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The physics of ‘snipers’ for gold

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There are three forces acting on debris. First, there is the gravitational force that pulls down (Fgram) due to interaction with 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 (Fb). When an object is submerged in water (or any other fluid), there is an upward buoyant force from the surrounding water. The magnitude of this force is equal to the weight of the displaced water, so it is proportional to the volume of the object. Note that both the gravitational force and the buoyant force depend on the size of the object.

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

Illustration: Rhett Allain

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.

We can now model the movement of a rock versus the movement of a piece of gold in moving water. However, there is a small problem. According second law of NewtonThe net force on an object changes the speed of the object, but as the speed changes, the force also changes.

One way to deal with this problem is to divide the movement 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 find the speed and position of the object at the end of the interval. So I just need to repeat this same process for the next interval.


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