Flat earth (another split)

[Robittybob1]

Sorry Strange, I thought I made myself clearer, but I guess not.

Things farther away actually weigh less because weight is a force due to gravity and changes with the inverse of the radius squared. In effect, the equatorial surface is feeling less gravitational force

Mass does not change, and the two are not the same.

Do you think that can be taken to the limit with the whole Earth as a spread out circular sheet will the gravity always be stronger at the center?

[swansont]

Do you think that can be taken to the limit with the whole Earth as a spread out circular sheet will the gravity always be stronger at the center?

It' not true for a sphere. Why would it be true for a sheet?

[Robittybob1]

Do you think that can be taken to the limit with the whole Earth as a spread out circular sheet will the gravity always be stronger at the center?

There doesn't seem to be an easy answer to this problem for any search for the answer seems to be directed to a Flat Earther situation. One thing I did learn is that if the sheet was wide enough there would be no drop off the gravitational strength with height.

But I wasn't intending it to be an infinite sheet Earth either but just on that got more and more oblate?

 

There was one YT which covered the topic in a humorous way.

Is Earth Actually Flat?

It' not true for a sphere. Why would it be true for a sheet?

For a sphere we'd say at the poles but on a disced shaped Earth we might just say center as in center of the circular disc. Not the Earth's center.

[swansont]

For a sphere we'd say at the poles but on a disced shaped Earth we might just say center as in center of the circular disc. Not the Earth's center.

Oh. A flat earth. That's a bit of a leap from rotational deformation.

[studiot]

Oh. A flat earth. That's a bit of a leap from rotational deformation.

 

Nah, it's the flattening effect of the vodka, stops things and people being upright.

 

[Robittybob1]

 

Nah, it's the flattening effect of the vodka, stops things and people being upright.

 

The words rotational deformation and vodka reminded me of the worst inebriation I inflicted on myself. That day the Earth started rotating upwards and I found it impossible to stand for I could no long tell which way was up. This was on gooseberry wine my father had brewed.

[studiot]

 

This was on gooseberry wine my father had brewed.

 

 

Why no smiley at the end?

 

Gooseberry wine

 

Mmmmm.

 

Reminds me of the nicest drink I have ever tasted.

[Robittybob1]

 

Why no smiley at the end?

 

Gooseberry wine

 

Mmmmm.

 

Reminds me of the nicest drink I have ever tasted.

It was so potent I found you didn't need to drink it, I'd just breath the fumes when it was in my mouth. It definitely wasn't pleasant but it was loaded with something. I had a hangover for the whole of the following week.

 

PS He might have added heaps of sugar to sweeten the gooseberries!

Edited April 14, 2015 by Robittybob1

[ydoaPs]

Do you think that can be taken to the limit with the whole Earth as a spread out circular sheet will the gravity always be stronger at the center?

 

Gravity, on a flat Earth, would be almost parallel to the surface instead of normal to it. Objects would be dragged along the surface towards the center until the reach the center (or stop due to friction) where the net gravitational force is zero. Assuming friction isn't enough to stop the object before it reaches the center, there will be damped harmonic motion back and forth across the center until the object comes to a rest.

[Robittybob1]

 

Gravity, on a flat Earth, would be almost parallel to the surface instead of normal to it. Objects would be dragged along the surface towards the center until the reach the center (or stop due to friction) where the net gravitational force is zero. Assuming friction isn't enough to stop the object before it reaches the center, there will be damped harmonic motion back and forth across the center until the object comes to a rest.

That was the point I was trying to lead to. If in your situation the center is where the gravity is the least, why is it currently stronger at the Poles? At what degree of oblateness would the effect be zero again? Logically there is no place or shape that will achieve this.

So why the difference between extreme oblateness and mild oblateness?

IMO it is because the core of the Earth is so much more dense it contributes to a greater proportion of the gravity. When we think of a flattened Earth we have also evened out the density of the material as well.

[pavelcherepan]

That was the point I was trying to lead to. If in your situation the center is where the gravity is the least, why is it currently stronger at the Poles? At what degree of oblateness would the effect be zero again? Logically there is no place or shape that will achieve this.

So why the difference between extreme oblateness and mild oblateness?

IMO it is because the core of the Earth is so much more dense it contributes to a greater proportion of the gravity. When we think of a flattened Earth we have also evened out the density of the material as well.

That if you assume a pretty much zero thickness of flat Earth. If it's not zero then center of mass will be below the surface and gravity would in fact be the strongest at the center and weaker closer to edges.

 

Currently it's stronger at poles because at poles you're closer to the center of mass and at the equator you're farther away.

[Janus]

Gravity, on a flat Earth, would be almost parallel to the surface instead of normal to it. Objects would be dragged along the surface towards the center until the reach the center (or stop due to friction) where the net gravitational force is zero. Assuming friction isn't enough to stop the object before it reaches the center, there will be damped harmonic motion back and forth across the center until the object comes to a rest.

Not quite. Basically what we have here is an Alderson disk with no center whole or star:

 

http://www.tp4.rub.de/~jk/science/gravity/chapt_alderson.html

 

The gravitational potentials of the non-rotating disk would be like this:

 

 

The line are of equipotentials. "down" would be perpendicular to these lines. Along the surface of the disk, this only becomes close to parallel to the surface near the outer edge and become closer to perpendicular to the surface as you move towards the center. .

 

The g force plots like this:

 

Purple is the radial component and red the component perpendicular to the surface. Green is the absolute value of the vector addition of the two.

 

Note that the total g force is never zero and stays fairly constant until you start nearing the edge where it starts to rise rapidly.

 

This "edge effect" can be minimized by rotating the disk at the right speed and you can something like this for the potential:

 

 

And this for the G-force:

 

 

Notice how you can get both the direction of gravity to be nearly perpendicular to the surface and the force of gravity to be nearly constant over the majority of the disk. The edge effects don't become noticeable until you get fairly near the edge.

 

The concept of an Alderson disk is an alternative to a Dyson sphere. You build the disk big with a hole in the center with a star in it. You then live on the disk at a distance that is in your habitable zone.

Edited April 15, 2015 by Janus

[Robittybob1]

That if you assume a pretty much zero thickness of flat Earth. If it's not zero then center of mass will be below the surface and gravity would in fact be the strongest at the center and weaker closer to edges.

 

Currently it's stronger at poles because at poles you're closer to the center of mass and at the equator you're farther away.

What I was suggesting is that if the Earth was even density the equatorial regions would have the higher gravity.

The Equatorial Regions are like the edge of the disc described by Janus.

Those graphs would be different if the disc had a varying radial density as does the Earth.

Edited April 15, 2015 by Robittybob1

[swansont]

That was the point I was trying to lead to. If in your situation the center is where the gravity is the least, why is it currently stronger at the Poles? At what degree of oblateness would the effect be zero again?

 

At zero deformation?

[Robittybob1]

At zero deformation?

Yes that is the starting position, but is there another point where the gravity at the center equals the gravity at the equator. It could do if the rate of rotation was chosen correctly and the substance of the Earth blended evenly.

[mathematic]

Do you think that can be taken to the limit with the whole Earth as a spread out circular sheet will the gravity always be stronger at the center?

Yes, the inverse square effect.

[Robittybob1]

Yes, the inverse square effect.

That is more for what happens above the surface, but I'll think about that.

[mathematic]

That is more for what happens above the surface, but I'll think about that.

At the surface also. Basically you are integrating 1/r^2 over te area.

Edited April 15, 2015 by mathematic

[Robittybob1]

At the surface also. Basically you are integrating 1/r^2 over te area.

So what is the integral of 1/r^2?

 

1/r^2 = r^-2 now what do you do?

Edited April 16, 2015 by Robittybob1

[Strange]

So what is the integral of 1/r^2?

 

1/r^2 = r^-2 now what do you do?

 

You take this course: https://www.coursera.org/learn/calculus1

 

It is not too advanced and will give you some great insights into all sorts of things.

[Robittybob1]

 

You take this course: https://www.coursera.org/learn/calculus1

 

It is not too advanced and will give you some great insights into all sorts of things.

I've been listening to YT videos with integration as the subject (integrals). I like it but I haven't learnt how to use it. Thanks.

[michel123456]

Somehow interestingly, flat earth proponents explain gravity by the Earth being in accelerated motion through space in a direction normal to the surface of the disk. Not gravity caused by the disk itself.

[Robittybob1]

Somehow interestingly, flat earth proponents explain gravity by the Earth being in accelerated motion through space in a direction normal to the surface of the disk. Not gravity caused by the disk itself.

Which is rather ridiculous as at that rate of acceleration we would be traveling through space at close to the speed of light.

[imatfaal]

Which is rather ridiculous as at that rate of acceleration we would be traveling through space at close to the speed of light.

 

 

Possibly if acceleration was linear. But as an alternative how large an orbit would the disc need to be on for the apparent magnitude of the pseudo-force to be equal to 9.8m/s^2 (ie consider the disc/planet as the bottom of the bucket being whirled around one's head).

[Janus]

Yes that is the starting position, but is there another point where the gravity at the center equals the gravity at the equator. It could do if the rate of rotation was chosen correctly and the substance of the Earth blended evenly.

No. For a non-rotating oblate spheroid of uniform density, the gravity at the equator will be less than that at the poles. Do not confuse an Alderson disk's gravity profile for that of an very oblate sphere as they have different mass distribution profiles..

With a disk, the thickness remains constant as you move away from the axis, and with an oblate spheroid, the thickness decreases as you move away from the axis this has an effect on how gravity behaves along the surface. (For example, if you revert the oblate spheroid to where its polar and equatorial radii are equal, you get a sphere. If you do the same to a disk you get a cylinder. The gravity along the surface of a cylinder is not going to match that along the surface of a sphere.)

 

Since any rotation has the additional effect of lessening the effect of gravity at the equator of an oblate spheroid, this will only increase this difference.