Receding a distant ship smoke on the horizon?

Question

Because the earth is round, there is observable horizon.

What should be the pressure of atmosphere to make the surface of the earth to appear flat, and horizon disappear?

Refraction index of standard air at sea level is
no = 1.00029

People,
the planar front of constant phase of horizontal ray remains always vertical when the ray bends properly,
right?

Answer 1

If air's refraction gradient enables you to see a little further, it's "flattening" the curvature. Increased pressure & density should increase the refractive index, but will it increase the gradient? Yes, according to the formula used to compute air pressure as a function of altitude and sea-level pressure (it's in the form p = p0 * e ^ (k * g * (h - h0)). But I'd suspect not reliably unless g increases also. One reason being that the pressure increase would expand the atmosphere and air would be lost at the fringes where molecular velocity >= escape velocity.
EDIT: OK, so the wavefront stays vertical, the velocity is proportional to radius R and the refractive index N inversely proportional. Then we need to use the pressure-vs-altitude equation, and the fact that the fractional part of N is proportional to pressure (ref.) to derive the desired SLP. This yields 440000 Pa giving an N value of 1.001259 and a gradient dN/dR of -1.575E-07/m. This gives (dN/N)/dR = 1/R. Of course the gradient is only correct at sea level for the desired flattening effect.

Answer 2

Although I understand Aviophage's point, I think maybe she is misinterpreting what Alexander is asking (and maybe he isn't stating it clearly).

It's a well-known fact that the atmosphere lets you "see" over the horizon a little bit farther than you could do in a vacuum. For example, you can still see the setting sun on the horizon, even after the moment when geometrical considerations tell you that its actual position is BELOW the horizon. This of course is because the light from the sun does not follow a straight line through the atmosphere, but refracts, following a curved path. The sense of the curvature is "downward" (i.e. the same as the curvature of the earth).

The curved refractive path is not caused by the air pressure per se, but by the changing density of the air at different altitudes. Less dense air has a lower index of refraction, so as the light travels across the "interface" from denser air to less dense air, it changes direction. (The change in density is continuous, so instead of abrubtly changing direction, the direction change is continuous; i.e., the light "curves" through the atmosphere).

So, I think what Alexander is asking is this: Under what hypothetical atmospheric conditions would the radius of the light's curvature, match the curvature of the earth? Under such conditions, if you were to shine a beacon horizontally (say from a lighthouse), the atmospheric refraction would cause the beacon's light to bend around and exactly follow the curvature of the earth. In theory (if the light were bright enough), you could literally be halfway around the world and still be able to see the beacon. Every place on the earth's surface would be (with sufficiently bright light) visible from every other place. The surface would have the APPEARANCE of being flat, rather than curved.

So, what atmospheric conditions could achieve that?

None that I can think of. As stated earlier, the light's curved path is not caused by pressure per se, but by CHANGES in pressure (hence in density). In other words, if the pressure (high or low) were uniform throughout the light's path, the light would go in a straight line rather than curving.

But generally speaking, in order to pass into a region of lower pressure/density, the light would have to pass into higher altitudes. Which means it's NOT traveling parallel to the ground, which means it's not following the earth's curvature. In the case of the setting sun phenomenon, the light in fact travels from higher elevations to lower elevations (this is apparent if you draw a diagram). One the one hand, this allows the light to curve (since it's going through a variable index of refraction); but at the same time, it prevent the light from exactly following the curvature of the earth (since the path of the beam is changing elevation).

Answer 3

You asked this before, and we explained that there is no relationship between atmospheric pressure and the curvature of the earth. The two phenomena are unrelated.

The horizon and the curvature of the earth are unchangeable physical features of the planet.

Atmospheric pressure is a function of the weight of the constituent gasses that make up the atmosphere and the depth of the atmosphere. Atmospheric pressure cannot affect the shape of the earth.

Even if you are only asking about the visual phenomenon of the receding ship over the horizon, you are asking about an effect of a physical geometrical relationship, that still has nothing to do with the pressure of the atmosphere.

Will you get it, this time? Please? There is no connection, and cannot be.

(Sorry, guys. I see what RickB is getting at, but I still can't read that in Alexander's question. If he actually is asking, "at what atmospheric pressure would the refractive index be unitary?" maybe someone has the physical chemistry to tell us. If there is a point, please go right ahead.)