Above: sum of frames from the video shot by the parachuters
Some eyeballing of the apparent motion of the object seemed to show that it was compatible with the approximate terminal velocity such a meteoroid would have. The fact that the video was two years old, and that the footage had been kept from being released for a while and it had generated a wide-ranging search for the fallen stone in the forests of Norway, after an accurate determination of its trajectory had been computed by a research team, implied that the hypothesis of an hoax was off the table. But then, a scientist should always be sceptical, and in cases such as this one it could be useful to apply some very basic statistical analysis to understand how steep the claim of a meteoroid caught on film in dark fall could be.
The data on which to base our estimate can be very sketchy but we can still obtain a reasonable probability for the observed effect if we try to structure our calculation a bit. A probability of the observation, given the hypothesis that the stone is indeed a meteoroid, needs first of all to be defined. In other words, what do we exactly want to estimate ?
I believe a good question to ask ourselves is: given the hours of video footage taken while airborne every year, and given an estimate for the rate of meteoroids falling on Earth, how many sightings similar to the one of the Norwegian clip should we expect in a year ? If we know the total footage time T in years, and the rate of meteoroids R per year and square kilometer, and if we assume that a camera will be able to capture meteoroids falling within a distance D, with an efficiency E and a fraction of horizon framed equal to F, the calculation is not difficult: the number of events we will catch is (assuming we are only counting footage taken in daylight in T)
N = R [km^-2 y^-1] * T [y] * pi * D^2 [km^2] * F * E
In the formula above R * T is the rate of meteoroids caught on video in a year if the camera observes a square kilometer around itself. The rest of the equation takes care of accounting for the limited range of the camera. I inserted a factor F to account for the limited fraction of angle observed by the camera. Note that I am considering a two-dimensional surface, implying that I assume a video camera will see the meteoroid if the latter intersects a certain observed area of the Earth's atmosphere. I also insert an efficiency factor E to account for the fact that I am considering a fixed "sensitive range" but this actually depends on the meteoroid's size.
So what values should we substitute for D, F, E, R in the formula for N ? Here I am as much in the dark as you are. But let's see. D can be at most 0.05 kilometers, I would say, as meteoroids in dark flight large enough to be seen at over 50 meters of distance are negligibly rare - we can ignore them. (Besides, much of the surprise of the video is the idea that the caught meteoroid really missed the parachute by just a few meters). For F we can take 0.2, implying a reasonably wide angle of 72 degrees. E and R are the hard part: I will make a guesstimate of E=0.1 and R=0.0001, implying a rate of one detectable-sized stone per year in an area of 10000 square kilometers (a 100x100km square), and an efficiency of detection of 10%. I might be off by one or two orders of magnitude here, but let's see what we get. (And after writing this, I found this page which tells me I am not quite far off!).
Now one ingredient is missing: the film time. I would say that ordinarily we do not spend our time filming the atmosphere when we are airborne, but there are exceptions; parachuters and wingsuit nutcases are just one category of people who do that routinely. Then there are cameras mounted on airplanes, and there is people taking footage from airplane windows. One could even count in surveillance cameras set on high points, and those work around the clock (although they usually take only a frame every few seconds, or less). I am making a wild guess here, and intentionally a very, very conservative estimate: a total footage of a thousand years ? It could be a lot less than that, while I doubt it can be too small by a factor of 100: even if we mounted a video camera on every freaking plane on Earth, we would not be underestimating T that much.
So what do we get for N using the above estimates ? The above formula gives
N = R [km^-2 y^-1] * T [y] * pi * D^2 [km^2] * F * E =
= 0.0001 * 10^3 * 3.14 * 0.05^2 * 0.2 * 0.1 = 1.5*10^-5
Interestingly, we obtain a rather small number! And we have probably been quite conservative in our estimates; in particular, it is quite doubtful that a video camera can spot the fall of a 10-gram meteoroid at tens of meters of distance; while if we take for the meteoroid rate that of larger objects, it is bound to become orders of magnitude smaller.
So what does the calculation above imply ? I think it shows that the hypothesis that by chance a parachuter or a tourist or somebody else with a video camera pointed at the horizon does record a falling meteoroid is too small to be considered even a "lucky" occurrence. We are led to conclude that some alternative explanation should be sought for the "falling stone" observed in the video.
And indeed, a much, much more likely explanation exists. It comes from a very detailed analysis, and it is explained in full in a blog post. The executive summary ? The stone was released when the parachute opened, and it soon "caught up" with the person holding the video as air slowed him down!