New answers tagged

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According to Mathematica, using Entity["Planet","Earth"]["HelioCoordinates"] and Entity["Planet","Mars"]["HelioCoordinates"], the current Earth-Sun-Mars angle is 4.35 degrees. But this does not assume that they are in coplanar orbits; it uses their actual non-coplanar orbits. If you want to impose ...


0

It looks overly large because its a closeup. The round shape of the photo makes you think you are seeing all of Jupiter but it is not. Its more like a view through a lens or fisheye, its rather misleading.


1

I did not read this book but I believe it will definitely suit you Extragalactic Astronomy and Cosmology: An Introduction by Peter Schneider Here is the first edition https://www.amazon.com/Extragalactic-Astronomy-Cosmology-Peter-Schneider/dp/3642069711 Here is the link for the second edition https://www.amazon.com/Extragalactic-Astronomy-Cosmology-Peter-...


0

comoving distance is not something that we can measure. We are actually measuring either luminosity distance or angular diameter distance. And they are related to the comoving and proper distances. By making observations we can determine the luminosity distance of an object $d_L$. And the $d_L$ is related to the transvverse comoving distance $r$ via $$d_L = ...


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I will tack on some background to benrg's excellent answer. The sun being in the classification known as "yellow dwarf" is an (almost intentional) misnomer. When the stellar classifications were being named, it was already well known that the sun and stars of similar temperature were white, but a name was needed to distinguish main-sequence stars ...


4

I think that this image is a fairly accurate rendition of blackbody colors, if you view it on a correctly calibrated monitor: You'll notice that there isn't any yellow in it. So I don't think that there actually are any yellow stars, traditional astronomical diagrams notwithstanding (though of course the spectra of stars are only approximate blackbodies). ...


0

The appearance of things to the human eye is almost always "wrong" in a physical sense. The human brain does a mighty amount of signal processing, and one of the things it does is "white balance" the scene. Our sense of color is constantly being adjusted by our brains. "Broad daylight" happens to be a situation our brains ...


-1

Never look directly into the sun but if you would it should appear white because of its high intensity. Also, sunlight may peak at yellow but contains a lot of red, green, and blue as well.


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I asked this question at a colloquium about the system. I think you understand it better than me but the answer involved atomic clocks at each site and satellite gps and synchronization


0

As G. Smith said, there are no exact solutions to GR describing orbiting bodies. That aside, the center of the orbit is absolute in general relativity in the sense that, for instance, the Doppler shift and aberration of distant objects shows a yearly variation consistent with circular/elliptical motion. Of course, you can always pick a coordinate system that ...


0

There isn’t even an exact solution for the two-body problem in General Relativity, much less for the $n$-body problem, even if you take the center of mass as the origin. The $n$-body problem doesn’t have an exact solution in Newtonian gravity.


1

It is easier to work in solid angle. If you have 5077 stars covering $4\pi$ steradian of solid angle, then the density of stars is 404 per steradian. The solid angle subtended by your picture frame of area $A$ would be $A/d^2$ if $d$, the distance to the frame were much larger than the frame dimensions. The more accurate formula is $$\Omega = 4 \arctan \left(...


1

The surface area of a sphere with a radius of $2$ metres is $4 \pi r^2 = 16 \pi \approx 50$ square metres. So a $1$ square metre frame held at a distance of $2$ metres covers about $\frac 1 {50}$ of the whole sphere. From this you can work out the number of square degrees covered by the $1$ square metre frame, and the maximum number of stars that you can ...


2

In the third century BCE Aristarchus of Samos estimated the ratio of the distance of the sun to the distance of the moon by observing the angle between the sun and the moon when the moon was exactly half full.


3

There is an article here that explains it where the normal "velocity $\times$ time" (velocity here is the speed of light which is multiplied by the time taken for the pulse to travel from the earth to the craft) calculations of distance are used but with more technicality. This is a snippet: "Voyager 1’s communication system has a high gain ...


0

A simple analogy based on the same physics. Have you ever bounced pebbles on water? If not see this video . Here is an explanation for the pebble, from which the angle and speed is the basic for the skipping meteorites. At high speeds and small angles the effective density of matter the meteorite scatters off is high enough to be like the water surface ...


0

An object, initially in an eliptical orbit, has a radial (due to gravity) acceleration. If there was no atmosphere it would continue following that orbit, unless if it intercept the earth surface. When the object enters in the atmosphere, the acceleration changes due to drag and bouyance. $\mathbf a = \mathbf g + \mathbf a_d + \mathbf a_b$ $\mathbf a_b = -\...


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