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Dec 26, 2014 · The same applies to the beams of light above them. The Sun is very far away and the beams are pretty much parallel, but they're pointing towards you, and perspective makes them appear to converge towards the vanishing point - which in this case is the Sun's location in the sky. The technical term for these beams is "crepuscular rays."
If you want your eyes to receive 32000~100000lux, you need to stare at the sun in a sunny day.If you stare at the ground in a sun-exposure area, your eyes receives much less than 32000 lux. If you stare at a shedded ground in a sunny day, your eyes receive slight less than 10000lux. human eyes could not tolerate 32000lux more than 5 secounds.
The Sun has actually set/risen and we see it due to the way light is bent across the atmosphere. Apparently due to coincidence of the size and distance of the sun, its exactly the same size - so if we see 50% of the sun, the sun is 50% below the horizon. So, I understand all this, so here is my question :
Mar 13, 2017 · The sun puts out about 8% of its energy in UV (which does the damage), about 44% in visible, and the rest in IR. A standard incandescent puts out effectively no UV, 10% visible and the rest in IR. Halogen lamps can be operated at higher temperatures with a reasonable lifetime, and produce some UV, with perhaps 15% visible.
May 31, 2015 · Stefan-Boltzmann: J = σT4 J = σ T 4. To lose 1 kW over 1 m 2 requires a temperature of. T = 1000 5.67 ⋅10−8− −−−−−−−−√4 ≈ 364 K T = 1000 5.67 ⋅ 10 − 8 4 ≈ 364 K. This assumes only the surface facing the sun loses heat by radiation: in other words this is only valid for a black surface mounted on a good insulator.
Feb 15, 2011 · Another way of calculating the earth - sun distance is to look at the centrifugal and the gravitational force. This solution assumes that one already knows the mass of the sun, but thats a different problem ;-).
Mar 20, 2019 · The sun is considered a black body, not only from the perspective of the Earth but from any perspective. The sun's intensity spectrum is roughly the one of a perfect black body minus absorption by the sun's atmosphere and the Earth's atmosphere depending of where we measure the spectrum. See the sunlight article on wikipedia.
May 22, 2018 · So, now you need to simply calculate the total power passing through a sphere whose radius is the same of the Earth's orbit, and this is. P = 4πR2F ≈ 3.8 ×1026W = 3.8 ×1026J/s P = 4 π R 2 F ≈ 3.8 × 10 26 W = 3.8 × 10 26 J / s. Since energy is conserved this number is the amount of power passing through any surface surrounding the Sun ...
But in Gerthsen Kneser Vogel there is an exercise where Sherlock Holmes estimated the temperature of the sun only knowing the root of the fraction of D and R. Lets say, he estimated this fraction to 225, so the square root is about 15, how does he come to 6000 K ?
Dec 4, 2012 · The light from the sun though spreads out with the area of the sphere, so the square of the distance. This means that at voyager it is only 1/ (122*122) = 0.007% as bright. Thats about the same as moonlight compared to daylight. All the other planets in the solar system are a lot closer to us than Voyager. Even the gas giants like Jupiter (5au ...
