What's the search area for a meteorite that lands in the ocean and sinks to the ocean floor?

Question

A meteorite survives entry and lands in the ocean. For various reasons, people would like to recover the core of this meteorite.

Observers of the landing are able to narrow down the location of its impact with the surface to an area of approximately 50 km² (about 19.3 mi²). The ocean in that area is a bit deeper than usual but not abyssal trench depths, 5000 to 6000 meters (16,400 to 19,700 feet). For comparison, the wreck of the Titanic is about 3,800 meters (12,500 feet) deep.

Assuming Earth-like oceans and the necessary magic for an ocean floor search is available, roughly how large would the search area on the ocean floor be, given that ocean currents at various depths could have moved the meteorite in various directions as it sank? Available ocean current charting is at a level roughly equivalent to Earth in 1800.

Assume the meteorite has a standard nickel-iron composition, and impacted at an angle of entry greater than 45 degrees but less than 80 degrees. Assume that the largest part of the meteorite to survive the impact with the surface (the piece that we are interested in) is roughly spherical and approximately 1 meter in diameter.

Un-specified: Assume a realistic size and velocity for the meteorite before impact with the surface that will not generate major tidal waves or other disasters and will produce a core or largest surviving chunk of around 1 meter diameter as mentioned.

Worst case (that is, largest) search area estimates are preferred.

Answer 1

This is basically a physics question. First, the terminal velocity of a sinking object is expressed by $v_t = \sqrt{\frac{2(m - p_fV)g}{Cp_fA}}$

Where $m$ is the mass of the object, $p_f$ is the density of the fluid, $V$ is the volume of the object, $g$ is the acceleration of gravity, $C$ is the drag coefficient, and $A$ is the cross-sectional area of the object.

So if we assume a density of ~8000$kg/m^3$ (nice round number and nickel-iron blend), and a 1$m$ sphere, the $(m-p_fV)$ expression evaluates to $\frac{4}{3}\pi (0.5m)^3 \times 7000kg \approx 3700kg$.

$p_f$ is $1000kg/m^3$

$C$ is 0.47 (for a sphere)

$g$ is $9.8m/s^2$

and $A$ is $\approx 0.785m^2$

So terminal velocity would be about 14 m/s.

If we assume the shallower end of your projected depth, 5000m, that means the meteorite would take about six minutes (358 seconds) to sink.

Ocean currents can be as rapid as 4m/s. If we assume the worst-case, that such a vigorous current is applied to the meteorite consistently in any given direction as it sinks, then that generates a radius of 4 * 358 $\approx$ $1.5km$ around a straight line down from its point of impact.

If we assume the original $50km^2$ approximate impact space is circular, that gives us a radius of $\approx 4km$. Again assuming worst-case, the drift would expand that radius by 1.5km uniformly.

$\pi \times (5.5km)^2 \approx 95km^2$, or about twice your original search area.

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Note: A bunch of assumptions were made in the course of this calculation (Earth gravity, ocean is made of water and not some other liquid, density of the meteorite) and a bunch of rounding was done because it's easier. You can tweak any number of these values for the world in which this occurs.