So for the bottle, the difference in gravitational pull from one side of the bottle to the other side of the bottle is extremely small because the distance is extremely small relative to the distance to the moon, and the tidal forces can not be observed.

How small? Let's work it out. A bottle has a radius of about 0.03 m.

$$F^{\text{lunar tides}} = Gm^{moon}\frac{2r^{bottle}}{d^3}$$ $$m^{moon} = 7.34 \times 10^{22} kg$$ $$r^{bottle} = 0.03m$$ $$d^{moon} = 3.84 \times 10^8 m$$ $$G = 6.674 \times 10^{−11} m^3⋅kg^{−1}⋅s^{−2}$$

Which is about $6 \times 10^{-15} m⋅s^{−2}$. This is an order of magnitude smaller than the smallest acceleration measured. Not just extremely small, unmeasurably small.

So for the bottle, the difference in gravitational pull from one side of the bottle to the other side of the bottle is extremely small because the distance is extremely small relative to the distance to the moon, and the tidal forces can not be observed.

How small? Let's work it out. The acceleration on a bottle of water due to the Moon is...

$$\text{Gravitational constant} \times \text{mass of the Moon} \times \frac{\text{diameter of the bottle}}{\text{distance to the Moon}^3}$$

Let's assume our bottle has a diameter of about 0.06m. The distance to the Moon varies, I'll use the semi-major axis. It won't make a difference.

$$\text{mass of the Moon} = 7.34 \times 10^{22} kg$$ $$\text{distance to the Moon} = 3.84 \times 10^8 m$$ $$\text{Gravitational constant} = 6.674 \times 10^{−11} m^3⋅kg^{−1}⋅s^{−2}$$

Which is about $6 \times 10^{-15} m⋅s^{−2}$. This is an order of magnitude smaller than the smallest acceleration measured. Not just extremely small, unmeasurably small.