I mentioned PRO failed to mentioned by what frame of reference will he be observing the fall, if that is not mentioned PRO's resolution already fails. 

A person is sitting still in a train moving at 60km/hr south, is he in motion or at rest ? Answer is both , and depends on the frame of reference 
If we look at the person from the frame of reference from the ground he is moving 60km/hr south. 
If we look at the person from the frame of refernce from the compartment he is sitting in he is not in motion as will be considered as such.


Even if PRO's arguments hold true, it is his BOP to prove the body will fall off but since the frame of reference is not mentioned, it cannot be proved conclusively where the body will go. Since the matter is ambiguous CON claims victory. CON made all possible attempts to ascertain what gravitational field will be associated with such a body and a voter  can decide from himself which side was more convincing.