New answers tagged

0

Since the effect on space-time curvature (gravity) of the Sun propagates through space at the speed of light, a observer beyond the Suns Cosmological horizon, or it's age in light years away, will never be able to feel it.


2

Yes this is a bit more of a challenge! The first scope in the photo is long and thin and before the big square cardboard shadow-maker is put on it can be easily pointed at the Sun by moving the telescope until its shadow becomes a small circle, the minimum outline of the scope. If it's off a little bit that long tube's shadow will immediately change shape. ...


-1

Take a simple case where we know the mass of two solar systems (M_1 and M_2) and the distance between their centers of gravity (x). We want to find the location between them where the two forces of gravity from each system cancel out. Where an object placed on one or the other side of that point would eventually fall into one or the other the star system. ...


3

Here's a partial answer: As mentioned in the comments, the transmission of the atmosphere depends quite a lot on local factors. But given a transmission $T_0(\lambda)$ at zenith, the transmission at an angle $\theta$ from zenith can be written $$ T(\lambda,\theta) = T_0(\lambda)\times X(\theta), $$ where $X$ is the air mass. The air mass can be approximated ...


4

If the Sun collided with another star about the same mass, then its mass would be slightly less than 2 solar masses, as some material would be ejected away from the merger. This would result in an A-type star, as the merger's mass is about 2 solar masses. A good example of a 2 solar mass star is Fomalhaut A, which is an A3V star. Therefore, this merger ...


1

First see the plethora of diverse answers to Where can I find the positions of the planets, stars, moons, artificial satellites, etc. and visualize them? Then if you can use Python or would like to learn, consider using Skyfield for everything! Also have a look at Astronomical Algorithms in this very long answer: This is something you'll have to dig into a ...


1

An approximation of the Sun’s declination for a certain date is: sin δ = 0.39795 ⋅ cos [ 0.98563 ⋅ ( N – 173 ) ] where N is the number of days since January 1. That, combined with the formula given by @user21, viz.: $ a_{min} = \lvert \phi + \delta \rvert - 90° $ will give you the dates where the Sun is at least 18° under the horizon. (Oddly enough, a simple ...


3

Many questions can be answered using the vis-viva equation: $$v^2 = GM\left(\frac{2}{r} - \frac{1}{a} \right)$$ which gives the velocity of an object in a Keplerian orbit at distance $r$ from a body of mass $M$ and with a semi-major axis $a$. $G$ is the gravitational constant. And for convenience and accuracy, the product $GM$ or standard gravitational ...


4

How much mass would have to be added to the Sun to significantly alter its characteristics Asking how much would be significant is inexact. The Sun is classified as a G2V main sequence star. Though the chart lists one solar mass as G4V, so there's some variation in there. The classifications seem to relate to temperature. To go 1 step up (and using ...


3

In principle this sundial is designed be used this way: Use the built-in compass to align the XII hour mark with geographic north. Look up the magnetic declination for your area to see how this differs from magnetic north. Use the hinge below the XII mark to tilt the dial for your latitude as indicated on the vertical protractor. The gnomon's angle above ...


2

The "stick" that casts a shadow is called a "gnomon" The gnomon should point at the celestial North pole (and so is angled at about 45 degrees for most people in temperate climates, or more precisely it is angled by your local Latitude. So if you are Scotland at 55 degrees North, you should angle the gnomon at 55 degrees to the ...


3

The point here is the "the solar spectrum" received from "the Sun" actually comes from different places depending on the frequency. The (stylised) smooth curve at optical and IR frequencies is pseudo-blackbody radiation from the photosphere. Pseudo-blackbody, because the radiation comes from different depths and temperatures in the ...


13

There are other ways of getting emissions than just direct thermal radiation. Most of it happens through plasma interactions in the solar corona and atmosphere than in the chromosphere. This review paper names bremsstrahlung, gyroresonance, cyclotronmaser, and plasma radiation as sources, each with their own brightness temperature way above 6000 K. (See also ...


2

Well, I did some digging and found a helpful chart here. The image itself is located at this link. Wikipedia defines the S band as the section of the electromagnetic spectrum from 2 to 4 GHz. To interpret the respective values the graph displays at these frequencies, we need to convert these two figures to wavelengths: $\lambda_1 = \dfrac{c}{\nu} = \dfrac{...


3

The picture is a mocked-up fake and is not an actual picture of the solar spectrum. You can easily see this because the black "Fraunhofer lines" extend beyond the spectrum and H alpha should have an appreciable width. The table is massively incomplete. It list only a tiny fraction (the strongest) absorption lines in the solar spectrum. There are ...


5

Low mass stars like the Sun do become very large prior to He ignition in the core. The exact value depends a bit on models for mass loss from the extended atmosphere (e.g. Guo et al. 2016), but estimates of 250 times its current size are possible (Schroeder & Smith 2008; Spiegel & Madhusudhan 2012). At this radius both Mercury and Venus are engulfed. ...


10

This answer is a supplement to the existing answers. I looked around for nice graphs showing the sunrise azimuth over the year for various latitudes, but I couldn't find anything suitable. So I just wrote a couple of small Python scripts, using Sage / Matplotlib to do the plotting. Sunrise azimuths for various latitudes Sunrise times for various latitudes ...


7

The Sun does indeed drift across the sky throughout the year, not only rising higher in the summer and lower in the winter, but also varying along an east-west axis. This can be shown by observing the Sun at the same time each day throughout the year, and seeing that it changes position. This shape is called an analemma, and is a result of the earth's axial ...


21

The Sun rises and sets at a different point on the horizon every day. The change is small, so without careful observations, it may take several days or weeks to be fully aware of the change. Mathematically, the position of rising/setting can be found from the following formula: $$\cos(\theta) = -\frac{\sin(declination)}{\cos(latitude)}$$ where $\theta$ is ...


3

I hope someone can come along and make this more precise, but I'm pretty sure the effect is greater the further you move away from the equator. So it would be more noticeable in Canberra (35° S) than in Darwin (12° S). Do you know the latitudes of the previous places you lived?


17

You're not silly1, it certainly swings back and forth (North and South) one full cycle every year. It's directly related to why days are longer in the summer and shorter in the winter. I am no expert, but some say that Stonehenge and other ancient "observatories" are supposedly set up to do exactly what it is you do, except much more carefully and ...


Top 50 recent answers are included