# Tag Info

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The radial velocity $V_r$ is the velocity resolved along the dashed line (the line of sight from the Sun to the star) with respect to the Sun. If $V_c$ is the velocity of the star, then the component along the dashed line is $V_c \cos (\alpha)$. We then have to subtract the velocity of the Sun resolved in the same direction. This is $V_{c,0} \sin (l)$. Hence ...

6

"before gravity stopped holding it together" is the same (pretty much) as "so that apparent gravity at the surface is zero". This means that $$m \omega^2 r = \frac{GM_{moon}m}{r^2}$$ with $G=6.7\cdot 10^{-11}$, $M_{moon}=7.3\cdot 10^{22} kg$, $r_{moon}=1740 km$, we find $$\omega=\sqrt{\frac{GM_{moon}}{r^3}}=0.0092 rev/min = 0.55 rev/hour$$ It is ...

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Luboš Motl addressed the reasoning very well, but for the mathematics: Richard Fitzpatrick's free e-book A Modern Almagest: An Updated Version of Ptolemy’s Model of the Solar System presents a modernized version of Ptolemy's Almagest. The following classic paper shows how epicyclic astronomy can be re-expressed in the modern mathematical idiom of complex ...

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The plane of the earth's orbit is extremely stable. Of course, the earth's orbit is affected by the other planets, especially Jupiter, but all the planets orbit in approximately the same plane, so the forces pulling the earth's orbit out of its plane are small. We can see that the planes of the planets' orbit are stable, because all the planets are in ...

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My answer is the same as Chris', but formulated in a different way (it's essentially the same as this wiki article): In polar coordinates, the position vector is $$\mathbf{r} = r\,(\cos\varphi,\sin\varphi) = r\,\mathbf{\hat{r}},$$ with $\mathbf{\hat{r}}$ the radial unit vector. The velocity is then  \mathbf{v} = \dot{r}\,(\cos\varphi,\sin\varphi) + ...

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The "associated scalar equation" is just the formula for the time evolution of the scalar magnitude of the displacement, $r$, rather than all its vector components. It really only makes sense to write such an equation if the right-hand side can be expressed in terms of $r$ only, and not $\mathbf{r}$. Then you can use it to analyze the evolution of $r$ in ...

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Thermally supported, self-gravitating bodies (those to which the Virial theorem applies) qualify if you are willing to neglect the radiative energy loss. Depending on the time-scales that interest you this can be quite a good approximation. Stellar nebulae, brown dwarfs and so on.

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