# How distant is the horizon on Venus?

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.")

• 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. Commented Oct 14, 2022 at 20:40
• 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. Commented Oct 15, 2022 at 14:20