Do we weight a tiny bit less due to the orbit of the Earth around the Sun? Of course the Earth is orbiting in a weightless state around the Sun so people on it too. 


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*Now there could also be some tidal force of the Sun on the Earth, but are they realy caused by the orbit of the Earth?

*But I've also heard that the acc due to the Earth motion around the sun
is 0.006 m/s², what could make the 9.81m/s² effect on our weight 0.06% less. But is this true and how to understand this lessening; is that just an effect of centrifugal forces?

 A: I do not think so. If you mean weigh due to earth's gravity, then note that we along with earth, are going around the sun at the same time, and the centrifugal effect is exactly countered by sun's gravity for all earthly things.
Moon is a different story. 
Anything that is at the center of the orbit (earth for moon, and sun for it's planets and grand planets), should not cause loss of the weight on bodies that are in orbit.
Other planets can have the weight effect you are thinking about, but sun - I do not think so.
A: Yes, there is a tidal force due to the Sun since its gravitational field on an extended body like Earth is not uniform. Moreover, the fact that this effect causes two tides per day (tides in both sides of the Earth) requires that Earth is orbiting or in free fall towards the Sun. If it was static, there would be only one tide (note that I am neglecting the Moon's influence which actually the most relevant). The existence of these tides proves that we are orbiting the Sun. You might be wondering that the Equivalence Principle should not allow any tidal force since all bodies on Earth are in free fall towards the Sun. The thing is that the Equivalence Principle regards only local effects whereas tides are non-local.
The translation as well as the rotation of the Earth have effect on the masses. These effects originate from the fact that Earth is not an inertial frame and they slightly deviate the gravity of the planet to an effective gravity. The deviation is of a vectorial character so the effective gravity also has its direction changed.
When Earth orbits around the Sun there is a centrifugal force (in our frame of reference). To estimate it, we just need the period of translation $T$ (1 year) and the radius $r$ of the orbit. For a body on Earth it gives
$$a=m\omega^2 r=\left(\frac{2\pi}{T}\right)^2r\approx 0.006,$$
which is just the acceleration you mentioned.
