Gravity acceleration?

How does the FE explain changes in "gravitational force"? I know that here in Denmark the acceleration of gravity is 9,82 m/s^2.  At equator it is 9,78 m/s^2 and at the poles it is 9,83 m/s^2?
If we are on a disc forever accelerating, "gravity" should be constant all over earth.
According to RE the earth rotates and centrifugal force counteracts gravity a little bit and explains the changes in gravity perfectly.

If centrifugal force were the factor for the mentioned variation in g, then we have the following identities:

because the centrifugal force is zero on the poles, according to the formula:

and r = 0 at the poles. At the Equator, r = R_E (the radius of the RE) and the centrifugal acceleration is directed out of the Earth's axis, i.e. opposite of the gravitational acceleration. That is why they subtract. Using these equations, one can determine the radius of RE if one knows the period of rotation of the Re. This is taken to be T = 24 h = 86,400 s and angular speed is connected with period of revolution through the formula:

Using the above formulas, we get:

Usually, it is customary to express the length of half a meridian. i.e. the distance from the North Pole to the Equator. This length is just a quarter of the circumference of the great circle (2*π*R_E), i.e.:

Why do we calculate L? Because the meter used to be defined through this quantity as ten millionth part of that distance, or, equivalently, this distance is exactly 10 million meters or 10,000 km. But, using the data you provided (g_pole = 9.83 m/s² and g_equ = 9.78 m/s²), we get the following value:


This is a 48.5% discrepancy from the accepted RE figure. Thus, we see that the data you provided is actually inconsistent with the RE model itself. How do you explain this?

Besides your math i wrong, i forgot to add in that there are alot of other factors that influence gravitational pull from the earth on you. The earth is thicker at equator than at the poles, meaning you are further away from the center of the earth, and also alitutide puts you even further away from the earths centermass, and distance weakens gravity. That was another major factor, many other smaller factors exists, but i dont feel it necesarry to mention them.
Anyways, even if i was wrong about the centrifugal force of the earth, how does FE explain the changes in gravity anyways? You did not answer my question.

Besides your math i wrong,

Please point out my mistake.

Quote from: brutsi on February 07, 2010, 09:01:59 AM

Besides your math i wrong,

Please point out my mistake.

You use too many equations and not enough common sense, that's your mistake.

Is there a mistake? I'm not sure there is, I certainly only skimmed it but it looked reasonably solid to me. I think it gave the answer one would expect for a planet of the assumed geometry. Still, I can't imagine such a world exists.

Quote from: parsec on February 07, 2010, 09:15:52 AM

Quote from: brutsi on February 07, 2010, 09:01:59 AM

Besides your math i wrong,

Please point out my mistake.

You use too many equations and not enough common sense, that's your mistake.

No one asked you.

Quote from: Thermal Detonator on February 07, 2010, 11:46:17 AM

Quote from: parsec on February 07, 2010, 09:15:52 AM

Quote from: brutsi on February 07, 2010, 09:01:59 AM

Besides your math i wrong,

Please point out my mistake.

You use too many equations and not enough common sense, that's your mistake.

No one asked you.

I gave my advice for free, be grateful.

Gravity simply doesn't work on a flat earth.  The force would also pull everything toward the middle, not straight down as we know it.

Gravity simply doesn't work on a flat earth.  The force would also pull everything toward the middle, not straight down as we know it.

Where is the middle of an infinite disc?

Quote from: Mr Pseudonym on February 08, 2010, 05:19:41 AM

Gravity simply doesn't work on a flat earth.  The force would also pull everything toward the middle, not straight down as we know it.

Where is the middle of an infinite disc?

Everywhere.....

Everywhere.....

So a place outside an infinite disc is in the middle of it?

I think Parsec showed that an infinite disk would be inherently unstable, although I can't seem to find the thread.

if it moves at all it would need to be infinitely dense

Quote from: Its a Sphere on February 08, 2010, 05:31:48 AM

Everywhere.....

So a place outside an infinite disc is in the middle of it?

Every point on the infinite disc is the center point.

Quote from: Parsifal on February 08, 2010, 05:40:42 AM

Quote from: Its a Sphere on February 08, 2010, 05:31:48 AM

Everywhere.....

So a place outside an infinite disc is in the middle of it?

Every point on the infinite disc is the center point.

Any point along the symmetrical plane in the center of the thickness of the infinite plane is its pseudo center, but clearly points on the surface are not the center.

Quote from: Its a Sphere on February 08, 2010, 11:01:48 AM

Quote from: Parsifal on February 08, 2010, 05:40:42 AM

Quote from: Its a Sphere on February 08, 2010, 05:31:48 AM

Everywhere.....

So a place outside an infinite disc is in the middle of it?

Every point on the infinite disc is the center point.

Any point along the symmetrical plane in the center of the thickness of the infinite plane is its pseudo center, but clearly points on the surface are not the center.

I had assumed a /parsifal semantic plane of differential(planar) thickness, as in created by intersecting a cylinder and 2D plane.

Quote from: Its a Sphere on February 08, 2010, 05:31:48 AM

Everywhere.....

So a place outside an infinite disc is in the middle of it?

There should be no outside if the disc is infinitely wide. The disk would fill up the space by its own definition. Therefore every point EVERYWHERE is the centre.

I guess the OP has decided to leave us. Maybe a mod should lock this thread.

EDIT:
lapsus calami. I changed OP to mod in the second sentence.

Maybe an OP should lock this thread.

wat

The real formula for calculating the gravity acceleration in your area would be at the end of this link http://www.formel.dk/fysik/g.htm

in the upper box type in your altitude above sea level in meters, and in the other box type in your latitude.

No i didnt leave you, i just have other stuff to do than sit in front of my computer invented by people who know that the earth is round.

And Parsec, you still didnt explain why gravity acceleration changes.

 

The real formula for calculating the gravity acceleration in your area would be at the end of this link http://www.formel.dk/fysik/g.htm

in the upper box type in your altitude above sea level in meters, and in the other box type in your latitude.

No i didnt leave you, i just have other stuff to do than sit in front of my computer invented by people who know that the earth is round.

And Parsec, you still didnt explain why gravity acceleration changes.

How is this the real formula???

According to the numbers you gave us, the radius of the Earth is still 46% bigger than the accepted RET value. My point is, if gravity really changes as claimed by the data provided, than it is inconsistent with the current RE model. So, your argument that RET is superior to FET in explaining the "gravity acceleration" variations is only a qualitative one. Once you punch in some numbers, you see it is equally unsatisfactory.

So now you are just denying that gravitational accelleration changes around the world?

Please read my posts.

Check out the GRACE mission. Yep gravity varies across the round earth.