They need to get Haswell out. They need to get Haswell out YESTERDAY. The touted massive battery life improvement will be the clincher in creating x86 tablets if that's their bag.
I'm rooting for Apple's choice, but it's silly to think that ARM designs will have any reason to get better if no one can compete against them. I'm rooting for Intel to compete as they have always done.
As a non imperial units user, I thought that pound was only a mass unit. I quick internet search showed me that there are two pound units...
As a reminder to anyone, mass and weight are two distinct concepts in physics. As in theory inertial mass and gravitational mass are two distinct things.
Well let's confuse it more and recognize that a pound is also a unit of currency! I don't think anyone disputes the separate natures of mass and force. There was just a lot of digging in over what pound refers to exactly, which is both of those and more. All contextually dependent.
I'm just trying to figure out why I have a couple of micro scratches in the back of my iphone. If a softer material really can't scratch a harder material (despite being sharp and a force being applied) then it must be a small rock or piece of sand in my pocket that made a scratch.
It's the bold underlined part, or some similarly hard material particle that made it into your pocket or onto some other item in your pocket which is itself softer.
Quote:
Originally Posted by bigdaddyp
I sometimes use our ottoman (footstool) as a desk. I will then sometimes charge my iPhone via the usb port on the unibody macbook to charge my iPhone. With kids and pets running around sometimes the computer gets bumped or nudged and it slides up and over the phones screen. I believe that the anodized coating on the Macbook or the Apple wireless keyboard is hard enough to scratch the gorilla glass.
Or like the above, there was some sand residue on the ottoman, which somehow got between the iPhone and the MacBook case. Now you have the proverbial rock between two hard places. The particles that can do the scratching can be really small, small enough to feel like smooth powder to a finger, if they are pinched down by another hard surface.
That depends where you are on the Earth. For example at the equator, on the top of a mountain you would weigh slightly less due to centrifugal force being greater than, let's say, The North Pole. Also the Earth's gravitational lines of force are not an exact vector pointing to the center of the Earth either. That also varies depending on several factors including latitude, altitude, the time of year and the position of the moon. I mean, we could be talking nearly several grams of difference if you want to get technical about it.
The centrifugal[centripetal] force has nothing to do with not at all, that would actually be trying to throw you off the mountain and would be higher on a high mountain. The gravity difference is simply that you would be farther from the Earth's center of mass, so your r^2 term is larger, making the force proportionally smaller. Latitude has the same effect because the Earth is an oblate spheroid bulging towards the equator, not a true sphere, so both distance and mass relative positions are affected. Other mass effecting properties of the Earth not being a homogenous constant density sphere hold true at the same time.
And the differences on a mountain would be in thousands to hundredths of a gram, weight-wise. Even in orbit a body still has a lot of "weight", it just isn't being pushed back-on by a fixed surface. The orbiting body is just traveling so fast sideways it is roughly canceling out the falling part of the motion and you never get to a place to stand or go splat.
The centrifugal force has nothing to do with not at all, that would actually be trying to throw you off the mountain and would be higher on a high mountain. The gravity difference is simply that you would be farther from the Earth's center of mass, so your r^2 term is larger, making the force proportionally smaller. Latitude has the same effect because the Earth is an oblate spheroid bulging towards the equator, not a true sphere, so both distance and mass relative positions are affected. Other mass effecting properties of the Earth not being a homogenous constant density sphere hold true at the same time.
And the differences on a mountain would be in thousands to hundredths of a gram, weight-wise. Even in orbit a body still has a lot of "weight", it just isn't being pushed back-on by a fixed surface. The orbiting body is just traveling so fast sideways it is roughly canceling out the falling part of the motion and you never get to a place to stand or go splat.
He meant that the earth's rotation affects gravity as a function of latitude, not altitude. At the equator, you are both further from the earth's center of mass and rotating around it. Those two actually produce roughly equal reductions in weight at the equator relative to the poles - of the order of 0.5%.
With altitude above the earth's surface at the equator, the effect is dominated by distance from CoM, since g varies inversely with r², whereas the centrifugal force increases only linearly with r.
He meant that the earth's rotation affects gravity as a function of latitude, not altitude. At the equator, you are both further from the earth's center of mass and rotating around it. Those two actually produce roughly equal reductions in weight at the equator relative to the poles - of the order of 0.5%.
With altitude above the earth's surface at the equator, the effect is dominated by distance from CoM, since g varies inversely with r², whereas the centrifugal force increases only linearly with r.
I know what he meant, and he was wrong, as are you. The rotation has NOTHING to do with the gravity. Ever.
You can't even construct the virtual centripetal[centrifugal] force vector because the solid surface is under the person standing, not on the outside radius as it is in a centrifuge. There simply is no force trying to throw things off the surface of the Earth because the earth is spinning, so there is no reduction in gravity because of it.
Differences in gravity between Poles and equator are due to the shape of the Earth and differences in local composition density contributions.
The only thing about escaping gravity that is good about the equator is that the speed for orbit is independent of the launch point. 17K MPH is 17K MPH whether you launch from Alaska or Canaveral, but closer to the equator you are moving faster and if you launch in the same direction as rotation you don't need as much fuel to accelerate to that 17K MPH, because you start with 1K MPH at the equator.
I know it is not really an authoritative source but... just saying.
It's a fine source.
Quote:
Originally Posted by Hiro
I know what he meant, and he was wrong, as are you. The rotation has NOTHING to do with the gravity. Ever.
You can't even construct the virtual centripetal force vector because the solid surface is under the person standing, not on the outside radius as it is in a centrifuge. There simply is no force trying to throw things off the surface of the Earth because the earth is spinning, so there is no reduction in gravity because of it.
Differences in gravity between Poles and equator are due to the shape of the Earth and differences in local composition density contributions.
Let me give you a question to consider. I assume that you would agree that on the equator, one is moving (relative to the center of mass, M, of the earth) in a circular path, of radius equal to the equatorial radius of the earth, R, and a period of one day. In which case, would you agree that your motion includes a radial acceleration, a, (normal to your instantaneous velocity vector) given by
a = - ω²R , [1]
and that to maintain that acceleration (circular motion) requires a centripetal force, F, given by
F = - mω²R, [2]
where m is your mass and ω your angular velocity?
Hopefully you would agree, because that is simple Newtonian physics. That centripetal force is, of course, provided by a fraction of your gravitational attraction to the earth, and your apparent weight, mg (the residual reaction between you and the surface you are standing on), is reduced by that amount. Or more rigorously, if G is the gravitational constant and g is the acceleration due to gravity in the local coordinate system (your initial acceleration if you jump off a building), for mechanical equilibrium between yourself and the surface of the earth:
mg = GMm/R² - mω²R , [3]
While there actually is no such thing as a centrifugal force - it is just Newton's 3rd law reaction to a centripetal force - there is indeed a centripetal force acting in this system.
So, I assume that you are drawing the distinction between pure gravitational attraction and actual acceleration due to gravity in a dynamic system. If the discussion is static gravitational attraction then you are correct, but when discussing actual acceleration due to gravity (i.e. the acceleration of an object in free fall in the local coordinate system), the angular velocity, ω, of the coordinate system has the effect described, leading to an acceleration given by dividing [3] though by your mass, m,
g = GM/r² - ω²r . [4]
The second term on the RHS is the effect of the earth's rotation. Does that explicit distinction reconcile the two views?
I don't disagree with it, but it is being horribly misinterpreted in this thread. Gravity is gravity. Apparent gravity (which is the part of that page which is being referred to without proper disambiguation) is an entirely different phenomena that takes everything into account including hiccups. It isn't gravity at all, but just a resultant force of all forces and accelerations currently acting on a body. And many shorthand the name as apparent gravity because the direction is almost the same as the local true gravity vector.
Quote:
Originally Posted by muppetry
Let me give you a question to consider. I assume that you would agree that on the equator, one is moving (relative to the center of mass, M, of the earth) in a circular path, of radius equal to the equatorial radius of the earth, R, and a period of one day. In which case, would you agree that your motion includes a radial acceleration, a, ...
No, I wouldn't agree with that. Bad assumptions. You are trying to measure from the wrong frame and wrong coordinate system to make that direct comparison. When the frame of reference is the center of rotation of the Earth there is no radial acceleration of an object on it's surface because the Earth is for all intent and purposes rotating at constant velocity (on a scale of billions of years). And an object on the surface remains in the same absolute & relative position on the surface. When you use the correct reference frame and a 3D polar coordinate system that is compatible with the 3D shape of the Earth there is no need for virtual forces to balance out weird interactions of a 2D projection of a 3D system. In the polar fixed system the answer are dead-simple vector-linear relationships to the cg of the Earth which follow the force over squared-distance rule.
The rest of the apparent gravity portion of the post was superfluous because it depended on the entering assumption which I completely disagree with. And it's a good thing I do because there is no skyhook to grab us and provide the centripetal force you want to find. If you want to measure from a non center of the earth frame and talk about earth surface gravity (not the fake apparent gravity) you have a hell of a lot more math to do to keep everything correct.
Quote:
Originally Posted by muppetry
So, I assume that you are drawing the distinction between pure gravitational attraction and actual acceleration due to gravity in a dynamic system.
Again I'll stop you right here. Gravity is gravity, there is no difference between static and dynamic gravity. The force and acceleration it imparts is the same whether you are talking static or dynamic. You are confusing the fact that when you measure gravitational force from two different reference frames the numbers will look different, because they are relative to some other dynamic thing, but the actual gravitational force and acceleration vector will be in the same direction and have the same magnitude. All because the vectors themselves are absolutes, independent of the frame. You just perceive the projections of these absolute vectors onto the alternate reference frames to be different from the relatively moving frame reference you are measuring from.
It is dangerous and incorrect to compare values measured in one frame to values measured in a different frame without applying a correction factor taking into account the difference in relative motion between the frames. Sometimes that is straightforward, sometimes it is an all-out-complete-cluster-mess. And the all-out-complete-cluster-mess is common enough that is is just a damn sight easier to remember to only compare inter-frame rather than intra frame.
Just keep it simple. Gravity is gravity, the attractive force between objects with mass. If you want to combine gravitational vectors with other force vectors, that works because D'Lambert proved so, but the resultant vector isn't gravity, it is just the resultant vector.
It isn't gravity at all, but just a resultant force of all forces and accelerations currently acting on a body.
That is exactly what the original thread is about. Since the gorilla glass has very little magnetic properties the force it can withstand is not affected by gravitation either on Earth or in space. The discussion was slightly diverted by the comment that 1 kg equals 9.8 N which is such a rounded average that it would be true even if a kg only had 995 g or as much as 1005 g which by definition is completely unscientific. I apologize for being overly terse and unnecessarily technical but when you examine physical properties from an analytical perspective you have to consider every possible implication that the laws of physics dictate.
When the frame of reference is the center of rotation of the Earth there is no radial acceleration of an object on it's surface because the Earth is for all intent and purposes rotating at constant velocity (on a scale of billions of years).
OK - well I guess we can stop right there. Your argument is that an object moving in a circle (represented by the motion at the earth's surface due to its rotation about its center of mass) has no centripetal acceleration because it has constant angular velocity? You really want to stick with that? Because if so then you are woefully ignorant of basic physics. That you choose to demonstrate that ignorance by arguing with a physicist is kind of unfortunate, because I can't let that kind of error go uncorrected. Actually "arguing with" is a bit of a stretch. You are simply contradicting me.
Centripetal acceleration is just ω²r - it depends on the square of the angular velocity, not the rate of change of angular velocity. All objects in constant circular motion have a constant and non-zero radial acceleration by definition. Since the centripetal (radial) acceleration vector is normal to the (tangential) velocity vector it results only in a change in direction (that which causes circular motion), with no change in the magnitude of the velocity vector (speed). Perhaps that is what is confusing you. They are separate.
And by the way - when I refer to "coordinate system" I don't mean projection or anything else related to maps - it is entirely shape independent and I'm referring to the frame of reference in Euclidean space.
In any case, attempting to dismiss the analysis by vague mis-application the laws of motion will not work. If you would like to take a more rigorous stab at proving your assertion then be my guest. You will fail though.
So we're arguing the very clear definition of gravity vs the effects of gravity of a given planetary body acting upon an object when accounting for effects that would negate the perceived effects of gravity at a given location?
So we're arguing the very clear definition of gravity vs the effects of gravity of a given planetary body acting upon an object when accounting for effects that would negate the perceived effects of gravity at a given point?
Well I thought that was exactly what we were arguing about until our friend started with the puesdo-scientific stuff and then massacred some of the basic laws of physics. Then it stopped just being two different viewpoints on the same thing and became a physics lesson.
Comments
I'm rooting for Apple's choice, but it's silly to think that ARM designs will have any reason to get better if no one can compete against them. I'm rooting for Intel to compete as they have always done.
Land.
African or European?
African or European?
Huh? I... I don't know that.
Huh? I... I don't know that.
It's a joke from Monty Python. I was making an oblique reference to it with sea or land slug. African and European is just a dead give away.
Since you don't know, you're going to be pushed off the cliff now.
It's a joke from Monty Python.
He's providing the in-movie response.
Since you don't know, you're going to be pushed off the cliff now.
Indeed.
It's a joke from Monty Python. I was making an oblique reference to it with sea or land slug. African and European is just a dead give away.
Since you don't know, you're going to be pushed off the cliff now.
Yes - sorry - I should have been clearer, but I assumed that you were expecting the Bridgekeeper's reply.
Some additional info in this BBC article: http://www.bbc.co.uk/news/technology-16523817
Nice article, thanks. Their stock has halved this past year; perhaps time to invest now ($14.32) GLW, NYSE.
As a non imperial units user, I thought that pound was only a mass unit. I quick internet search showed me that there are two pound units...
As a reminder to anyone, mass and weight are two distinct concepts in physics. As in theory inertial mass and gravitational mass are two distinct things.
Well let's confuse it more and recognize that a pound is also a unit of currency! I don't think anyone disputes the separate natures of mass and force. There was just a lot of digging in over what pound refers to exactly, which is both of those and more. All contextually dependent.
I'm just trying to figure out why I have a couple of micro scratches in the back of my iphone. If a softer material really can't scratch a harder material (despite being sharp and a force being applied) then it must be a small rock or piece of sand in my pocket that made a scratch.
It's the bold underlined part, or some similarly hard material particle that made it into your pocket or onto some other item in your pocket which is itself softer.
I sometimes use our ottoman (footstool) as a desk. I will then sometimes charge my iPhone via the usb port on the unibody macbook to charge my iPhone. With kids and pets running around sometimes the computer gets bumped or nudged and it slides up and over the phones screen. I believe that the anodized coating on the Macbook or the Apple wireless keyboard is hard enough to scratch the gorilla glass.
Or like the above, there was some sand residue on the ottoman, which somehow got between the iPhone and the MacBook case. Now you have the proverbial rock between two hard places. The particles that can do the scratching can be really small, small enough to feel like smooth powder to a finger, if they are pinched down by another hard surface.
That depends where you are on the Earth. For example at the equator, on the top of a mountain you would weigh slightly less due to centrifugal force being greater than, let's say, The North Pole. Also the Earth's gravitational lines of force are not an exact vector pointing to the center of the Earth either. That also varies depending on several factors including latitude, altitude, the time of year and the position of the moon. I mean, we could be talking nearly several grams of difference if you want to get technical about it.
The centrifugal [centripetal] force has nothing to do with not at all, that would actually be trying to throw you off the mountain and would be higher on a high mountain. The gravity difference is simply that you would be farther from the Earth's center of mass, so your r^2 term is larger, making the force proportionally smaller. Latitude has the same effect because the Earth is an oblate spheroid bulging towards the equator, not a true sphere, so both distance and mass relative positions are affected. Other mass effecting properties of the Earth not being a homogenous constant density sphere hold true at the same time.
And the differences on a mountain would be in thousands to hundredths of a gram, weight-wise. Even in orbit a body still has a lot of "weight", it just isn't being pushed back-on by a fixed surface. The orbiting body is just traveling so fast sideways it is roughly canceling out the falling part of the motion and you never get to a place to stand or go splat.
The centrifugal force has nothing to do with not at all, that would actually be trying to throw you off the mountain and would be higher on a high mountain. The gravity difference is simply that you would be farther from the Earth's center of mass, so your r^2 term is larger, making the force proportionally smaller. Latitude has the same effect because the Earth is an oblate spheroid bulging towards the equator, not a true sphere, so both distance and mass relative positions are affected. Other mass effecting properties of the Earth not being a homogenous constant density sphere hold true at the same time.
And the differences on a mountain would be in thousands to hundredths of a gram, weight-wise. Even in orbit a body still has a lot of "weight", it just isn't being pushed back-on by a fixed surface. The orbiting body is just traveling so fast sideways it is roughly canceling out the falling part of the motion and you never get to a place to stand or go splat.
He meant that the earth's rotation affects gravity as a function of latitude, not altitude. At the equator, you are both further from the earth's center of mass and rotating around it. Those two actually produce roughly equal reductions in weight at the equator relative to the poles - of the order of 0.5%.
With altitude above the earth's surface at the equator, the effect is dominated by distance from CoM, since g varies inversely with r², whereas the centrifugal force increases only linearly with r.
He meant that the earth's rotation affects gravity as a function of latitude, not altitude. At the equator, you are both further from the earth's center of mass and rotating around it. Those two actually produce roughly equal reductions in weight at the equator relative to the poles - of the order of 0.5%.
With altitude above the earth's surface at the equator, the effect is dominated by distance from CoM, since g varies inversely with r², whereas the centrifugal force increases only linearly with r.
I know what he meant, and he was wrong, as are you. The rotation has NOTHING to do with the gravity. Ever.
You can't even construct the virtual centripetal [centrifugal] force vector because the solid surface is under the person standing, not on the outside radius as it is in a centrifuge. There simply is no force trying to throw things off the surface of the Earth because the earth is spinning, so there is no reduction in gravity because of it.
Differences in gravity between Poles and equator are due to the shape of the Earth and differences in local composition density contributions.
The only thing about escaping gravity that is good about the equator is that the speed for orbit is independent of the launch point. 17K MPH is 17K MPH whether you launch from Alaska or Canaveral, but closer to the equator you are moving faster and if you launch in the same direction as rotation you don't need as much fuel to accelerate to that 17K MPH, because you start with 1K MPH at the equator.
I know what he meant, and he was wrong, as are you. The rotation has NOTHING to do with the gravity. Ever.
Maybe we should start by reading the wikipedia page on gravity.
http://en.wikipedia.org/wiki/Gravity_of_Earth
I know it is not really an authoritative source but... just saying.
Maybe we should start by reading the wikipedia page on gravity.
http://en.wikipedia.org/wiki/Gravity_of_Earth
I know it is not really an authoritative source but... just saying.
It's a fine source.
I know what he meant, and he was wrong, as are you. The rotation has NOTHING to do with the gravity. Ever.
You can't even construct the virtual centripetal force vector because the solid surface is under the person standing, not on the outside radius as it is in a centrifuge. There simply is no force trying to throw things off the surface of the Earth because the earth is spinning, so there is no reduction in gravity because of it.
Differences in gravity between Poles and equator are due to the shape of the Earth and differences in local composition density contributions.
Let me give you a question to consider. I assume that you would agree that on the equator, one is moving (relative to the center of mass, M, of the earth) in a circular path, of radius equal to the equatorial radius of the earth, R, and a period of one day. In which case, would you agree that your motion includes a radial acceleration, a, (normal to your instantaneous velocity vector) given by
a = - ω²R , [1]
and that to maintain that acceleration (circular motion) requires a centripetal force, F, given by
F = - mω²R, [2]
where m is your mass and ω your angular velocity?
Hopefully you would agree, because that is simple Newtonian physics. That centripetal force is, of course, provided by a fraction of your gravitational attraction to the earth, and your apparent weight, mg (the residual reaction between you and the surface you are standing on), is reduced by that amount. Or more rigorously, if G is the gravitational constant and g is the acceleration due to gravity in the local coordinate system (your initial acceleration if you jump off a building), for mechanical equilibrium between yourself and the surface of the earth:
mg = GMm/R² - mω²R , [3]
While there actually is no such thing as a centrifugal force - it is just Newton's 3rd law reaction to a centripetal force - there is indeed a centripetal force acting in this system.
So, I assume that you are drawing the distinction between pure gravitational attraction and actual acceleration due to gravity in a dynamic system. If the discussion is static gravitational attraction then you are correct, but when discussing actual acceleration due to gravity (i.e. the acceleration of an object in free fall in the local coordinate system), the angular velocity, ω, of the coordinate system has the effect described, leading to an acceleration given by dividing [3] though by your mass, m,
g = GM/r² - ω²r . [4]
The second term on the RHS is the effect of the earth's rotation. Does that explicit distinction reconcile the two views?
It's a fine source.
I don't disagree with it, but it is being horribly misinterpreted in this thread. Gravity is gravity. Apparent gravity (which is the part of that page which is being referred to without proper disambiguation) is an entirely different phenomena that takes everything into account including hiccups. It isn't gravity at all, but just a resultant force of all forces and accelerations currently acting on a body. And many shorthand the name as apparent gravity because the direction is almost the same as the local true gravity vector.
Let me give you a question to consider. I assume that you would agree that on the equator, one is moving (relative to the center of mass, M, of the earth) in a circular path, of radius equal to the equatorial radius of the earth, R, and a period of one day. In which case, would you agree that your motion includes a radial acceleration, a, ...
No, I wouldn't agree with that. Bad assumptions. You are trying to measure from the wrong frame and wrong coordinate system to make that direct comparison. When the frame of reference is the center of rotation of the Earth there is no radial acceleration of an object on it's surface because the Earth is for all intent and purposes rotating at constant velocity (on a scale of billions of years). And an object on the surface remains in the same absolute & relative position on the surface. When you use the correct reference frame and a 3D polar coordinate system that is compatible with the 3D shape of the Earth there is no need for virtual forces to balance out weird interactions of a 2D projection of a 3D system. In the polar fixed system the answer are dead-simple vector-linear relationships to the cg of the Earth which follow the force over squared-distance rule.
The rest of the apparent gravity portion of the post was superfluous because it depended on the entering assumption which I completely disagree with. And it's a good thing I do because there is no skyhook to grab us and provide the centripetal force you want to find. If you want to measure from a non center of the earth frame and talk about earth surface gravity (not the fake apparent gravity) you have a hell of a lot more math to do to keep everything correct.
So, I assume that you are drawing the distinction between pure gravitational attraction and actual acceleration due to gravity in a dynamic system.
Again I'll stop you right here. Gravity is gravity, there is no difference between static and dynamic gravity. The force and acceleration it imparts is the same whether you are talking static or dynamic. You are confusing the fact that when you measure gravitational force from two different reference frames the numbers will look different, because they are relative to some other dynamic thing, but the actual gravitational force and acceleration vector will be in the same direction and have the same magnitude. All because the vectors themselves are absolutes, independent of the frame. You just perceive the projections of these absolute vectors onto the alternate reference frames to be different from the relatively moving frame reference you are measuring from.
It is dangerous and incorrect to compare values measured in one frame to values measured in a different frame without applying a correction factor taking into account the difference in relative motion between the frames. Sometimes that is straightforward, sometimes it is an all-out-complete-cluster-mess. And the all-out-complete-cluster-mess is common enough that is is just a damn sight easier to remember to only compare inter-frame rather than intra frame.
Just keep it simple. Gravity is gravity, the attractive force between objects with mass. If you want to combine gravitational vectors with other force vectors, that works because D'Lambert proved so, but the resultant vector isn't gravity, it is just the resultant vector.
It isn't gravity at all, but just a resultant force of all forces and accelerations currently acting on a body.
That is exactly what the original thread is about. Since the gorilla glass has very little magnetic properties the force it can withstand is not affected by gravitation either on Earth or in space. The discussion was slightly diverted by the comment that 1 kg equals 9.8 N which is such a rounded average that it would be true even if a kg only had 995 g or as much as 1005 g which by definition is completely unscientific. I apologize for being overly terse and unnecessarily technical but when you examine physical properties from an analytical perspective you have to consider every possible implication that the laws of physics dictate.
When the frame of reference is the center of rotation of the Earth there is no radial acceleration of an object on it's surface because the Earth is for all intent and purposes rotating at constant velocity (on a scale of billions of years).
OK - well I guess we can stop right there. Your argument is that an object moving in a circle (represented by the motion at the earth's surface due to its rotation about its center of mass) has no centripetal acceleration because it has constant angular velocity? You really want to stick with that? Because if so then you are woefully ignorant of basic physics. That you choose to demonstrate that ignorance by arguing with a physicist is kind of unfortunate, because I can't let that kind of error go uncorrected. Actually "arguing with" is a bit of a stretch. You are simply contradicting me.
Centripetal acceleration is just ω²r - it depends on the square of the angular velocity, not the rate of change of angular velocity. All objects in constant circular motion have a constant and non-zero radial acceleration by definition. Since the centripetal (radial) acceleration vector is normal to the (tangential) velocity vector it results only in a change in direction (that which causes circular motion), with no change in the magnitude of the velocity vector (speed). Perhaps that is what is confusing you. They are separate.
And by the way - when I refer to "coordinate system" I don't mean projection or anything else related to maps - it is entirely shape independent and I'm referring to the frame of reference in Euclidean space.
In any case, attempting to dismiss the analysis by vague mis-application the laws of motion will not work. If you would like to take a more rigorous stab at proving your assertion then be my guest. You will fail though.
So we're arguing the very clear definition of gravity vs the effects of gravity of a given planetary body acting upon an object when accounting for effects that would negate the perceived effects of gravity at a given point?
Well I thought that was exactly what we were arguing about until our friend started with the puesdo-scientific stuff and then massacred some of the basic laws of physics. Then it stopped just being two different viewpoints on the same thing and became a physics lesson.