Stan Liou
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 2d comment On a log-log plot of surface gravity to planet mass, what is the meaning of the y-intercept? That $4k_2$ should be $2k_2$ in my comment (typo). 2d comment On a log-log plot of surface gravity to planet mass, what is the meaning of the y-intercept? @RobJeffries The OP is far-extrapolating gas giants, so there $\rho$ is also strongly dependent on $M$--but that shouldn't change the correspondence to mean density for any particular fixed $M$, which was actually OP's question as stated. Really, though, the OP's extrapolation is physically inappropriate, and for light exoplanets ($M\lesssim 4$ in units of $R_\oplus = M_\oplus = 1$), Seager et al. (2007) $$\renewcommand{\lg}{\log_{10}} \lg R = k_1 + \frac{1}{3}\lg{M}-k_2M^{k_3}\text{,}$$ and so $\lg g = \frac{1}{3}\lg M - 2k_1 + 4k_2M^{k_3}$. 2d comment On a log-log plot of surface gravity to planet mass, what is the meaning of the y-intercept? Shouldn't that be $g = (\tfrac{2}{3}\sqrt[3]{6\pi^2})(G\sqrt[3]{\rho^2M})$? Note that the OP's question is actually mostly about $\log M\to 0$, but otherwise, I agree, the most straightforward physical thing it corresponds to is mean density, fit to the data. May1 reviewed Close Faster than light May1 comment Faster than light @slw_ Mass does not increase due to speed. There is a concept of 'relativistic mass' that's E/c^2, but that's different from 'mass' and not used in modern presentations of relativity. More generally, $m^2c^4 = E^2 - p^2c^2$, where $p$ is momentum, and $m$ is mass. Thus, $E = mc^2$ is only valid in the object's rest frame, i.e. at zero speed. May1 comment Faster than light I'm not sure what you mean by 'you could calculate that the ship exceeds the speed of light'. ... By the way, a ship starting at rest with uniform acceleration $g$ for $\tau$ of ship time will have velocity $v = c\tanh(g\tau/c)$ relative to its original inertial frame; an easy way to think of this is that a uniform acceleration is like a uniform rotation in spacetime. As you say, this is strictly below $c$. May1 comment Faster than light That said, it's really hard for me to understand what you're even trying to ask, and I can see very little connection to astronomy. ... But whatever you intend to ask, I really think you should try to condense your question into something shorter and more specific. May1 comment Faster than light @slw You never "feel" that time has stopped, at least not for reasons of basic physics (psychological reasons are irrelevant here). And no, you're mistaken about looking forward or behind, for various reasons, but immediately because there is no physically distinguished "forward" or "behind". Fundamental physics is the same in every direction. Apr30 comment Faster than light Could you perhaps reduce this to more specific questions? But the short answer is that you are indeed wrong about there being no problem in exceeding the speed of light, although you are right in that there is a sense in which it is misnamed in relativistic physics--there is an invariant speed that's fundamentally part of how spacetime is structured. This speed happens to match the speed of light in vacuum because photons are massless, but conceptually it's not really about light itself. Light being massive would not actually falsify relativity. Apr29 comment Could someone see anything while being inside black hole? crap, you're right. Just considering this problem in reverse from the origin and at escape velocity makes it trivial. My thinking was in a rut formed by cycloidal relations for the finite-apsis case. Apr29 comment Could someone see anything while being inside black hole? ah, thanks for the info. That would be a cute homework exercise for me to do. ;) Apr28 comment Could someone see anything while being inside black hole? +1, but as a side note, I get about $67\,\mathrm{s}$ as an upper bound for the horizon-to-singularity time for Sgr A*-massed Schwarzschild black hole (freefall from rest near the horizon; all other times will be shorter because of higher inward velocity at the horizon, but it's unclear what initial condition would give $26\,\mathrm{s}$). Apr20 awarded Enlightened Apr20 awarded Nice Answer Apr19 comment Why are there no green stars? @HyperLuminal It means that a green blackbody is literally impossible, and since the spectra of stars is that of a blackbody plus some relatively spectral line corrections, green stars should be impossible too. I'm not entirely sure how to quantify a deviation from greenness, though. In any case, you can see from the color space that there's nothing special about green in this way, as you stars actually miss almost all the colors. Apr19 answered Why are there no green stars? Apr17 answered Years, Months, Day, and Weeks? Apr16 answered Does the Sun's light travel fast enough to have a straight path to Earth? Apr15 comment Time according to the gravity of Sagittarius A? For a hypothetical Sgr A*-massed Schwarzschild black hole, the tidal forces across $1.8\,\mathrm{m}$-tall human near the horizon should be on the scale of $10^{-4}\, \mathrm{gee}$ or so. Supermassive black holes don't spaghettify until well past the horizon. The time dilation calculations are misleading because the Schwarzschild radial coordinate does not straightforwardly correspond to a radial distance. For example, if $r_\text{ft} = r_\text{S}+1\,\mathrm{m}$, then $r_\text{hd} = r_\text{ft} + 16\,\mathrm{\mu m}$ for the human. That's one way to think about why the tidal forces are small. Apr12 comment Is the sun too small to self-ignite? Sounds like a somewhat garbled explanation of quantum tunneling through the Coulomb barrier, which is a probabilistic effect that, in this case, allows fusion at lower energies than would be required without tunneling. Regardless, it's plainly not true that the Sun is too small to self-ignite, and it does actually have sufficient gravity.