[saspinski (OP)]

I was puzzled for a while, and then found an explanation, but it is tricky at a first sight.

Using the Newton's equation: a_s = GM_s / r² for the  gravity acceleration of a body by the Sun, and replacing the constants by its values, here at Earth, a_s is 0,006 m/s².

As g (only due to Earth) is 9.8  m/s², that means that at midnight it is a net 9.806 m/s², because any body is being pulled by Sun and the center of the Earth.  At noon, the effect of the Sun is opposite and it is a net 9,794 m/s².

That effect could be detected by a kitchen balance, because the difference (9.806 - 9,794 ) / 9.8 = 0,12% is not so small. An object of 3000 g would show at least 3g of difference.

But there is no difference at all. Why?

[chiralSPO]

Balances measure mass, not weight, so the magnitude of the gravitational field is irrelevant (as long as it is nonzero).

[wolfekeeper]

Your equation for the tidal force is wrong.

[chris]

Balances measure mass, not weight, so the magnitude of the gravitational field is irrelevant (as long as it is nonzero).

You should probably explain the subtlety in that answer, and why this is the case...

[chiralSPO]

Balances measure mass, not weight, so the magnitude of the gravitational field is irrelevant (as long as it is nonzero).

You should probably explain the subtlety in that answer, and why this is the case...

Indeed...

As their name suggests, balances work by balancing an unknown mass against that of a known mass. It will accurately determine the unknown mass long as both the known and unknown masses are experiencing the same gravitational acceleration (which is a safe bet on earth, tides or no, as the two are typically well within 50 cm of each other, but tidal forces near a small black hole could mess the balance up).

In many cases the balancing act depends on torque--it is easy to change the torque exerted by a mass by sliding the mass along the length of the lever, and then by comparing how far from the fulcrum the known mass must be moved to equal the torque of the unknown mass (at known distance from the fulcrum), one can calculate the ratio of the known and unknown masses from the ratio of the two distances from the fulcrum.

[chris]

Thanks @chiralSPO - so what would be the outcome of doing the experiment with a spring set of scales, where the work is done by the mass against a spring...?

[chiralSPO]

Thanks @chiralSPO - so what would be the outcome of doing the experiment with a spring set of scales, where the work is done by the mass against a spring...?

I would expect a spring scale to be sensitive to the weight (gravitational force exerted on the object of interest) rather than the mass. To first approximation, the gravitational force on the object will extend the spring by an amount proportional to the gravitational force and fundamental constant of the spring (this approximation does no account for the mass of the spring itself, or the fact that in reality the "spring constant" is not actually independent of the degree of extension.)

[Janus]

I was puzzled for a while, and then found an explanation, but it is tricky at a first sight.

Using the Newton's equation: a_s = GM_s / r² for the  gravity acceleration of a body by the Sun, and replacing the constants by its values, here at Earth, a_s is 0,006 m/s².

As g (only due to Earth) is 9.8  m/s², that means that at midnight it is a net 9.806 m/s², because any body is being pulled by Sun and the center of the Earth.  At noon, the effect of the Sun is opposite and it is a net 9,794 m/s².

That effect could be detected by a kitchen balance, because the difference (9.806 - 9,794 ) / 9.8 = 0,12% is not so small. An object of 3000 g would show at least 3g of difference.

But there is no difference at all. Why?

Your basic assumption is flawed.  The Earth is in orbit around the Sun, thus it is in free-fall.  This is no different than the astronaut in the ISS.  At the altitude of the ISS,  acceleration due to gravity from the Earth is still 8.67 m/s^2 or 88.6% of that at the surface.  Yet the astronauts float around weightless with respect to the ISS.  Both the ISS and the astronauts are freely falling in the same orbit.     In the same way, both objects on the Earth and The Earth are freely falling around the Sun in the same orbit.   In order to be able to measure that 0.006m/s^2 by the Sun on the Earth, You would have to stop the Earth in its orbit and suspend it motionless with respect to the Sun.

Now that is not to say that there is no gravitational difference measurable on the Earth that is caused by the Sun. Wolfekeeper mentioned tidal force.   Tidal force is due to the fact that the gravitational pull of the Sun varies with distance and since for example, at noon you are some 6478 km closer to the Sun then the center of the Earth is, the pull of the Sun is slightly stronger on you.   Since the orbit of the Earth is determined by the Earth's center of mass,  and you are being forced to orbit with the Earth, you are not quite in free-fall with respect to the Sun when the Sun is directly over head.  The net effect is due o the difference in acceleration at the center of the Earth and the acceleration at your distance from the Sun.  Across the radius of the Earth, this works out to be 0.00000051 m/s^2   
This less than 1/10000 of the 0.006m/s^2 in your original post and not something you are going to easily differentiate with a kitchen scale.
At midnight, you are further from the Sun than the center of the Earth, and there will be a like effect.   But unlike your inference in the OP, this will not be added to your weight, but like at noon, subtracted from it. (you are now further from the Sun than the Earth's center, so the pull of the Sun on you is weaker.)
This differential pull acting across the body of the Earth is referred to as tidal force. (since it produces tides). 

[jeffreyH]

Now that is what I call a best answer.