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Of course the Earth is orbiting in a weightless state around the Sun so people on it too.

  1. 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?
  2. 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?
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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.

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  • $\begingroup$ Ok, but can you perhaps explain the answer of Walter Lewin on quora.com/… $\endgroup$
    – Marijn
    Feb 16, 2017 at 21:13
  • $\begingroup$ @Marijn: Well, .006 is total/average acceleration due to sun in earth's orbit. The tidal forces are due to difference in acceleration between near side of earth and far side of earth. That difference, even though causes the tides (shape change), it would be much smaller than .006 (negligible) when it comes to weighing stuff on earth's surface. For example, the mass center of earth may move due to the tides and the weight depends upon where the center of mass of the earth is. In any case, it is not even close to .006. $\endgroup$
    – kpv
    Feb 16, 2017 at 22:05
  • $\begingroup$ This is wrong: "the centrifugal effect is exactly countered by sun's gravity for all earthly things". It is only exactly countered for one point in a planet. $\endgroup$ Nov 7, 2021 at 10:34
  • $\begingroup$ @DavidJonsson: Earth and the thing that you weigh are all in free fall around sun. So, effective weight is only the mutual attraction among bodies other than sun. $\endgroup$
    – kpv
    Nov 7, 2021 at 15:41
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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.

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  • $\begingroup$ I'm not sure if I understand it. Are there two kinds of tidal forces of the Sun? One of the Suns mass and one of the Eath orbit around the sun? $\endgroup$
    – Marijn
    Feb 16, 2017 at 21:24
  • $\begingroup$ @Marijn Please have another look. I tried to clarify my answer. $\endgroup$
    – Diracology
    Feb 16, 2017 at 21:34

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