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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.

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  • $\begingroup$ Is the size, angle of entry, speed of entry, or composition known? $\endgroup$
    – Nosajimiki
    Aug 16, 2023 at 21:30
  • $\begingroup$ In addition to the comment from @Nosajimiki we would need to know what the interaction of it hitting the water was like. E.g. did it stay relatively intact because it was super dense or did it shatter into pieces? $\endgroup$ Aug 16, 2023 at 22:42
  • $\begingroup$ satellite tracking should be able to narrow it down by a lot more. big meteorites like that garner a lot of attention. $\endgroup$
    – John
    Aug 17, 2023 at 0:11
  • $\begingroup$ @John - I am skeptical of that assertion. We have enough trouble determining where larger (in terms of volume) pieces of space junk have impacted. Most metorite-hunters searching for the shards of a meteorite in a land impact are dependent on ground-level weather radar, not satellite tracking. $\endgroup$
    – jdunlop
    Aug 17, 2023 at 1:11
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    $\begingroup$ OP said "presumably at terminal velocity". I don't believe that such a meteorite weighing around 4.3t would be moving that slowly when it hits the water. $\endgroup$
    – Monty Wild
    Aug 17, 2023 at 8:04

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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.


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.

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    $\begingroup$ "First, we assume a spherical cow..." $\endgroup$
    – jdunlop
    Aug 17, 2023 at 0:07
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    $\begingroup$ keep in mind a meteorite of that size is going to plow through a lot of water ballistically before it starts to sink. meteorite are falling at a LOT more than terminal velocity. $\endgroup$
    – John
    Aug 17, 2023 at 0:15
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    $\begingroup$ @John - the OP indicated that the meteorite is falling at terminal velocity by the moment of impact. And the incompressibility of water decelerates things quickly. A small meteorite is likely to be well, well below orbital velocities by the time it hits the ground - most metorites hit at well under the speed of sound. $\endgroup$
    – jdunlop
    Aug 17, 2023 at 0:23
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    $\begingroup$ Sorry missed the OP's mention of terminal velocity, that makes it a very strange meteorite. Keep in mind the largest part is a meter across after landing, that makes the initial meteorite much larger, 1 meter across is already among the top 10-20 largest meteorites on earth. $\endgroup$
    – John
    Aug 17, 2023 at 1:38
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    $\begingroup$ Using Impact Depth approximations, the meteorite would make it less than 10m below the surface of the water before it was at the terminal velocity of 14m/s. $\endgroup$
    – IronEagle
    Aug 17, 2023 at 14:07

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