The Physics of ‘Sniping’ for Gold

I’m not exactly sure how the YouTube algorithm finds videos for me to watch, but now that I’ve stumbled into videos about people searching for gold, I can’t stop. There are a bunch of prospecting videos, but I like the ones where people wade knee-deep into rivers and search for tiny bits of gold stuck in the cracks of rocks. If you want to check them out, take a look at Tassie Boys Prospecting or Pioneer Pauly. Both are great. (But be careful or YouTube will just give you more gold videos.)

One way to look for these flecks of gold is to use the “sniping” method. Here’s how it works, according to my extensive analysis of YouTube: Find a river that could have gold in it. Put on your wet suit, mask, and snorkel. Dig around in the rocks looking for the places most likely to harbor the flecks. Swish the water around with your hand to stir up debris, which will include a lot of small rocks and dirt, but maybe also some gold. Most of the debris will get swept away in the river current, but the gold will start to sink. Use a little squeeze bottle and suck up those tiny pieces. Profit! (Or, at least enjoy some entertainment.)

But why doesn’t the gold get washed away along with the flowing water? It seems strange to me, but I suspect it has to do with the very high density of gold, around 19.3 grams per cubic centimeter—much higher than rock, which is about 2.7 grams per cubic centimeter. You know what this means? I have to build a model of debris and gold pieces in a moving river.

(Please note: This article is only about the physics of gold sniping. If you want to give it a try, you’ll need to check out regulations that govern gold prospecting in your area. Prospecting is illegal in some places, or there may be limits on the devices you can use or how much material you can gather.)

Let’s start with modeling a random chunk of debris released into a moving river. (It could be rock, gold, or whatever.) I’ll assume the piece is spherical with a radius (r) and density (ρ) that will give it some mass (m). Now, let’s consider the forces acting on this object.

There are three forces acting on the debris. First, there’s the downward-pulling gravitational force (F_g) due to the interaction with the Earth. This force depends on both the mass (m) of the object and the gravitational field (g = 9.8 newtons per kilogram on Earth).

Next, we have the buoyancy force (F_b). When an object is submerged in water (or any fluid), there is an upward-pushing force from the surrounding water. The magnitude of this force is equal to the weight of the water displaced, such that it’s proportional to the volume of the object. Notice that both the gravitational force and the buoyancy force depend on the size of the object.

Finally, we have a drag force (F_d) due to the interaction between the moving water and the object. This force depends on both the size of the object and its relative speed with respect to the water. We can model the magnitude of the drag force (in water, not to be confused with air drag) using Stoke’s law, according to the following equation:

In this expression, R is the radius of the spherical object, μ is the dynamic viscosity, and v is the velocity of the fluid with respect to the object. In water, the dynamic viscosity has a value of about 0.89 x 10^-3 kilograms per meter per second.

Now we can model the motion of a rock versus the motion of a piece of gold in moving water. There is one small issue, though. According to Newton’s second law, the net force on an object changes the object’s velocity—but as the velocity changes, the force also changes.

One way to deal with this issue is to break the motion of each object into small time intervals. During each interval, I can assume that the net force is constant (which is approximately true). With a constant force, I can then find the velocity and position of the object at the end of the interval. Then I just need to repeat this same process for the next interval.

But if I used an interval of 0.001 seconds, I would need to do 1,000 of these calculations to get the motion of the object during a single second. No one wants to do all that—so instead I will write a Python program.

Here is a quick test of this calculation. Suppose I have two small spherical objects, each with a radius of 0.5 millimeters—one is a rock and the other is gold. Both are released in a stream of water that is moving at 0.1 meters per second, from a position 10 centimeters above the bottom. This is a plot of the vertical position (y) as a function of time (t):

Notice that the gold object (the blue curve) sinks down faster than the rock (the red curve). That’s basically what you want as a gold sniper. You want the rocks to get swept away and the gold to sink down.

Let’s consider how far downstream an object moves once it’s released. The downstream distance doesn’t just depend on the density of the object, but also on its size. Suppose I model the motion of a gold sphere compared to a rock sphere released at the same height in a moving stream. How far downstream does each object travel before hitting the bottom? Here’s a plot of downstream travel distance versus object radius:

There can be other materials mixed in with river debris, too. Sometimes you can find tiny pieces of iron (with a density of 7.87 grams per cubic centimeter) or even lead (11.34 g/cm³). These other materials would have similar-shaped curves, but they would be in between the ones for gold and rock. The gold pieces would sink to the bottom first.

There’s something else to see from this plot. The smaller the stuff, the greater the downstream separation between rocks and gold. If the two pieces each have a radius of just 0.2 millimeters (that’s pretty tiny), they will end up about 5 centimeters apart after sinking in the water. That’s exactly what you want: Get the rock out of there, leave the gold. But as the rocks and gold pieces get bigger, the downstream separation is quite small. Still, that should be OK, because with a bigger object, a gold sniper should be able to clearly see the difference between something that’s gold and something that’s not.

This is a great example of the physics of scale. We often like to think that big things (like large rocks) will behave just like small things (like pebbles). I mean, if you drop a small rock and a large rock, they are going to fall with essentially the same motion. So it seems reasonable to assume small and large rocks would be affected by the water in the same way. But that’s not the case. A difference arises when you have two different influences that have different relationships to size, which physicists also call scale.

Let’s look at the example of a sphere sinking in a moving river. Just to make things simpler, I’m going to look at a sphere that’s only moving vertically in water, so I don’t have to deal with two dimensions. In this case, we can calculate the acceleration of the object as the sum of the forces divided by the mass. (This is straight from Newton’s second law.)

Notice that the gravitational force (F_g) is negative, or downward, but the drag force (F_d) is positive, or upward, since it is in the opposite direction of motion.

Of course, we are going to need the mass (m) of the object. If it’s a sphere, the mass is proportional to the volume, which depends on the radius (r) raised to the third power. But the drag force also depends on the object’s size. The magnitude of this force is proportional to the radius of the object. Let’s just rewrite the acceleration with the radius terms in the expression.

Now suppose we double the size of the sphere. This will double the drag force. (Just put in 2R instead of R.) But what about the gravitational force? Since this depends on R³, a doubled radius would increase the mass by a factor of 8 (which is 2³). So, as the size of the object increases, the gravitational force will get much larger than the drag force. Eventually, you would get to a point where the magnitude of the drag force is insignificant compared to the gravitational force. At that point, a large rock and a large piece of gold would move through the water in a very similar manner.

There are tons of great examples of the physics of scale. For example, the Earth has a molten core, but the moon doesn’t, and that’s because the Earth is larger and takes longer to cool. In general, small things cool off faster than large things because the ratio of surface area to volume is greater. The larger the volume, the more thermal energy an object has, but you need to radiate this energy through a relatively smaller surface.

Another example: Large birds don’t look like small birds because they need huge wings to fly. A flying bird experiences two equal forces, the downward gravitational force and the upward lift from its wings. The gravitational force is proportional to the volume of the bird, but the lift depends on the area of the wings. So, if you doubled the size of a hummingbird without changing its shape, its weight would increase by a factor of 8 (its size cubed), but the lift only increases by 4 (its size squared). The only way to fix this problem is by giving larger birds much larger wings. That’s why you can’t have an eagle-sized hummingbird.

The physics of scale even explains why large hail is so much more dangerous than small hail. Hail is just like a flying bird, except that it’s cold and can damage your car. If you double the radius of a hail ball, you increase its volume (and thus its weight) by a factor of 8. However, the surface area only increases by a factor of 4. This means that larger hail can fall at greater terminal velocity before hitting your car. And on top of that, it has more mass because it’s bigger. That’s why hail might not just dent your car but could even break the windshield.

And of course for gold snipers, the physics of scale is the difference between finding a tiny piece of gold or just a dumb old rock.