Let's say we have two people viewing the moon who are 10300km apart along the same line of latitude at the same time. How much of a shift against the more distant background stars should the moon appear to shift?

This approximation assumes that the sublunar point is roughly between the two observers.

Well, first we need to find their actual linear distance (rather than the distance over the curvature of the Earth, which is what you get from Google Earth).

Let's define our variables:

\[ R = 6371393 m \;\; \text{Earth Average Radius} \] \[ d = 10300000 m \;\; \text{Earth distance along curvature} \] We can find the angle in radians from the arc length using:

\[ \theta = d/R \] and we can find the length of a chord using the half-sine rule:

\[ \text{crd} \, \theta = 2 \sin \frac{\theta}{2} \] Plugging that in we find that the straight-line distance is slightly shorter than around the curvature:

\[ 6371393 \times 2 \sin(\frac{1}{2} \frac{10300000}{6371393}) \approx 9214000m \]

Next we can find the angular size of a shift of some distance \( g \) from a distance \( r \):

\[ \alpha = 2 \arctan (\frac{1}{2} \frac{g}{r}) \] and :

\[ r = 384400000m \;\; \text{distance to the moon} \] \[ g = 9214000m \;\; \text{distance between observers} \] And we find that the expected angular shift is about:

\[ \alpha = 2 \arctan (\frac{1}{2} \frac{9214000m}{384400000m}) \] \[ \approx 1.373° \nonumber \] \[ \approx 2.7 \times \text{angular diameter of the moon} \nonumber \]

This seems to be a good match to explain this:

This approximation assumes that the sublunar point is roughly between the two observers.

Well, first we need to find their actual linear distance (rather than the distance over the curvature of the Earth, which is what you get from Google Earth).

Let's define our variables:

\[ R = 6371393 m \;\; \text{Earth Average Radius} \] \[ d = 10300000 m \;\; \text{Earth distance along curvature} \] We can find the angle in radians from the arc length using:

\[ \theta = d/R \] and we can find the length of a chord using the half-sine rule:

\[ \text{crd} \, \theta = 2 \sin \frac{\theta}{2} \] Plugging that in we find that the straight-line distance is slightly shorter than around the curvature:

\[ 6371393 \times 2 \sin(\frac{1}{2} \frac{10300000}{6371393}) \approx 9214000m \]

Next we can find the angular size of a shift of some distance \( g \) from a distance \( r \):

\[ \alpha = 2 \arctan (\frac{1}{2} \frac{g}{r}) \] and :

\[ r = 384400000m \;\; \text{distance to the moon} \] \[ g = 9214000m \;\; \text{distance between observers} \] And we find that the expected angular shift is about:

\[ \alpha = 2 \arctan (\frac{1}{2} \frac{9214000m}{384400000m}) \] \[ \approx 1.373° \nonumber \] \[ \approx 2.7 \times \text{angular diameter of the moon} \nonumber \]

This seems to be a good match to explain this:

That's a good first approximation anyway. There would also be some (usually small) refraction effects if the view of the Moon was lower on the horizon for one observer, and, of course, you need to know the current Earth-Moon distance and not just some average value.No takers for my little quiz?— FrĂ©dĂ©ric Marchal (@badibulgator) December 14, 2017

Here’s a hint: simultaneous views from Stockholm and Cape Town (very different latitudes, nearly same longitude), both oriented North up đź‘‡.

(Among other things note the curious undulating shape of the path of the moon) pic.twitter.com/gqeOVh3ayv