Post #9by Spaceman Spiff » 10.11.2004, 22:07
Some comments to help, I hope...
You have to distinguish between a 'clean/dry' and 'dirty/wet' atmosphere. A clean, dry atmosphere is composed only of 'air' molecules (from monatomic noble gases like argon, to maybe larger gaseous hydrocarbons like butane). As such, typical particle sizes are only some 10's of nanometres across. A dirty or wet atmosphere would include also aerosols, or water vapour as in clouds and fog, that are condensed droplets of molecules around the micrometre size for fog or smog, up to millimetre size for rain drops.
The vital factor in what scattering effect each has on visible light is the ratio of the scatterer size to the wavelength of light scattered. For visible light, wavelengths are 600nm for red light to 450nm for blue light. As such, we can see the air molecules are smaller than such wavelengths, and vice versa for aerosols.
The scatterer/wavelength ratio is divided into three regimes to note different physical effects:
- < 1 (e.g. air) : Rayleigh scattering. Scattering cross-section increasingly smaller compared to physical cross-section at smaller ratios. That is, light is hardly effected by atomic/molecular scatterers.
- ~ 1 (e.g. fine mist): Mie scattering. Scattering cross-section compares to physical cross-section, but stronger scattering occurs at some precise ratios, weaker scattering at other ratios.
- > 1 (e.g. rain): Optical scattering. Scattering cross-section equals physical cross-section, that's it.
To cut a long explanation short, air and fog have totally different effects. A clean atmosphere at 100 atmospheres would be extremely transparent, the light waves hardly notice the air molecules.
Venus has a very clean atmosphere at the surface, it's the right planet to use for considering the original question. The Venera pictures do show the horizon quite clearly, there is little muting.
It's correct that a denser atmosphere would clear less quickly of dust or aerosols that formed, and that would cause obscuration. For truly dense atmospheres, the gas giants are the places to think of.
Remember people, Venus' surface is not visible in visible light from space, not because of the atmosphere being dense, but because there is a very thick cloud deck composed of sulphuric acid smog between 50km and 100km above the surface. The air underneath is clear, and you would see the cloud deck as a high overcasting canopy. The smog is thought to condense into sulphuric acid rain, but the atmosphere is so hot underneath that the rain evaporates in mid air at 30km up. So, there's no rain obscuration near Venus' surface.
It's the cloud deck that forces us to use radar to image the surface. If it wasn't there, we'd have had visible light pictures with no problem.
On Venus' refraction, a few myths need clearing out. It was put about before the space era that the atmospheric refraction on Venus would be so strong that a person standing on the surface would see it rise up about, such that they'd appear to be at the bottom of a bowl. It's not true. Neither is the converse, that they'd appear to stand on a dome.
The Venera pictures do not show this refraction effect. Unfortunately, the pictures did appear to show each craft sitting on top of a dome, and some people wrongly jumped to the conclusion that the strong refraction just operated in the other direction. In fact the reason for the twist was that each craft often landed tilted, and the camera's panoramic field of view arced below the horizon, giving the 'on top of a dome' illusion. Once corrected, the ground is seen to be largely horizontol out to the horizon.
Now to explain why this refraction is not strongly seen. Refraction does not depend on density of the atmosphere, but on rate of change of density. It's changes in the speed of light within media that does this.
I've mentioned this before as a better way of defining atmospheres for Celestia, but it's really worth understanding about 'scale heights' of atmospheres. Here's some numbers for Earth: Assuming a constant temperature throughout the atmosphere, the pressure and density of Earth's atmosphere drops by 1/e (e=2.71828...) for every 8 km you ascend in it. Ground pressure is 1013mBar or 101.3kPa, so at 8km high, pressure drops to 101.3/2.71828 = 37.3 kPa (373mB), a bit higher than atmospheric pressure at the summit of Mt Everest (330mB at 10km high). See, it works! Also, the reason that the ground pressure is 101.3kPa is because of the weight of the Earth's atmosphere under Earth's gravity. Earth's surface gravity causes a free fall acceleration of 9.81m/s?, or a force of 9.81 N/kg. Turning this round, each square metre has 101,300 N/m? / 9.81 N/kg ? 1m? = 10,300 kg of air above it (a Pascal (Pa) is 1 N/m?). Another way to think about scale height is that it is the height that the atmosphere would have if it was squashed and made to have uniform density from ground to top. Taking that, we can work out the density of air at the Earth's surface, it is: 10,300 kg / ( 8,000 m ? 1 m? ) = ~ 1.3 kg/m?.
Now, it is the change in air density that causes refraction in Earth's atmosphere, including the effects of flattened rising/setting sun and moon.
Scale height is inversely proportional to surface gravity. If the surface gravity is halved, the atmosphere would be relieved of half its weight, pressure would half, and by the laws of an ideal gas (physics), the density would have to halve if the temperature remains the same. To do this, the atmosphere would expand to twice its height, thus scale height is doubled. This is one reason that Titan's atmosphere is so 'high up'. How high? Well, if we take the Earth's atmosphere as 100km high (X prize territory), then that corresponds to a pressure of 101.3 kPa / ( e ^ (100 km / 8 km ) ) = 0.378 Pa. How high is this limit in Titan's atmosphere? First, we adjust the scale height for Titan's lower surface gravity: 1/5 that of Earth, so it's scale height is 5 ? 8 km = 40 km. Then we must remember that surface pressure is 1.5 times Earth. Titan's atmosphere height H must be such that 1.5 ? 101.3 kPa / ( e ^ ( H / 40 km) ) = 0.378 Pa. Solving, H = 40 km ? ln [ ( 1.5 ? 101.3kPa ) / 0.378 Pa ] = 516 km - not far from the 800-900km put about the Celestia forum.
Now, because the scale height is inversely proportional to surface gravity, only surface gravity determines the strength of refraction per unit distance through the atmosphere. In fact, the rate of change of atmosphere density would be inversely proportional to scale height, which means the strength of refraction is proportional to surface gravity. For Venus, surface gravity is slightly less than Earth: 0.9. So, Venus has weaker refraction than Earth.
Then, what total refraction could occur to distort the landscape. Well, because Venus is smaller than Earth ( 12,100 km / 12,756 km = 0.95 ) the horizon is only 0.95 as far away as on Earth. So, the accumulated refraction effect over the landscape to the horizon is smaller still.
Finally, what about the highly flattened sunsets of Venus if the clouds weren't in the way? Well, this is true. Although strength of refraction is less, the sunlight would still have to travel though 35 times more refracting air than Earth. ( I'll skip a detailed explanation that when you account for the hotter Venus atmosphere ( 730K versus 300K ) the density of Venus atmosphere is found by dividing the 89 atmosphere surface pressure by 730 / 300 = 2.43... to get 35 'air masses' ).
Hope this helps.
Spiff.
. So foggy in fact that I realised that I couldn't actually tell where the sun was in the sky - there was no bright patch, the sky was (and still is) pretty much uniform in brightness. I suspect there's some directional light scattering going on that one can see if one's eyes are sensitive enough and looking in the right direction, but mine aren't 