How would the volume of a drop of water (from a dropper) on the Moon and other bodies compare to one on Earth? (indoors of course!)

Question

Searching for "volume of a drop of water from an eyedropper", I ran across this answer on Quora:

It depends on the size of the dispensing tip, but generally for a Pasteur pipette it is about 1/20'th of a mL. That is on Earth, at or near sea level.

(Incidentally, other values from other sources put it at 1/17 or 1/18 mL. It will probably be somewhat dependent on the shape of the dropper tip and material.)

Assuming we are "indoors" at standard temperature and pressure, how would the volume of a drop of water vary with surface gravity?

Limits:

• Experiments in space suggest that in microgravity water will likely remain stuck and creep up your arm, so if you call that a drop, the volume could be infinite. (And with you in the center of it! Wringing out Water on the ISS - for Science!)
• At extremely high gravity it may be possible that the water is simply pulled out of the dropper by overwhelming surface tension effects.

But if one were doing this on Mars, or the Moon, or on a hypothetical 2g surface gravity body, how would the volume of a drop of water from a dropper compare to that on Earth? How would it scale with the magnitude of gravity?

Answer 1

As per your comment below, there could be two aspects to your question, so I'll address both.

According to this article,

In the pendant drop test, a drop of liquid is suspended from the end of a tube or by any surface by surface tension. The force due to surface tension is proportional to the length of the boundary between the liquid and the tube, with the proportionality constant usually $\gamma$. Since the length of this boundary is the circumference of the tube, the force due to surface tension is given by, $$F_\gamma=\pi d\gamma $$

This means $$d=\frac{mg}{\pi\gamma}$$ and the size (diameter) of the drop can be computed to be $$s\propto (\rho g)^{-\frac{1}{2}}$$ So it is indeed the case that the size of the drop changes as the strength of the gravitational field changes. To that, you can estimate how this will vary on earth, mars and the moon.

As for the case where we investigate the profile of the drop as it is falling, I found this study, Effect of gravity on the spreading of a droplet deposited by liquid needle deposition technique, 2023, where the researchers investigate the effects of gravity on the "spreading dynamics of a water droplet":

Referring to the classical theory of capillarity, if the characteristic length of a drop is less than a capillary length, gravitational effects can be neglected and hydrostatic pressure rapidly stabilizes across the droplet profile. It leads to a spherical shape being adopted by the droplet in order to obey the Laplace law.

The following graph relates the strength of the gravitational field (varying from $0$ to two times that on earth $2g$) versus time. It shows the effect of gravity and how the shape of a $10 μl$ sessile (stationary) drop evolves over time.

Note that "at the onset of the deposition, the drop grows vertically faster than the spreading...Once the spreading begins, the base diameter and the maximum height of the drop grow until the volume is added to the drop. After the deposition of a required volume, the drop remained stable during the microgravity period..."

In this diagram (where theory is compared to experiment) we observe how a droplet changes shape over time for both earth and a $\mu$-gravity (parabolic flight) environment:

The variations in volume seem to be relatively small regardless of the strength of the gravitational field, though the shape of the drop varies. For the relatively small water volumes tested (water droplets), the effects of surface tension dominate.

Answer 2

The size of the water drop squeezed from a small orifice is determined both by surface tension and gravity (see Wiki). To obtain the functional form of the dependence you can use dimensional arguments. In the simplest case, the diameter $d$ of the squeezed drop depends on gravitational acceleration $g$, surface tension of the liquid $\gamma$ and its density $\rho$. Then dimensional consistency requires: $d \sim \sqrt{\gamma/\rho g}$ (the proportionality constant must be evaluated empirically). Therefore, if the gravity is four times stronger then the squeezed drop will be half as big.