Strong atmospheric refraction can make the horizon on Venus much closer than expected, but how close is it? My naive calculation is very different from reported observations.

A source cited in How far can you see on the surface of Venus? says that the horizon for the Venera landers 0.9 m high camera appeared to be much less than a kilometre away, and a NASA website says the horizon was only 100 m distant "possibly due to a mirage".

The radius of curvature ($r$) of a horizontal light ray in a planetary atmosphere is given by

$$\frac{1}{r} = \frac{n-1}{n T} \left[ \frac{g M}{R_{gas}} + { \mathrm{d} T \over \mathrm{d} h } \right]$$

where $n$ is the refractive index, $g$ the local gravity, $T$ and $M$ are the atmosphere's temperature and mean molecular weight, $R_{gas}=8.31446\,\textrm{J/K/mol}$ is the universal gas constant, and $\frac{dT}{dh}$ is the lapse rate describing how the atmospheric temperature varies with height $h$. The index of refraction for ideal gases as a function of temperature and pressure ($P$) is

$$n(P,T) = 1 + \delta n_0 \times {T_0\over P_0}{P\over T}$$

where $n_0=1+\delta n_0$ at some reference temperature ($T_0$) and pressure ($P_0$). For $n\approx 1$, this gives

$$\frac{1}{r} = \delta n_0 \times {T_0\over P_0}{P\over T^2} \left[ \frac{g M}{R_{gas}} + { \mathrm{d} T \over \mathrm{d} h } \right]$$

where $\frac{dT}{dh}$ is the lapse rate describing how the atmospheric temperature varies with height $h$.

For Venus,$R=\textrm{6052 km}$, $g=\textrm{8.87 m/s}^2$, $P=\textrm{93 bar}$, $T=\textrm{737 K}$ and $\frac{dT}{dh}$ =-7.7 K/km. Ignoring the few percent of$N_2$ in Venus's $CO_2$ atmosphere, $M=0.04401\,\textrm{kg/mol}$, $\delta n_0 = 0.00045$ at $T_0=273.15\,{K}$ and $P_0=1\,\textrm{bar}$.

This gives $r=\textrm{1211 km}$, which should make the horizon farther away since light from distant objects will partially follow the curvature of the planet. Why do the Venera landers see such a close horizon at 0.1-1 km?

P.S. That the horizon should be farther away is backed up by the amusing and often thought-provoking book "The Inventions Of Daedalus: A Compendium Of Plausible Schemes", where it was facetiously claimed that if the earth's atmosphere was sufficiently dense, "a ray at the surface would follow its curvature exactly, and it would appear flat. Departing ships would not sink below the horizon but merely dwindle into the distance, and people would not have realized the Earth was round until they discovered that, with a good telescope, you could see the back of your own head.")

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    $\begingroup$ I'm not sure what equation you need, but for an ideal gas, density is proportional to molecular mass. So a box of Venus air at the same temperature & pressure as a box of dry Earth air has ~1.53 times the density. $\endgroup$
    – PM 2Ring
    Commented Oct 14, 2022 at 20:40
  • $\begingroup$ Thanks @PM-2Ring. The reduced index of refraction (n-1) of CO2 is about 1.5 times that of air, and I agree that would need to be factored in to any precise calculation, but I haven't worried too much about ti since I don't see how it could explain the 3 orders of magnitude discrepancy between my expectation and the observation. $\endgroup$ Commented Oct 15, 2022 at 14:20

1 Answer 1


A problem with my calculation appears to be in my assumption of a uniform small lapse rate.

According to the 1969 paper "Some Consequences of Critical Refraction in the Venus Atmosphere", we would indeed naively expect the horizon to be very far away. Sitting on the surface, it would look like we were in "a very flat bowl with its edges extremely far away and just at the level where we are sitting."

That the Venera landers did not see this "bowl" was, in fact, turned around to tell us about the Venusian atmosphere's vertical temperature profile. According to the 1976 paper, "Inferences on the structure of the surface layer of Venus' atmosphere", "the visibility of the horizon in the panoramas photographed by the Venera 9 and 10 landers" implies that " In a 1-m ground layer of the atmosphere of Venus, the temperature decreases with height by about 0.1-1 K". This is 1-2 orders of magnitude larger than the value I used of 7.7 K/km, which is presumably only valid over larger scales above the surface.

The 1976 paper also notes the the lapse rate "will probably range from a few hundredths of a degree to a few tenths of a degree or more, depending on local conditions. As a result the distance of the horizon may differ greatly in different places (and even in different directions at the same place). Evidently there should also be strong time variations."

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    $\begingroup$ Yes, so Venusian mirages depend on localized abnormally steep temperature gradients, just like Earthly ones. This is presumably why NASA used the term mirage in the first place, rather than simply declaring that that is how things usually look on Venus. $\endgroup$ Commented Oct 22, 2022 at 16:17

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