We can perform an experiment like this right here on Earth. Let's say a photon is emitted directly along the surface of the Earth at b = 6.371*10^6 m. First let's find the angle of curvature for the photon to travel from a radius of b to b + 1 micrometer. Integrating to four orders to find the angle, I get
θ = [-atan(B / sqrt(R^2 - B^2)) + pi/2]
+ [M sqrt(R^2/B^2-1) (1/(B+R) + 1/R) - 0]
+ [1.5 M^2 ((sqrt(R^2/B^2-1)/6) (3/R^2 - 2 (5 B + 4 R) / (B (B+R)^2)) - (5 / (2 B^2)) atan(B / sqrt(R^2-B^2)))
+ 1.5 M^2 (5 / (2 B^2)) (Pi/2)]
+ [(2.5 M^3 / (15 B^3)) (45 atan(B / sqrt(R^2-B^2)) + (B sqrt(R^2/B^2-1) / (R^3 (B+R)^3)) (5 B^5 + 15 B^4 R + 70 B^3 X^2 + 247 B^2 R^3 + 306 B R^4 + 122 R^5))
- (2.5 M^3 / (15 B^3)) 45 (Pi/2)]
Using UBasic (prog"lense170" for my own notes) and plugging in b = 6.371*10^6 and r = b + 1 / 10^6 with m = G M / c^2 = 4.43589170257*10^(-3), I get an angle of θ = 5.60287838354*10^(-7) radians. So if the original coordinates of the photon were x1 = 0, y1 = 6.371*10^6, they are now
x2 = (sin θ) r = 3.56959381816 m
y2 = (cos θ) r = 6370999.99999999999999791121093462 m
So let's say that we design an immensely strong metal tube, out in free space away from any gravitation, so that we can use a light beam while inertial to be sure that it is perfectly straight. Its dimensions will be 7.1391876363 m long with a radius of 2.0887890653751*10^(-15) m. Then we will bring it to the surface of the Earth and align it perfectly level with its center at b, lying perfectly along y = b. We will then use a laser to cut it in half carefully and remove one half, so that one end of the tube will now be centered on xA = 0 and yA = b while the other end is centered at xB = 3.5659381816 m and yB = b. From end A we will fire a laser perfectly tangent down the center of the tube. From earlier, we see that the beam will bend so that it strikes right at the edge of the tube at the other end, having fallen a distance in the y direction of dy = 2.0887890653751*10^(-15) m, the radius of the tube.
Now let's say we only fire a single photon down the tube. The time it will take to traverse the tube is approximately dt = dx / c = 1.19068833218*10^(-8) sec. Let's also say that at the same time that we fire the photon at end A, we drop a massive particle from rest at the center of end B. The particle will just fall due to Earth's gravity at a rate of g = G M / b^2 = 9.82216285185 m/sec^2 (for the values used). In the time dt, then, the particle will fall a distance dh = g dt^2 / 2 = 6.96263021792*10^(-16) m. This is only about 1/2.999999999999607595510104 the distance the photon falls in the same direction over the same time. So then, if we were to release the tube at the moment we fire the photon, allowing it to freefall, the tube will fall only about a third of the distance the photon will, while if it were to be considered inertial, the tube should fall at the same rate so that the photon would continue to travel along the center of the tube, but that is not what occurs according to the metric.