# Tag Info

## Hot answers tagged size

14

The confusion comes from the difference between the nucleus and the coma. The nucleus is a small icy body, only a few km across. The coma is the cloud of gas and dust released from the nucleus as it warms up. With not much gravity, the coma spreads out into space, and it can be hard to say exactly where the edge of the coma lies, however, a coma "the size ...

4

...despite having more than a quarter the 2-dimensional size of Earth. I think herein lies the problem; diameter is a 1-dimensional measurement, it's units are distance. Let's rewrite 27.3% as 0.273. If that's the ratio of diameters, then the ratio of 2-dimensional areas should be (0.273)2 and the ratio of the volumes should be (0.273)3. Those numbers are 0....

3

It is all just geometry and mathematics. The volume of a sphere is calculated according to this formula: Volume = $(4/3) \times \pi \times r^3$ where $\pi$ = ‎3.14159..., and $r$ is the radius of the sphere. The Earth radius is 6,371 km. The Moon radius is 1,737 km. We put the numbers into the formula and we get: • The volume of Earth is 1,083,206,916,845....

3

The volume isn't well-defined, because the radius isn't. There are practical values for the radius such as scale length/height, half-light radius, $R_{200}$, etc. that are commonly used, depending on what you're interested in. But the densities never reach zero, so any value for the radius will be arbitrary. Moreover, you always only observe a 2D projection ...

3

Your maths looks ok, bar the fact that $1/$parallax is a biased estimate of the distance (but that can be forgiven so long as you are using data where the parallax uncertainties are much smaller than the parallax). Your main problem is that stars do indeed have a vast range of sizes. Thus if you really do want to show the relative sizes of the stars you ...

2

The SDSS exercise shows how to estimate a star's actual radius. If you use this radius, you should also use different model materials for different color index values, since luminosity per unit area is a function of temperature. If you prefer to avoid that complexity, base your model stars' radii on visual magnitude alone. If you put the model stars at a ...

2

I think the main reason the answer doesn't come out correct is that the simple Hubble's law equation: $v = dH_{0}$ Is only applicable in the local universe (Hence $H_{0}$, the value of the Hubble constant today). The rate of expansion of the Universe (the Hubble parameter) is determined by the composition of its energy density. Today, the composition is ...

2

There have been direct measurements, but they are quite delicate -- this is all the interferometry and so on. The simplest way though is based on spectroscopy, brightness and parallax. The argument goes a follows: The spectrum (how much light of each colour it gives off) is not hard to observe -- you basically just need a prism. The spectrum of most stars ...

1

After a bit of searching in Aladin. I found that the other star is V* V2170 Ori. By comparing their respective Simbad pages one can see the following: LL Ori is actually slightly further away. V2170 Ori has a GAIA parallax of 2.84 while LL Ori has one of 2.55. Simply using $D=1/p$, this translates to a distance of 0.35 kpc and 0.39 kpc LL Ori and V2170 Ori ...

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