I am going to quote verbatim from Phil Plait along with the image he cites and then comment on it after his post:

http://www.slate.com/blogs/bad_astronomy/2015/11/18/astrophoto_space_station_photo_gets_a_seismic_jiggle.html

Take, for example, Tim Ashby-Peckham, who saw that the International Space Station was going to make a nice pass of his location in the early evening of Nov. 1, 2015. He set up his Canon 70D camera on a tripod, aimed it in the right direction, and waited. Once the ISS came into view, he started snapping away.

Take a look at the photo above. You can see the ISS as a long streak, since it moves a lot during the 30-second exposure. But take a closer look. See the wiggle in the streak at the bottom?

Ashby-Peckham took this shot in South Auckland, New Zealand, around 8:57 p.m. local time on Nov. 1, 2015. At 8:58, a magnitude 2.2 earthquake hit a couple of hundred kilometers to the southeast.

My strongest doubt was that a quake that weak could be felt that far away. However, in an email discussion about it, Ashby-Peckham told me that another earthquake about the same distance away was indeed felt in Auckland.

So this is a new one on me. Using the ISS as a virtual seismograph! It’s a pretty funny idea. I suppose if you live in a place with enough earthquakes you could actually calibrate your photography equipment to them, measuring how much the image wiggles versus size/distance of the quake. You need a moving target; stars don’t move quickly enough, and all you get is smeared-out disks for them (note that in the shot, the stars do appear to wiggle a bit). Of course, you could put the camera on a motor, letting it slowly scan the sky over and over again.

My comments:

ISS orbital period = 90 minutes which corresponds to 360°. 

So 30 secs corresponds to an angular displacement of (0.5/90)(360) =  2°

More precisely:

ISS orbital speed is 7.66 km/s. So in 30 secs, ISS moves 229.8 kms.

Assuming Earth radius as 6400 kms and the ISS average altitude as 380 kms, so the total distance from the centre of the Earth is: 6780 kms.

The angle traversed is thus: 229.8/6780 =  0.0339  rads = 1.94°

https://en.wikipedia.org/wiki/Peak_ground_acceleration

According to wiki[edia (link above), a Richter 2-3 corresponds to a peak ground acceleration (PGA) of 0.0017 to 0.014 g’s.

Assume a value of 0.0017 for Richter 2. This means a Dq = 0.0017 rads or 0.097°.

 In the image above, if the whole track is FOV is 0.0339 rads, and its corresponding length is 147 mm, while the displacement above the track is about 1 mm. This gives an angle of about 2.31x10^-4 rads (i.e. 0.013°).

Richter 2 should give 0.097°. This means that the shaking at Ashby’s location was closer to Richter 1, because the angular displacement of 0.013° is smaller by 7.5X than for Richter 2.

http://www.iitk.ac.in/nicee/wcee/article/13_480.pdf

According to the authors, T.Trombetti et al (13^th World Conference on Earthquake Engineering, Vancouver, Canada,2004), the Sabetta-Pugliese law (of 1987) states:

Log (A) = - 1.845 + 0.363 (MS) – log (R²+25)^1/2+0.195s

Where A is the PGA in g’s, MS is the quake intensity in the Mercalli scale, R is the distance in kms and s is a parameter that is 1.0 for soft soil and 0.0 for bedrock.

According to the S-P law at a low R = 0.001 kms and assuming MS = 2, PGA = 0.0019 for s = 0 and PG = 0.0029 for s = 1. The 1^st value is close to that mentioned in wiki.

For s = 0, MS = -2 gives PGA = 6.6x10^-5.

Instead we can use the total duration time of earthquake t (in secs) given by Tsimura’s empirical formula:

ML = -2.53 +2.85 log₁₀(t) + 0.0014R

Where ML is the local quake intensity and R is the distance from the epicenter.

The image shows that the quake lasts for about (3/147)(30) = 0.6 secs.

According to Tsimura, for ML = -3, the time t is 0.7 sec at R = 0.001 kms and 0.85 sec at R = 200 kms.

The duration would suggest that the quake was ML= -3 which is very feeble indeed.

On the other hand, as Plait points out the distance is R = 200 kms. With a ML = 2.2, the S-P law gives a PGA = 4.5x10^-7 at R = 200 kms. This is much smaller than the actual on-site PGA of 2.31x10^-4 from the angular displacement. At that distance, even with ML=6, the PGA is only 1x10^-5!

The two arguments from PGA vs R and from duration t are completely inconsistent.

At the end, one would have to invoke – as Phil Plait did – another completely different quake, much closer by to the observation point, to get better consistency. But not a lot, as it happens…

So what values could one try out?

Suppose we go for ML=-3 so that the duration is about right (0.7 sec). The PGA at R = 0.001 kms is 2.9x10^-5. So there should have been a mini-quake literally under Ashby’s feet! But this PGA is an order of magnitude too small.

Or a somewhat stronger one a few kilometers away…

If we choose, ML=-0.5, we get PGA=2.3x10^-4. Unfortunately, this gives t = 5 secs!

What this means is that we are unable to get consistency between the PGA and the time. What is clear, however, is that it was a mini-quake (weaker than ML = -0.5) and its epicenter was pretty close to where Ashby and his tripod were located.