Is it possible that the gravity that keeps our feet planted on the Earth is..

I see how you could make certain conceptual mistakes and come to that conclusion, but it's not a reasonable conclusion at all. The law of inertia means that every chunk of mass on the surface of the Earth will keep moving at a constant speed in a constant direction, and will go flying off into space as the Earth rotates, unless a gravitational force is applied. Inertial objects cannot rotate around something unless an actual force is holding them in place. Imagine whipping a ball around on a tethered string- if you cut the string, the ball does not come flying back to you.

Its momentum holds it into place. I don't know how else to describe it. the resultants of momentum of the center of the Earth and the objects on its surface intersect, drawing them together as they move around the galaxy.
 
I think the problem is he spent a bunch of time on his youtube presentation and cannot bear the thought that it was a complete waste of time, so he is now in the mode of making it up as he goes along. Jiveabillion - it is time to cut your losses and move on to something else. Your ideas are demonstrably wrong and just plain nonsensical - sorry.

I don't understand how you can ignore the effects of momentum just because there is already an explanation of gravity. Don't you think that it would throw things off?
 
Its momentum holds it into place. I don't know how else to describe it. the resultants of momentum of the center of the Earth and the objects on its surface intersect, drawing them together as they move around the galaxy.

What I'm saying is that there's a mathematical way of showing that even if you added a billion different velocities together to get your resultant velocities, without any net forces you would find that any object on the surface of the Earth should go flying away off into space. Inertia alone never, ever holds you to a rotating surface. No different than a ball swinging on a tether when you cut the tether.
 
What I'm saying is that there's a mathematical way of showing that even if you added a billion different velocities together to get your resultant velocities, without any net forces you would find that any object on the surface of the Earth should go flying away off into space. Inertia alone never, ever holds you to a rotating surface. No different than a ball swinging on a tether when you cut the tether.

Please show me the math and I will try to read it.
 
Will you just go away if you have nothing of value to add to the conversation?
There is no value in this thread to begin with. You've been corrected more times than I can count, by everyone who posts, and you still don't have a clue. You're a true crank, knowing nothing and convinced you're right.
 
I don't understand how you can ignore the effects of momentum just because there is already an explanation of gravity. Don't you think that it would throw things off?
As I pointed out before, there are no major errors in our current theory of gravity. So no, it could not be throwing things off. If what you predict doesn't match what the current theory predicts, then your predictions are wrong.
Its momentum holds it into place. I don't know how else to describe it. the resultants of momentum of the center of the Earth and the objects on its surface intersect, drawing them together as they move around the galaxy.
If that were true, then Earth's perceived gravity would be substantially different at different locations, times and times of the year. And it isn't. And Newtonian mechanics would already account for it.

Newton's first law says that an object will continue in motion unless acted upon by an outside force. So there must be a force pulling objects into their circle around earth, not the other way around (moving in a circle causes the force).
 
There is no value in this thread to begin with. You've been corrected more times than I can count, by everyone who posts, and you still don't have a clue. You're a true crank, knowing nothing and convinced you're right.

You, sir, are a disgusting individual. Wouldn't be surprised if you had very few real friends. You have no social skills, you put people on the defensive, and you're just a jerk. You are no longer welcome in this thread. If I could force you out, I would.
 
Please show me the math and I will try to read it.

Alright, let's say we denote the centre of Earth's velocity relative to the sun by the vector $$\vec{v_1}=\left(v_{1_x},v_{1_y},v_{1_z}\right)$$. And the sun has velocity $$\vec{v_2}$$ relative to the centre of the galaxy. And our galaxy has velocity $$\vec{v_3}$$ relative to whatever, and so on... Denote the vector sum of these velocities as $$\vec{v_e}=\vec{v_1}+\vec{v_2}+\vec{v_3}+\ldots$$. So the centre of Earth is moving with velocity $$\vec{v_e}$$ with respect to some reference point somewhere in the universe.

Place an observer at point $$A$$ on the Earth's surface, they will be rotating about Earth's centre with velocity $$\vec{v_o}$$, and thus have a net velocity with respect to our chosen reference point which is simply $$\vec{v_e}+\vec{v_o}$$, where $$\vec{v_o}$$ is tangent (parallel) to the Earth's surface and perpendicular to the radial vector pointing from point $$A$$ to the Earth's centre. Imagine the observer at $$A$$ is hanging from a high bar above the surface and then lets go so they're not touching anything.

Since the observer at $$A$$ isn't touching Earth at the beginning, and we imagine gravity doesn't exist so nothing else is affecting the Earth, the Earth doesn't undergo any acceleration: $$\frac{\Delta\vec{v_e}}{\Delta t}=0$$ ($$t$$ here is obviously time, and $$\Delta$$ refers to changes). So what happens to the observer? We want to calculate the change in the observer's velocity as well: $$\frac{\Delta\left(\vec{v_e}+\vec{v_o}\right)}{ \Delta t }=\frac{\Delta\vec{v_e}}{\Delta t}+\frac{\Delta\vec{v_o}}{\Delta t}=0+\frac{\Delta\vec{v_o}}{\Delta t}=\frac{\Delta\vec{v_o}}{\Delta t}$$.

So we see that any acceleration by the observer can be determined entirely in terms of $$\Delta\vec{v_o}$$, the change in the velocity of the observer with respect to Earth, regardless of how Earth is moving relative to any other reference points. Like the Earth, the observer does not feel any forces from the rest of the universe, nor are they touching Earth at the start, so without gravity, they feel no forces and hence no accelerations either: $$\frac{\Delta\vec{v_o}}{\Delta t}=0$$. Therefore the observer maintains a constant velocity $$\vec{v_o}$$ relative to the centre of the Earth, until such time as they come into contact with it. But since $$\vec{v_o}$$ was tangent to the surface at $$A$$, the observer's starting point, this means the observer never collides with Earth, but rather keeps drifting off into space.

A similar argument shows that if I jump off the Earth's surface and there's no such thing as gravity, I won't come back down no matter how the Earth is moving relative to any other reference points.
 
Imagine whipping a ball around on a tethered string- if you cut the string, the ball does not come flying back to you.

Its momentum holds it into place. I don't know how else to describe it. the resultants of momentum of the center of the Earth and the objects on its surface intersect, drawing them together as they move around the galaxy.

No, its momentum is what would cause it to go flying away. Momentum wants things to continue in a straight line.
 
Alright, let's say we denote the centre of Earth's velocity relative to the sun by the vector $$\vec{v_1}=\left(v_{1_x},v_{1_y},v_{1_z}\right)$$. And the sun has velocity $$\vec{v_2}$$ relative to the centre of the galaxy. And our galaxy has velocity $$\vec{v_3}$$ relative to whatever, and so on... Denote the vector sum of these velocities as $$\vec{v_e}=\vec{v_1}+\vec{v_2}+\vec{v_3}+\ldots$$. So the centre of Earth is moving with velocity $$\vec{v_e}$$ with respect to some reference point somewhere in the universe.

Place an observer at point $$A$$ on the Earth's surface, they will be rotating about Earth's centre with velocity $$\vec{v_o}$$, and thus have a net velocity with respect to our chosen reference point which is simply $$\vec{v_e}+\vec{v_o}$$, where $$\vec{v_o}$$ is tangent (parallel) to the Earth's surface and perpendicular to the radial vector pointing from point $$A$$ to the Earth's centre. Imagine the observer at $$A$$ is hanging from a high bar above the surface and then lets go so they're not touching anything.

Since the observer at $$A$$ isn't touching Earth at the beginning, and we imagine gravity doesn't exist so nothing else is affecting the Earth, the Earth doesn't undergo any acceleration: $$\frac{\Delta\vec{v_e}}{\Delta t}=0$$ ($$t$$ here is obviously time, and $$\Delta$$ refers to changes). So what happens to the observer? We want to calculate the change in the observer's velocity as well: $$\frac{\Delta\left(\vec{v_e}+\vec{v_o}\right)}{ \Delta t }=\frac{\Delta\vec{v_e}}{\Delta t}+\frac{\Delta\vec{v_o}}{\Delta t}=0+\frac{\Delta\vec{v_o}}{\Delta t}=\frac{\Delta\vec{v_o}}{\Delta t}$$.

So we see that any acceleration by the observer can be determined entirely in terms of $$\Delta\vec{v_o}$$, the change in the velocity of the observer with respect to Earth, regardless of how Earth is moving relative to any other reference points. Like the Earth, the observer does not feel any forces from the rest of the universe, nor are they touching Earth at the start, so without gravity, they feel no forces and hence no accelerations either: $$\frac{\Delta\vec{v_o}}{\Delta t}=0$$. Therefore the observer maintains a constant velocity $$\vec{v_o}$$ relative to the centre of the Earth, until such time as they come into contact with it. But since $$\vec{v_o}$$ was tangent to the surface at $$A$$, the observer's starting point, this means the observer never collides with Earth, but rather keeps drifting off into space.

A similar argument shows that if I jump off the Earth's surface and there's no such thing as gravity, I won't come back down no matter how the Earth is moving relative to any other reference points.

What about the other momentum the body would have from the earth moving and why would they fly off if the earth would move into them?
 
What about the other momentum the body would have from the earth moving and why would they fly off if the earth would move into them?

The momentum from the Earth is already there in the body at the start. The Earth starts off with velocity $$\vec{v_e}$$ and momentum $$m_e\vec{v_e}$$, whereas the observer starts off with velocity $$\vec{v_e}+\vec{v_o}$$ and momentum $$m_o\left(\vec{v_e}+\vec{v_o}\right)=m_o\vec{v_e}+m_o\vec{v_o}$$.
 
The momentum from the Earth is already there in the body at the start. The Earth starts off with velocity $$\vec{v_e}$$ and momentum $$m_e\vec{v_e}$$, whereas the observer starts off with velocity $$\vec{v_e}+\vec{v_o}$$ and momentum $$m_o\left(\vec{v_e}+\vec{v_o}\right)=m_o\vec{v_e}+m_o\vec{v_o}$$.

Why doesn't the Earth rotating a body on its surface cause any normal force on it. The body has momentum in a direction perpendicular to the rotation of the Earth. Doesn't momentum and inertia make it more difficult to change the body's direction?
 
Why doesn't the Earth rotating a body on its surface cause any normal force on it.

Because I specifically said the observer at $$A$$ starts off suspended just above the Earth's surface, so they're not touching Earth at the start and can't experience any normal force. If they did feel a normal force from the Earth, it would simply push them away from the surface out into space even faster.

The body has momentum in a direction perpendicular to the rotation of the Earth. Doesn't momentum and inertia make it more difficult to change the body's direction?

The body can have momentum in any direction whatsoever. I made no assumptions whatsoever about the direction or magnitude of $$\vec{v_e}$$. What matters is that $$\vec{v_o}$$, defined as the initial velocity of the observer relative to Earth's centre, is perpendicular to the line connecting point $$A$$ to the centre of the Earth (as is always the case for objects moving with a rotating surface). Earth has velocity $$\vec{v_e}$$, the observer has velocity $$\vec{v_e}+\vec{v_o}$$, and they never touch nor change direction or speed.
 
Because I specifically said the observer at $$A$$ starts off suspended just above the Earth's surface, so they're not touching Earth at the start and can't experience any normal force. If they did feel a normal force from the Earth, it would simply push them away from the surface out into space even faster.



The body can have momentum in any direction whatsoever. I made no assumptions whatsoever about the direction or magnitude of $$\vec{v_e}$$. What matters is that $$\vec{v_o}$$, defined as the initial velocity of the observer relative to Earth's centre, is perpendicular to the line connecting point $$A$$ to the centre of the Earth (as is always the case for objects moving with a rotating surface). Earth has velocity $$\vec{v_e}$$, the observer has velocity $$\vec{v_e}+\vec{v_o}$$, and they never touch nor change direction or speed.

I guess my problem with this is that I don't see the rotation of the earth acting upon a body the same way you do.

Did you calculate the size and shape of the earth?It isn't the center of the Earth that is going to collide with the body. Even if the body moves tangent to the surface of the Earth, I think it will collide with it because of the way it moves and the fact that its a sphere. I've spent time working with those vector addition calculators and the resultant always intersects with where the Earths surface will be unless I add a vector with a great amount of force. Do you know of any interactive tools that will let me plot these vectors in 3D space?

Objects don't just magically become near the Earth in space. They have to either come from the Earth or be traveling faster than the Earth when "behind" it, or slower than the Earth or heading towards it when in front of it. They also have to have a trajectory that will intersect with the Earth at some point.
 
The momentum from the Earth is already there in the body at the start. The Earth starts off with velocity $$\vec{v_e}$$ and momentum $$m_e\vec{v_e}$$, whereas the observer starts off with velocity $$\vec{v_e}+\vec{v_o}$$ and momentum $$m_o\left(\vec{v_e}+\vec{v_o}\right)=m_o\vec{v_e}+m_o\vec{v_o}$$.
I didn't realize that arc's could be expressed so simply.
 
You, sir, are a disgusting individual. Wouldn't be surprised if you had very few real friends. You have no social skills, you put people on the defensive, and you're just a jerk. You are no longer welcome in this thread. If I could force you out, I would.
Tough. I have no sympathy for willful ignorance.
 
I guess my problem with this is that I don't see the rotation of the earth acting upon a body the same way you do.

Did you calculate the size and shape of the earth?It isn't the center of the Earth that is going to collide with the body. Even if the body moves tangent to the surface of the Earth, I think it will collide with it because of the way it moves and the fact that its a sphere. I've spent time working with those vector addition calculators and the resultant always intersects with where the Earths surface will be unless I add a vector with a great amount of force. Do you know of any interactive tools that will let me plot these vectors in 3D space?

No need to worry about the Earth's size and shape. If a giant mountain came spinning around and smacked into you, you'd feel a sideways force, not a downward force pulling you into the Earth. You could also take your "observer" to be a molecule of air floating around high in the upper atmosphere and calculate how fast it goes flying out into space.

I think I see where your misconception arises. If the Earth were at rest relative to your chosen reference point, and the observer had initial rotational velocity $$\vec{v_o}$$ with respect to the centre, directionally tangent to the surface, you would probably agree that it should keep flying off into space. Now if in fact the centre of the Earth is moving with velocity $$\vec{v_e}$$, and the observer thus has initial velocity $$\vec{v_o}+\vec{v_e}$$, you seem to think you could find scenarios where the observer ultimately collides with the Earth's surface... If that's what you're saying, I'm saying it doesn't happen.

I'm saying the surface observer drifts away from the Earth's centre and surface in the same way for either scenario, i.e. all that matters is the relative velocity between observer and Earth's centre, our choice of $$\vec{v_e}$$ makes absolutely no difference (otherwise we'd be able to tell when the Earth is moving and when it's at rest, which even Isaac Newton said couldn't be done). You could calculate positions with respect to time in order to see that there's no collision, but there's some math and geometry involved that I'd rather leave to you as an exercise. For software packages, I really don't know of anything specifically geared for this kind of work. You could probably do it in programs like Mathematica, Matlab or Maple, but they all have learning curves and cost a lot of money if you don't have the right connections.

Edit: Try Googling for free physics software packages, maybe search for kinematics software as well. I'm sure there has to be something out there you can use, but you'll probably find a lot of junk cluttering the search as well.
 
Why doesn't the Earth rotating a body on its surface cause any normal force on it. The body has momentum in a direction perpendicular to the rotation of the Earth. Doesn't momentum and inertia make it more difficult to change the body's direction?
As said, for an object sitting on earth, Newton's first law tells us that the object will continue in a straight line unless acted upon by gravity, so the inertia of the object causes a normal force away from the surface of the earth, not towards it.
 
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