New answers tagged

15

It wouldn't be too apparent, but there are a few objects that you can see in good viewing conditions with the naked eye that would disappear. Here they are in order of brightness. I marked the objects in a reddish color. The Large Magellanic Cloud, apparent magnitude 0.9, located in the constellation Dorado. Only visible from the southern hemisphere. ...


23

I feel sure this is a repeat, but couldn't immediately find it. The only things in the night sky we can see (with the naked eye) that are not part of our own Galaxy are (on a good night) the Andromeda galaxy and the Large and small Magellanic clouds. Every individual star brighter than $V=6$ and visible to the naked eye is in the Milky Way. So it would ...


0

Comoving separations go with the scale factor. Comoving volumes scale as $(1+z)^{-3}$. At $z=3$, that volume was 64 times smaller.


2

The Earth moves in an elliptic orbit around the sun (or around the barycenter). If, in helocentric coordinates the Earth is at position (x,y), then in Geocentric coordinates the position of the sun is in position (-x,-y) So the locus of the Sun in Geocentric coordinates exactly matches the locus of the Earth in Heliocentric. The path of the sun in ...


2

The magnitude scale is a logarithmic scale. An increase of 1 magnitude corresponds to a decrease in brightness of about 2,5 times dimmer. Vega, a bright star has a magnitude of 0, so any star that is brighter than Vega would have a magnitude that is less than 0. This is an odd system; the reason for it is historical. The ancient Greeks ordered stars by ...


2

Apparent magnitude is measure of how bright an object appears to an observer on Earth, meaning it's a function of both the object's intrinsic luminosity and its distance from us. The concept of magnitudes dates back to the Ancient Greeks, when stars in the sky were categorized into six magnitudes (the brightest being 1 and the faintest being 6). Each ...


2

Sounds like homework (correct me if that assumption is wrong, as you may want a somewhat different answer then.), so here is a hint: Clean up the equation first, to something that looks more like an equation for mass: $$\frac{L}{L_{sun}} = \left(\frac{M}{M_{sun}}\right)^{3.5}$$ $$\left(\frac{L}{L_{sun}}\right)^{\frac{1}{3.5}} = \frac{M}{M_{sun}}$$



Top 50 recent answers are included