/sci/ - Science & Math

>I am pretty sure the earth is flat
Lol

>>5431377
This is ridiculous.
People have known that the earth is round for hundreds of years.

People have circumvented the globe; this is not possible on a flat earth.

NASA has even taken pictures from space, showing the whole globe as a slightly squashed ball shape.

It is certainly not flat.

>>5431388
>>5431392
God damn it just do the equations

They are all lying to you, its all there in area 51, the illuminati teamed up with the scientologists to rule the earth with their lizard men by telling everyone it is round

>>5431377
Mount camera to a helium filled balloon and send it to atmosphere.

Seems nobody else has the balls to stick it to the lizard man, so OP will just have to delivar what he can

We need to get the gravity at any set height, h, what we will need for this is the radius of the flat earth, r, the thickness, t, and the mass, m

To simplify things down lets say you are standing in the very centre of the flat earth so that all the side gravity will cancel, furthermore we are going to say the flat earth is equally dense at all points

So now it is really just integrating some triangles

>>5431398
I'm guessing you're using your friends computer and trying to get him banned for pranks n shits n niggles. Brilliant I might say.

As a quick refresher the Newtonian equation for gravity is:

F=(G.M1.M2)/R^2

>>5431432
We can kick off the M2 to the side of the F so that we can more easily swap out masses when intigrated

>>5431437
So now we need to find what we are intigrating
For each infinitesimal point of the earth's flat body we take the FM, by using the direct line distance from the point to the object

However this includes, for all points that are not directly below, the horizontal force, witch cancels due to the assumptions made, so now we need to split it into component vectors

>>5431425
Oh and we are also saying the mass is a point mass

>>5431450
now then this is where it gets complicated, the point attracting is at distance R, R = SQRT(h^2 + l^2), where l is the horizontal distance to the point

>>5431463
What we can do here is then use the R in the equation for FM and then use the principle of similar triangles to find the downward pointing component of force

This is where I get stuck, somebody else take the reigns

Responding just for the sake of argument here. Why would there be components going off to the sides? Just take the center of mass of the new earth (a rectangular prism is what you're saying?) and use newtons formula. Even when considering spherical earth it is locally linear yet there are no components going off to the sides. You are pulled to the center of mass and the shape makes no difference. If this were an e&m problem with a field it would be a different story.

>>5431482
The reason you are only being pulled down is because the components cancel, but each point to the side has got a downward pulling component.

E.G. if you cut the earth down all the way through so it was just a pillar you were standing on, there would be far less gravity

Nope. Check gauss law for gravity (or calculate using calculus for force about a mass distribution of sphere) and Newton's law for gravity: <span class="math">\bar{F}=-\frac{mM}{r^2}\hat{r}[/spoiler]

>>5431392
>hundreds of years

pre-pyramid egyptians knew the earth was round

http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/sphshell.html

Good guestion.

>>5431414
Actually, most of those videos only display the curvature of the lense, not the earth.