... Even though I guess effectively certain spots (I.e. front of the axis vs back of the axis) are moving inside and outside from the orbit they would naturally take if they were single points, as a whole spherical object, gravity doesn't give a crap if the sphere is spinning or not: it's just a sphere. It's acting on the entire sphere. If it rotates or doesn't rotate there's no force pulling one side greater than the other in order to make it turn as it orbits.
I suggest you draw some force vectors to prove your point
Yes, I see my friends start to accept my talk. Please reread the link in the thread: using our helpful discussion I have modified the manuscript. SiDawg, it makes no special sense to draw some force vectors, because all the vectors from gravity are the tidal forces, which are minimized by taking a small test-body.
Tidal forces can be drawn with vectors (see below)... I'm not sure i understand where you say they're "minimized": even if the forces are small, it's your choice how large or small you draw the vectors... Or if you want to show a mix of hugely different forces, just write "not to scale". I think the important part is that you describe an origin and direction of the forces you're talking about. I don't think the earth is a small test-body
To add to my previous post, the way i'm conceptualising your argument: if you have two bodies in orbit (around the sun) at the same speed and the same mass, then they will orbit at the same distance. If they moved to a bigger orbit (say they were hit by an object) then assuming it doesn't escape orbit or change speed, it will naturally return to the same orbit equilibrium. Same if it goes to a smaller orbit: the speed will "push" it back to it's original orbit. In other words, if the speed stays constant, the objects will maintain their orbital distance (in reality orbits are usually elliptical, but lets just use a perfect circular orbit for arguments sake)
So if you then connect those two objects together, say with a solid magic rod, then those rules still apply right: it's now one mass, but the forces of gravity combined with the imaginary centrifugal force from it's speed, will result in a force differential if one end of the mass is outside its normal orbit compared to the other side. So the object will effectively "rotate": the rod connecting the two will stay perpendicular to the sun, or point in the direction of travel, as both ends will be pushed or pulled back to their normal orbit.
But then you're trying to apply that logic to a sphere: although it has an axis of rotation, it has an essentially even distribution of mass... For the sake of this argument let's consider it a perfect sphere with density increasing consistently towards the centre. If the mass in a small area at the north and south pole was drastically bigger than the rest of the sphere, then yeah, your theory might hold, but it's not. But you're not saying that: you're just saying that because my first example of a rod object stays perpendicular, then the sphere must also be perpendicular.
Because the only difference in density in the sphere is in relation to distance to the centre, then there is no "special" force applied that would cause a differential. i.e. if it rotates due the orbit as you say, then it would need an area of increased density offset from the centre in order for the centrifugal and gravity forces to be different at that point, and thus rotate the sphere.
Now, we know the earth DOES have variations in density... but we also know that it spins on it's axis 365.15 faster than it orbits, so any orbital forces acting on different areas of density would surely be balanced as it rotates. Probably contributes to a slight "wobble"? I guess that's further homework!