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5

No, in a vacuum light sources will appear dimmer as you move further from them because of the inverse square law. In a medium, a light source suffers from both the inverse square law and absorption/scattering. Below is a diagram illustrating the inverse square law: As you move further away from a light source, your pupil (assume it remains the same size, ...

1

Phase-contrast imaging with x rays has been achieved at the LCLS. This is used to probe extreme matter states, such as the ones created by a shock launched by high-intensity laser-pulses incident on metal slabs. See this or this for instance.

0

Well just what do you mean by "Brightness." There is no such unit of measurement in the field of Photometry. A more appropriate and meaningful term is "Luminance" measured in Lumens per steradian, per square meter, or alternatively, in Candela per square meter; since the candela is one lumen per steradian. And Luminance is a property of a light source; so ...

5

It's because the amount of area "covered" increases as the square of the distance. Imagine a sphere, centered on the source, at a radius equal to your eyeball's location. If the source generates X watts (or whatever unit you like) total, the brightness, i.e. the percent of light which hits your eyeball, is X divided by the ratio of your eyball's area to ...

0

It doesn't make sense that satellites cannot detect radiation on the ground. They can detect visible light which is far less penetrating than gamma radiation from a decaying radioisotope. Remember, we are talking about MASSIVE amounts of radioactive isotopes in the atmosphere. They have fingerprints, I'm confident, which can be detected from space. Also, ...

3

If you look carefully at the figure, you will see that $r$ is the length of the hypotenuse (point $P$ to the center of the source), not the length of point $P$ to the tangent. Thus, we have that $\sin^{-1}R/r\equiv\theta_c$.

4

The Earth receives approximately $6.8\text{mW/m}^2$ of reflected sunlight from the moon (see below for details of how I calculated that). However, the sunlight is also absorbed by the moon and this raise the surface temperature. So the moon also emits thermal radiation towards the Earth (assuming the highest day time temperature of 400K, see comments below ...

2

Assume the moon is at roughly the same distance from the Sun as the Earth and so receives the same amount of solar energy / area. Find the area of the moon facing the Earth (hint, it's roughly the area of the moon's disk). Multiply by the reflectivity of the moon (about 12%). But that power is reflected from the moon in all directions, so you need to ...

8

How do I stay alive to be killed by neutrinos? You wouldn't. The point is being made that even the beam of neutrinos with a supernova at one astronomical unit distance would be intense enough that enough of them would interact with the matter of your body to be lethal. So even the neutrinos would get you if all the other stuff - notably $\gamma$s didn't. ...

0

I don't know if you have already solved the problem, but the same think happened to me when I ran the experiment. It turned out that whenever using a 90Sr source, you actually have a 90Sr/90Y radioactive source and they both undergo beta decay, but with different energies (2.28 MeV and 0.546 MeV). So for short distances you have the sum of the two spectra ...

0

The charge accelerated by the Earth gravity does not emit any radiation,follows from transforming to a frame of reference in which the charge is stationary and applying relativistic requirement that the behavior of the charge including whether or not it radiates,cannot depend on the frame of reference from which it is viewed.

8

In fact, an electric charge at rest on the Earth's surface is accelerated and this actually poses a challenge to the idea that uniformly accelerated charge radiates. I believe this is still an open question. For example: One of the most familiar propositions of elementary classical electrodynamics is that "an accelerating charge radiates". In fact, ...

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