# Tag Info

5

I spot two pairs of related quantities in your list. Speed (I assume you mean orbital speed, not how fast the planet is spinning about its own axis) and distance. The orbital period and the semi-major axis of the orbit are related by Kepler's third law. For a circular orbit, the semi-major axis is the radius (distance) and the period is trivially related to ...

1

This is just a sidenote to Nijankowski's nice answer: This twinkling of stars caused by atmospheric turbulence was a major problem for the earlier reflecting telescopes when astronomers tried to look deep into the sky. Placing the telescope over mountains solved only a part of the problem. A good solution was brought up in two ways in the 1990s. First, ...

9

Here is a nice answer, taken from http://www.enchantedlearning.com/subjects/astronomy/stars/twinkle.shtml The scientific name for the twinkling of stars is stellar scintillation (or astronomical scintillation). Stars twinkle when we see them from the Earth's surface because we are viewing them through thick layers of turbulent (moving) air in the Earth's ...

10

In most cases you are right, stars are pointlike. They are spread over multiple pixels not because of charge overflow (this can be overcome with shorter exposures and/or better equipment; there really should never be overflow in the image) but because the point spread function (PSF) of the telescope is larger than a single pixel (which is a good thing; you ...

2

Yes, though the only star that appears as a disk rather than a point is Betelgeuse. Even though it's 520 light years away, about 100 times farther than the closest star, it can just be resolved because it's a red supergiant and absolutely enormous!

1

I think no. The diameter of Riegel Kentaurus A has an average diameter of $1.4 \cdot 10^6\,\mathrm{km}$. Say it is only four lightyears away. Than the viewing angle would be $3.6995 \cdot 10^{-8}\,\mathrm{rad}$. If you want a picture with some 100 Pixels of that star, you will need a resolution of $10^{-11}\,\mathrm{rad}$. I think $\mathrm{\mu rad}$ is ...

5

Yes, as this link (suggested by Will) shows. They're expected to basically always be stars that used to be in galaxies (they formed there), but have gotten kicked out somehow. You won't really see many stars forming between galaxies because there just isn't enough matter there to collect into stars. Even the ones you do find will be in messy regions ...

1

The crucial concept behind the formation of planar orbital systems like spiral galaxies and our solar system is that the transition from a vague cloud with some orbital motion to a planar system dissipates some of the kinetic energy in the orbital motion (which it can, by turning it into heat) but cannot get rid of the angular momentum. As a cloud of gas ...

8

The color of a star depends on its mass and temperature. The distribution of these also depends on the age of the galaxy. When the galaxy is very young, there are large amounts of gas still available for star formation, and many young, heavy, hot stars mean that the galaxy is very bright and shines in bluer light. This can be seen as an analogy of the quick ...

5

It is certainly possible. The human eye will be able to resolve objects in starlight sky and modern sensors are more sensitive than the human eye (permitting single photon detection even). The illuminance will be dominated by a phenomenon known as airglow, an atmospheric illumination that exists even in the absence of light pollution If you have a ...

2

The apparent magnitude of the sun is -26.75, so 2.512^26.75 = 50,000,000,000 times brighter than the brightest star. There are around 50 stars with a brightness within 1 magnitude (ie 50%) of the brightest star so roughly 1/billion-th the light from the sun for starlight.

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