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-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})/...


1

Mass extinctions have been linked with the Solar System's oscillations up and down through the galactic plane (the Galactic Cycle). You could also take a look at this question.


6

To simplify, I used Starry Night to generate a visualization. The yellow arc is a piece of the ecliptic. You can see (in faint gray) the center of the yellow arc is labeled 'Dec'. The Sun will be in this part of the sky in December. You can also see planets of the inner solar system and in the background, part of Sagittarius (and the Teapot asterism) is ...


16

No, although there are times when it can't be seen, it isn't true that it is visible for 183 days of the year. The general question could be "If I take an arbitrary location on the sky (say a randomly chosen star) will it be visible for half the year? The answer is "it depends on the star!" Polaris and many other stars are circumpolar for ...


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