# Tag Info

0

According to the wiki entry, the error runs up to roughly 15 minutes. One can use the "equation of time" , for which there appear to be published tables, to generate a correction curve. (The main contributors to the error are eccentricity and obliquity of Earth's orbit.

5

In theory, perhaps. It is possible, using multilayer dielectric coatings, to produce a surface which is reflective in very narrow bands (in this case, the Sun's dark lines)and transmissive (or absorptive) elsewhere. In practice, the spectral "blurring" caused by atmospheric transmission/absorption/re-emission effects would make this effect pretty much ...

3

The sun's spectrum is very complex, and indeed there are a lot of "lines" both light and dark (emission and absorption) amidst a sea of what looks to be continuous frequencies. Note that the atoms you study in a textbook are idealizations. In a hot object such as the sun, some photons come to us by way of atomic emissions, but the speeds of the atoms that ...

-3

Certain wavelengths of electromagnetic spectrum is emitted when the electrons in an atom to move from a higher level to a lower. The wavelength that is emitted depends upon the number of shells the electrons move down. When an electron/electrons move into certain number shells, white light is emitted i.e. when two hydrogen atoms fuse into an Helium atom, ...

1

As pointed out in the comments, the method is correct aside from a careless error of using the diameter, when radius was clearly intended to be used by the argument. I calculate the "area" of the earth, as seen from the sun, and then divide that by the surface area of a sphere of radius 1 AU, to get the portion of the sun's rays we absorb, and then I ...

3

The area of a circle is calculated using its radius instead of its diameter: $$A = \pi r^2 = \pi \left(\frac{d}{2}\right)^2 = \pi \frac{d^2}{4}$$ which is where your missing factor of 4 reappears.

1

Let's go about this from a different angle and compare answers. The radius of the earth is about 6.4 Mm, so the area of its disk is 130 x 1012 m2. Figure about 1.2 kW/m2 of incident sunlight power at earth's distance, so that yields 1.5 x 1017 Watts. That's close enough for such a quick back of the envelope calculation to the 2 x 1017 value you show that ...

3

From your point of view it would be instant. Sun is there, then poof! The sun is gone and so is its influence. We only know this event happened ~8.5 minutes before we saw it because we're so clever :-). However there's no way to detect it early and warn ourselves because no information could reach us any faster than the sun's extinguished light and ...

2

Gravitational waves travels at the speed of light, thus you would feel it at the same exact moment you saw the sun disappear. By general relativity, spacetime acts like a trampoline being bent by a central mass. When the mass is removed the trampoline does not go back to the unbent state instantaneously.

0

The temperature of our sun is determined primarily by the generation of heat (the fusion process) and the rate of heat loss by various means. When the process reaches equilibrium, a given average temperature is maintained. Any other process that has a higher generation rate or smaller energy loss, or both, would be hotter than our sun. In the case of ...

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