New answers tagged

-1

If you have a ray from an origin point $\mathbf{x}_0$ with unit direction $|\mathbf{v}|=1$, $\mathbf{x}(s)=[x(s),y(s),z(s)]=\mathbf{x}_0+\mathbf{v}s$ and look at the density times distance $\rho(s)=\rho_0 e^{-R(\mathbf{x})/h_R}e^{-z(\mathbf{x})/h_z}s^2$ the derivative is $\rho'(s) =\rho_0\left\{ [-e^{-R(\mathbf{x})/h_R}R'(\mathbf{x})/h_R]e^{-z(\mathbf{x})/...


2

Not yet, as far as I can tell. The comet looks very different from Mars and while I haven't done the math, I don't think it has an apparent size from the rovers' perspectives to make it look like anything more than a dot. Since Neowise is moving away from the Sun, the length and brightness of its tail is diminishing rapidly.


2

This is an incomplete answer—a placeholder for a more complete answer that I'm adding here at the request of the OP (who, I will say, has been very patient with me!). Although the precise reasons aren't well understood, I don't think people find the observed brightening of Neptune to be especially anomalous. We have accurate photometry for Neptune for only ...


2

The introduction of the Wyrzykowski & Mandel paper gives the following information about estimating the lens mass. In order to obtain the mass of the lens (Gould 2000a), it is necessary to measure both the angular Einstein radius of the lens ($\theta_\mathrm{E}$) and the microlensing parallax ($\pi_\mathrm{E}$) $$M = \frac{\theta_\mathrm{E}}{\kappa \pi_\...


3

Allow me to coin a phrase. Phonygraphic. Without getting into specific details, the first two images are the result of camera tricks and image processing. The two photos 25.07.2020 22:32 and 24.07.2020 22:00 are in keeping with what I could see of the comet with my unaided eyes from a dark site between July 15 and July 23 2020. By the twenty third the comet ...


7

There's one main reason - and maybe some other reasons as well: The primary reason is the date the photos were taken you compare your photo with. Two weeks ago (10th July 2020), the brightness of the comet was around 1 - 2 magnitudes - that is comparable to the brighter stars and it was readily visible with the unaided eye on onset of nautical twilight. A ...


5

As @ELNJ answer pointed out, fully ionized the atoms at the star surface are not. It is not hot enough. Star cores are another case, but we usually don't see them. There, both pressure and temperature make impossible the existence of the usual atoms. Atoms and molecules usually emit their characteristic wavelengths because of the electrons' energy levels... ...


7

You are correct that the characteristic emission and absorption lines we see in stars' spectra are from electrons that are bound to atoms making transitions between different energy levels. That is possible because the elements in a star's photosphere are not fully ionized. Hydrogen - the easiest element to fully ionize because its nucleus only has a ...


4

You cannot have free protons without electrons. Plasmas, in general, are electrically neutral. It is usually electrons that dominate the scattering (note that a point-like charge cannot absorb a photon and conserve energy and momentum) in a plasma at low photon energies. That is basically due to their much lower masses (classically you can think of the ...


2

I suspect that that is simply a statement about where you would have the best visibility of each meteor shower. The name of each shower tells you where the radiant is, the point in the sky from which the trails of the meteors appear to originate. Perseus in the northern sky (with the specific radiant point of the Perseids at +58 degrees declination), ...


3

I agree with @JamesK's answer that it doesn't really matter. The direction of longitude = 0 is the same in both cases, and it's really a matter of taste if you display in a plot or table 0 to $2 \pi$ or $-\pi$ to $+ \pi$ (0 to 360 degrees or -180 to plus 180 degrees). For example in Python when I type np.arctan2(-5, 5) I get $-\pi/2$ (-90 degrees) but if I ...


5

It is purely a matter of convention. Longitude and latitude are normally expressed in degrees, longitude from 0 to 360⁰ and Latitude from -90 to 90⁰. If you express these in radians, longitude would be from 0 to $2\pi$ and latitude from $-\frac\pi2$ to $\frac\pi2$. If you are writing software, your functions should be robust enough to accept any value out ...


1

tl;dr nothing is oscillating now, but it was in the early universe. Now we just see the frozen relics This is just an attempt at a simplified explanation of the Wikipedia article linked in the question. There were pressure waves in the very early universe. They came from a balance of gravity and radiation pressure (the matter at the time was hot enough to be ...


0

The verbal description in the NYT article is correct but not precise enough to locate the comet in practice. Here is a finder chart generated by Stellarium for 90 minutes after sunset at New York City and similar latitudes. The Big Dipper is at the top, just to the right of center. The cyan dots indicate the comet's position at 21:50 EDT from July 22 to 31. ...


5

Horizon The horizon is Earth's horizon ... absent any obstructions such as buildings, trees, etc. Or another way to think of it is the horizon you would observe looking out across the ocean while standing on a beach. The notion that it is "10° above the horizon" means that from the horizon line, it would be 10° up. You closed fist (as if you are ...


1

I've been watching the comet for the last few nights from 50+ deg North, around midnight. A simple way to find it is to imagine the comet has been poured out of the end of the Big Dipper and is falling headfirst toward the horizon. It wouldn't hurt to imagine a strong wind has blown from behind the Dipper's handle and over cup to push the comet a little ...


1

The comet will be roughly 10 degrees above the Earth's horizon. But the horizon is a pretty vague indication as to where to look. To know what direction on the horizon to look towards, find the big dipper. The comet will be between the horizon and the big dipper.


1

The article just refers to the earth's horizon. Neowise can be seen about 10 degrees above where the sky meets the earth.


4

tl;dr No additional data of interest, but I explain how I searched, and I can explain the green color. It seems that there have not been any additional observations of that object, which perhaps isn’t too surprising given how faint it is - 21st magnitude is possible only with big telescopes. Simbad lists that object under the name of “EQ J103712-274051” but ...


5

tl;dr they're a bit too small for SOHO and Stereo and only visible in EUV The "campfires" are described as a few hundred km across, The smallest of those campfires are about the size of a European country, according to Berghmans. which means that from Earth (or SOHO or Stereo) they will subtend from one to a few seconds of arc (a few micro-...


20

For that specific E-mode map we have applied a Wiener filter to highlight the high SN modes (those "rings"). I also further apply the following filter: $((1 + (kx/5)^{-4})^{-1}) * ((1 + (k/150)^{-4})^{-1})$. This second filter gives the "hole" and a "thin" vertical line in your 2D PS. The image above is just for PR purposes. In ...


22

Having now looked at the paper by Aiola et al. (2020), it emerges that for that map, they filtered the data to exclude low frequency multipoles with $|l|<150$, corresponding to about 1 degree. This filtering was done to all the maps in the paper and will be responsible for the dramatic "hole" in your Fourier transform. As for the high frequency ...


4

For both conversions, you need a bit more information about the data. How can we switch from Jy/pixel to MJy/sr? The Jy -> MJy part is just a factor of $10^{6}$ Jy/MJy. The pixel -> sr (steradian) part requires that you know the angular size of a pixel on the sky. For PACS it looks like the pixel size is different for different wavelengths; see ...


-4

Of course; how would we find it otherwise? This is how we look for it: image one part of the sky with a telescope, about where the planet is presumed. At another time we image the same part again and see whether a star moved or is no longer there. This star is the wandering star = planet we're looking for. The hypothetical object has a very elliptical orbit, ...


5

Yes, radio spectra have been used extensively to find the distances and locations of HI regions and molecular clouds inside the Milky Way. Observations of the 21 cm hydrogen line and/or several carbon monoxide lines (in particular, $\text{CO}(1\to0)$) enable us to make radial velocity measurements of clouds within the galaxy. From there, some geometry (see ...


5

I assume that the diagram indicates what the observer sees (if they had a big enough telescope!). i.e. The viewpoint is nearly in the orbital plane but not quite. Why then are the eclipses asymmetric, with the secondary eclipse being shallower than the primary? Well probably because the surface brightness of the two stars is different - i.e. they have ...


1

Your intuition is correct - a moving source emitting wavefronts periodically will be closer to the previously emitted wave in the direction of motion, and farther from the previously emitted wave in the opposite direction - see the simulation here. You are also correct that the size of the effect depends on the speed of the observer relative to the speed of ...


4

Adding to the other excellent answers here, if there were an unobserved "Planet 9" the possibility of it being a difficult to observe primordial black hole has been raised. According to Phys.org's Scientists propose plan to determine if Planet Nine is a primordial black hole if Planet 9 were a primordial black hole it could be detected by the Vera ...


1

The point of the paper is not that no calibration has been used previously. It points out that they now use a more refined method and set of parameters to calibrate the data. That in general is usual procedure: the more data you gather, and the longer a mission runs, the better you understand your instrumentation and all external factors which influence the ...


30

C means "comet": it has a coma which means that volatiles are being released due to solar heating. Other possible letters are "A" for asteroid, "P" for (short) periodic comet, "D" for disappeared comet, "I" for interstellar object and "SN" for supernova. 2020 is the year of discovery F represent ...


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