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

[origin]

There is one in the Hudson Bay Area exactly where the resultant vector of the Earth's momentum is.

How did you determine that?

What do you mean how did I figure that out? I used the directions the Earth is moving both relative to the Sun and the center of the Galaxy and the speed it is moving in those directions. How would you have done it?

Great could you please show the calculations that gave the results:

"..the Hudson Bay Area exactly where the resultant vector of the Earth's momentum is."

Frankly, I would not know how to do the calculations that could pinpoint an area like you did.

Thanks

 

[jiveabillion]

Great could you please show the calculations that gave the results:

"..the Hudson Bay Area exactly where the resultant vector of the Earth's momentum is."

Frankly, I would not know how to do the calculations that could pinpoint an area like you did.

Thanks

I used various vector addition calculators online. A couple of them are visual, so I have to estimate the directions and magnitude, but some of them are just numbers you punch in and it gives you the resultant and angle. I'll find the links and post them on here when I get to work.

 

[jiveabillion]

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.

Right, but the Earth does not move in a straight line. Are any of you looking at a model of the Earth to try to see what I am saying. Download Celestia (just google it), its free and pretty awesome for observing the motion of the solar system.

At any point on the Earth, if you throw something straight up, the momentum the Earth had exerted on that object will determine where it goes, but the Earth will move that point it was thrown from right back into its path because it moves it "away" from the direction it was thrown in, but "towards" it at another angle, so it "falls" straight back down. You have to see it to believe it.

 

[Beer w/Straw]

It all makes sense now!

 

[jiveabillion]

It all makes sense now!

You actually made me laugh.

 

[Russ_Watters]

Right, but the Earth does not move in a straight line. Are any of you looking at a model of the Earth to try to see what I am saying. Download Celestia (just google it), its free and pretty awesome for observing the motion of the solar system.

At any point on the Earth, if you throw something straight up, the momentum the Earth had exerted on that object will determine where it goes, but the Earth will move that point it was thrown from right back into its path because it moves it "away" from the direction it was thrown in, but "towards" it at another angle, so it "falls" straight back down. You have to see it to believe it.

Sure, you can do exactly the same centripedal acceleration calculation that you should have done for Earth's rotation, for the Earth's motion around the sun and see just how tiny, not to mention cyclical it is. For an object on earth's surface between the earth and sun, the centripetal acceleration causes a miniscule increase in weight and for an object on the other side, a miniscule decrease. Both of these effects are quite different from and much smaller than gravity.

Remember, you said this would explain all of gravity, not just tiny anomalies.

 

[jiveabillion]

Here are some links

Here is a very old java applet that you can use to add vectors and drag the arrows around and then calculate the resultant. You can continue to drag the arrows around after the resultant has been calculated and it will update in real time. Since this one doesn't give you the ability to set the exact magnitudes or angles of the vectors, you'll have to approximate. http://www.walter-fendt.de/ph14e/resultant.htm

Here is another visual one, but this one will let you see the numbers representing your vectors. However, the scale isn't exactly what we need in order to display the vast differences in magnitude of the vectors. It's still a bit useful because you can put the vectors anywhere on the grid. http://phet.colorado.edu/en/simulation/vector-addition

Here is the calculator I used to get the actual numbers http://www.1728.org/vectors.htm


Since these are all 2D, you'll probably want to pick a point on the Earth at a time when the Earth, Sun, and Galactic center line up, which happens sometime every December. I chose the Equator at midnight so that it is facing away from the galactic center

Here are the angles and magnitudes of the major vectors relative to the Earth's orbit on ecliptic plane.

~225km/s @ 120 degrees - Earth around the Milky Way.

29.7km/s @ 0 degrees - Earth around the Sun.

0.465km/s @ 336.56 degrees - Rotation of the Earth.

Calculating these should give you 211.38km/s at 112.94 degrees.

Edit: I used this acceleration calculator to, well, calculate the acceleration http://www.smartconversion.com/unit_calculation/Acceleration_calculator.aspx

Initial speed (v0) = 0
Final speed (v1) = 211.38km/s
The Time (t) = 359.0170848 minutes

Calculation result = 9.81290347773444m/s^2


I used this online trigonometry thing to help me visualize the changes in speed of a point on the surface of the Earth as it rotates as well as the whole Earth as it revolves around the Sun. http://www.touchmathematics.org/topics/trigonometry

I found that if I move the "dial" to 45 degrees, it gives me the best point to start from (which is still going to be the same point on the Equator I mentioned above, I just shift it in my mind so I have a good visual representation). I use the COS and SIN lines to visualize the changes in speed in the different directions. I also use this to help me imagine which directions I am slowing down or speeding up a ball in when I throw it straight "up" or outward from the center. I see that I am slowing it down about half of the magnitude of my throw in the COS direction and speeding it up about half the magnitude of my throw in the SIN direction. I don't slow it down at all in the TAN direction because it is perpendicular to my throw. I also slow the ball down a little bit in the direction the Earth travels around the Galaxy and speed it up and down in the SIN and COS directions of the Earth traveling around the Sun. So, when you throw something straight up or when something falls straight down, it will still impact the Earth because of the way the Earth moves.

If you need help visualizing how this could create the normal force on your body against the surface, think of the Earth constantly "throwing" you tangent to its rotation. This is where the mass of a planet vs the mass of your body come into play.
The more massive a planet is, and the faster it rotates, the more momentum it is going to have when it collides with your body after this "throw". The inertia of the planet and the inertia of your body create the normal force during this collision.
The speed of a planet's rotation is also what helps it grow to a great mass. The faster a planet rotates, the harder the collision will be as it "throws" less massive bits of space debris into its surface while it "clears its neighborhood". It also makes it harder for those bits to fly off the surface into space. This works best with inelastic collisions for obvious reasons, but will also work with elastic collisions as the object will lose momentum every time it collides with the surface.
The more massive the planet is, the more momentum it has, and the more massive the bits of debris it can keep on its surface. It's like a Katamari Damacy.

Now do you see it the way I see it?l

 

[billvon]

If you need help visualizing how this could create the normal force on your body against the surface, think of the Earth constantly "throwing" you tangent to its rotation.

That works to provide a normal force - but only if you are inside a hollow Earth. The force you describe will act outwards, not inwards.

 

[jiveabillion]

That works to provide a normal force - but only if you are inside a hollow Earth. The force you describe will act outwards, not inwards.

No, did you not read the rest of that post at all? The Earth throws you into its path and you collide with it. You can't think if the Earth as just spinning around like a merry go round that isn't moving in any other direction.

 

[billvon]

No, did you not read the rest of that post at all? The Earth throws you into its path and you collide with it.

That would require you to slow down again so the Earth can "catch up." If the Earth throws you anywhere, and gravity is not a factor, you just keep going.

You can't think if the Earth as just spinning around like a merry go round that isn't moving in any other direction.

Right. Its center of mass has a fixed vector velocity at any given time, which depends on which reference you use. That means that some people will be on the "front" of the earth, along the direction of its travel. Some people will be on the "back" of the earth, opposite its direction of travel. Some people will be along the circumference of travel. Any assumptions you make that require a certain direction of motion will not apply to most people on the planet.

 

[jiveabillion]

That would require you to slow down again so the Earth can "catch up." If the Earth throws you anywhere, and gravity is not a factor, you just keep going.



Right. Its center of mass has a fixed vector velocity at any given time, which depends on which reference you use. That means that some people will be on the "front" of the earth, along the direction of its travel. Some people will be on the "back" of the earth, opposite its direction of travel. Some people will be along the circumference of travel. Any assumptions you make that require a certain direction of motion will not apply to most people on the planet.

This is why you have to calculate the resultant vector from the point you are on the Earth. Its direction changes depending on where you are. Because the Earth rotates at neither the same angle it revolves around the sun, nor the direction it revolves around the galaxy, the direction that is tangent to its rotation will never be against both of its directions of travel. In fact, the angle of rotation is about 45 degrees relative to the resultant of the center of the Earth's momentum.


Edit: Let me just clarify how this works on the "bottom" of the Earth. When the Earth "throws" you, it moves into your path and YOU catch up with it. When you're on the "top" of the Earth, it catches up with YOU. It works the same way for the "front" (facing the Earth's path around the Sun) and "back" (facing away from that path) of the Earth. All of these work together in different proportions depending on where you are on the surface.

 

[billvon]

When the Earth "throws" you, it moves into your path and YOU catch up with it.

So you start out with a positive velocity relative to the Earth (i.e. being "thrown") and end up with a negative velocity relative to it (i.e. it "moves into your path.") How does that happen?

When you're on the "top" of the Earth, it catches up with YOU. It works the same way for the "front" (facing the Earth's path around the Sun) and "back" (facing away from that path) of the Earth.

But you can make "front" and "back" mean anything you like simply by choosing a different coordinate frame (i.e. the frame of the galactic center, the frame of the solar system.) Thus any system that relies on motion relative to an assumingly "fixed" frame is going to not work very well.

 

[jiveabillion]

So you start out with a positive velocity relative to the Earth (i.e. being "thrown") and end up with a negative velocity relative to it (i.e. it "moves into your path.") How does that happen?

The acceleration you exert on an object changes its momentum in the directions it is already moving depending on the direction you accelerate it. When you throw something from the surface of the Earth, you both speed it up and slow it down relative to the different directions in which it already had momentum from being a "passenger" on the Earth. The point on the surface of the Earth from which you threw your object will also slow down and speed up in different directions because it is a point on a sphere that is both rotating and revolving. This allows the object to collide with the earth again at the same point you threw it from. It doesn't move "up and down" It moves straight in the direction you thew it, but it continues to move in the other directions it had momentum in, which makes it move diagonally. Technically, the object and the point you threw it from catch up to each other.

But you can make "front" and "back" mean anything you like simply by choosing a different coordinate frame (i.e. the frame of the galactic center, the frame of the solar system.) Thus any system that relies on motion relative to an assumingly "fixed" frame is going to not work very well.

I don't really understand your argument. It's not about any sort of "fixed" frame. It's about the way objects move relative to each other. The "fixed" frame is just a point of reference so that you can measure the speeds and angles.

 

[billvon]

The acceleration you exert on an object changes its momentum in the directions it is already moving depending on the direction you accelerate it. When you throw something from the surface of the Earth, you both speed it up and slow it down relative to the different directions in which it already had momentum from being a "passenger" on the Earth.

Right. Although from any one coordinate frame it only speeds up or slows down, not both.

the point on the surface of the Earth from which you threw your object will also slow down and speed up in different directions because it is a point on a sphere that is both rotating and revolving. This allows the object to collide with the earth again at the same point you threw it from. It doesn't move "up and down" It moves straight in the direction you thew it, but it continues to move in the other directions it had momentum in, which makes it move diagonally. Technically, the object and the point you threw it from catch up to each other.

Correct! Now let's take two cases - throwing an object away on the sunward side of the planet and on the antisunward side.

Sunward side - the rotation of the Earth causes the Earth to move away from the object. Thus separation increases. However, the motion of the Earth around the Sun causes the Earth to "move back in the way" and thus get closer to the object. Which one will dominate? Using basic physics, the contribution from the Earth's rotation on a 10kg object will be .3 newtons away from the Earth. The contribution from the orbit around the Sun will be .06 newtons TOWARDS the Earth. Thus the object departs.

Antisunward side - the rotation of the Earth causes the Earth to move away from the object. Thus separation increases. The motion of the Earth around the Sun causes the Earth to move farther and farther away from the object. Thus the object moves away even more quickly.

You can choose any point you like, but since the Earth is rotating, eventually you will get your object on the anti-sunward side and it's going to go flying into the night. (The poles are not stable either, since the Earth's orbit will attempt to fling them outwards in a straight line.)

(All of the above neglects gravity of course.)

 

[jiveabillion]

Right. Although from any one coordinate frame it only speeds up or slows down, not both.



Correct! Now let's take two cases - throwing an object away on the sunward side of the planet and on the antisunward side.

Sunward side - the rotation of the Earth causes the Earth to move away from the object. Thus separation increases. However, the motion of the Earth around the Sun causes the Earth to "move back in the way" and thus get closer to the object. Which one will dominate? Using basic physics, the contribution from the Earth's rotation on a 10kg object will be .3 newtons away from the Earth. The contribution from the orbit around the Sun will be .06 newtons TOWARDS the Earth. Thus the object departs.

Antisunward side - the rotation of the Earth causes the Earth to move away from the object. Thus separation increases. The motion of the Earth around the Sun causes the Earth to move farther and farther away from the object. Thus the object moves away even more quickly.

You can choose any point you like, but since the Earth is rotating, eventually you will get your object on the anti-sunward side and it's going to go flying into the night. (The poles are not stable either, since the Earth's orbit will attempt to fling them outwards in a straight line.)

(All of the above neglects gravity of course.)

I like this argument, but I don't think you're taking everything into account. When an object is on the surface of the Earth between the center of the Earth and the Sun, the resultant is going to have it going towards the Earth in its clockwise revolution around the Sun, but the rotation is 45 degrees relative to its revolution around the Galaxy. The momentum the object has tangent to the rotation of the Earth will put the object right into the Earth's path around the galaxy, even though the Earth is moving away from it as it moves around the Sun.

This is actually really hard to describe in words. Download that Celestia program and watch the Earth move. You'll have to imagine the direction it is moving around the galaxy. You can right click on the Earth and choose refrence marks and select the Body Axis and the Frame Axis and the Planetographic Grid. The vector of travel around the galaxy is in the same direction as the Frame axis relative to the body (polar) axis, but it's at about 45 degrees north. The resultant vector of the Center of the Earth's movement is about 66 degrees North. You can go to Render > View Options and check the box for rendering orbits up in the top right. You can increase and decrease the speed of time using the "L" and "K" keys and reverse time with "J". "Help" has a list of controls and you can just poke around with it to figure out the rest.

 

[billvon]

I like this argument, but I don't think you're taking everything into account. When an object is on the surface of the Earth between the center of the Earth and the Sun, the resultant is going to have it going towards the Earth in its clockwise revolution around the Sun, but the rotation is 45 degrees relative to its revolution around the Galaxy.

Right. But just as the centrifugal force from the Earth's orbit is an order of magnitude less than the force generated by the Earth's own rotation, so the effect of any circular path the Earth describes around the center of the Galaxy is going to be lesser still. Thus the previously mentioned two dominate - and everything flies off the Earth.

 

[jiveabillion]

Right. But just as the centrifugal force from the Earth's orbit is an order of magnitude less than the force generated by the Earth's own rotation, so the effect of any circular path the Earth describes around the center of the Galaxy is going to be lesser still. Thus the previously mentioned two dominate - and everything flies off the Earth.

Nothing flies off the Earth. It has nothing to do with acceleration from the rotation of the Earth or the Earth's orbit. Those forces are minuscule and add up to only about 0.039m/s^2 of acceleration. I thought you understood what I meant, but I guess not. It's my fault though, I lack the proper vocabulary and math to convey my thoughts to you in a way that you understand. Right now, I only speak "broken physics".

 

[origin]

Here are the angles and magnitudes of the major vectors relative to the Earth's orbit on ecliptic plane.

~225km/s @ 120 degrees - Earth around the Milky Way.

l

Let's start with the first one. The speed is OK. Why did you use the direction as 120 degrees?

 

[jiveabillion]

Let's start with the first one. The speed is OK. Why did you use the direction as 120 degrees?

Because that's where the galactic plane intersects with the Earth. Its 60 degrees from the ecliptic, but that is if 0 degrees is on the left. If 0 degrees is on the right, its 120 degrees. http://lifeinthemix2.co.uk/ESW/Images/galactic_cross.jpg?xcache=3511
http://i215.photobucket.com/albums/cc42/Edward_Wolf/EclipticvsEquator-1.png

It's pretty hard to find pictures representing it. I had to research it for a long time to find out which way it intersects. The Earth travels in the direction it intersects. and that intersection stays about 45 degrees from the north pole. Edit: sorry, not 45 degrees from the north pole, but about 51.56 degrees from it.

Edit2: As funny as this photo is, it was actually the most informative that I found http://www.infinitelymystical.com/essays/images-2012/3d-model.gif

Edit3: Here is another good one http://www.infinitelymystical.com/essays/images-2012/intersection-of-two-planes.jpg

 

[billvon]

Nothing flies off the Earth.

Right, due to gravity. But neglect gravity and you'd have almost everything unsecured departing very quickly.

It has nothing to do with acceleration from the rotation of the Earth or the Earth's orbit. Those forces are minuscule and add up to only about 0.039m/s^2 of acceleration.

Agreed. They are ordinarily overwhelmed by gravity and only become significant if you postulate the absence of gravity.