What is the temperature inside a Black Hole?

Temperature of a black hole is determined by 'black body radiation temperature' of radiation which comes from it (if something is hot enough to give off bright blue light,it is hotter than something that is merely a dim red hot.) For black holes the mass of Sun, radiation emitted from it is so weak and so cool that temperature is only one-millionth of a degree above absolute zero. some black holes are thought to weigh a billion times as much as the Sun and they would be a billion times colder, far colder than what scientists have achieved on Earth.
Even though these things are very cold, they can be surrounded by very hot material. As they pull gas and stars down into their gravity wells, material rubs against itself at a good fraction of speed of light. this heats it up to hundreds of millions of degrees. Radiation from this hot, in-falling material is what high-energy astronomers study.
A black hole behaves as though its horizon has a temperature and that temperature is inversely proportional to the hole's mass: T = (6 x 10^-8)M. Here M is in units of solar mass (2 x 1033 grams). Temperature is in degrees Kelvin.
This means that a hole recently formed by gravitational collapse of a star (which has to have a mass larger than about 2 suns) has a temperature less than 3 x 10^-8 centigrade above absolute zero which is very cold.
Mass with a finite temperature radiated energy. Anything which radiates energy is also losing mass.As black hole loses mass, emission of energy from black hole increases and its temperature increases and thus rate of mass loss increases.As mass of black hole gets small,one has unstable 'runaway effect'. Black hole gets hotter and hotter which causes M to decrease rapidly.When the hole is reduced to a fraction of the size of an atomic nucleus, it will be trillions of degrees. Hole will burn up and disappear. Lifetime of a black hole is greater than age of the universe.

vision - How do we know the brain flips images projected on the retina back around?

The basis of this question is a common misconception, and unfortunately the accepted answer by @CHM is also based on this common misconception. The misconception is based on the homunculus falacy and the tendency for people to think that the image that lands on the retina is somehow 'assembled' and presented for something (the 'consciousness') to view. This is not the case.

As the comment by @mgkrebbs expains, there is no orientation (up or down) in the brain, there is only neural firing. The infromation of the visual scene is distributed over the brain, and information does not have physical properties like orientation. Although as @nico pointed out the neurons that process the information do have a spatial structure that mimics that of the retina, this is a topological property (i.e. simuli that are close on the retina are processes by neurons that are close in V1) and such a topological property does not induce an orientation.

The root of the problem is really that the question "How do we know the brain flips images projected on the retina back around?" is a pseudo-question. Although it is grammatically well-formed, it makes no semantic sense. When the image is 'in the' (i.e. being processed by the) brain it no longer has physical properties like orientation. Thus you cannot ask if it has been flipped or not.

radial velocity fitting of a binary

The radial velocity curve of a star in a binary system (with another star or a planet) is defined through 6 free parameters
$$V_r(t) = Kleft(cos(omega + nu) +e cos omega right) + gamma,$$
where $K$ is the semi-amplitude, $gamma$ is the centre of mass radial velocity, $omega$ is the usual angle defining the argument of the pericentre measured from the ascending node and $nu$ is the true anomlay, which is a function of time, the fiducial time of pericentre passage $tau$, the orbital period $p$ and the eccentricity $e$.

To proceed you estimate what all these parameters are - i.e. an initial guess.

Then, for each time $t_i$ of a data point in your RV curve you:

1. Calculate the mean anomaly
   $$M(t) = frac{2pi}{p}(t - tau),$$




2. Solve "Kepler's equation"
   $$M(t) = E(t) - e sin E(t)$$
   numerically (its a transcendental equation, you could use Newton-Raphson or similar) to give $M(t_i)$, the eccentric anomaly.




3. Use
   $$tan frac{E(t)}{2} = left(frac{1+e}{1-e}right)^{-1/2} tan frac{nu(t)}{2}$$
   to calculate the true anomaly $nu(t_i)$.




4. Calculate $V_r(t_i)$

You then calculate some figure of merit (e.g. chi-squared) for how closely the model and data agree and go through an iterative process to adjust the parameters and optimise the fit of model to data.

A more sophisticated discussion can be found in this paper by Beauge et al.

If you have the RV curves of both stars, then you can fit them both simultaneously. Obviously, they have $p$, $e$, $gamma$ and $omega$ in common, but their RV amplitudes $K_1$ and $K_2$ will be different. The ratio of $K_1/K_2$ gives you the ratio of the two stellar masses.

If you only have one RV curve you are limited to estimating the mass function of the binary system.
$$ frac{M_2^{3} sin^{3} i}{(M_1 + M_2)^2} = frac{p K_1^{3}}{2pi G},$$
where $i$ is the inclination of the orbit with respect to the line of sight.
This can only give you a lower limit to $M_2$ unless $i$ is known.

Taking your specific case study. If you know $M_1$ and $i$ (this could be the case for a transiting exoplanet, or maybe a binary featuring an eclipsed black hole candidate), then the primary radial velocity curve gives you $K_1$ and hence $M_2$. If the masses and $p$ are known then Kepler's laws give the orbital separation.

There are a number of options if you want an off-the-shelf solution to fitting RV curves. Perhaps the best free one is Systemic Console.

There is no fundamental difference between analysing the RV curves of stars with exoplanets and stars with unseen (stellar) companions.

Plausible? Brown Dwarfs are rogue celestial bodies because they absorb all light due to their thick layer of matte black soot

They are so dark they absorb almost every bit of light from any star and receive no photon velocity momentum or electromagnetic radiation.

The momentum that a planet receives or does not receive from photons is negligible for its orbit.

These dark masses are extremely high in mass and do not interact with light like other planets and moons.

A dark object very much interacts with light: it absorbs light. An object that does not interact with light would be transparent, not dark.

I believe different levels of light interaction on a mass causes planets to move away from stars or stay at bay.

Planetary orbits are governed by gravitational effects. The effect of light interaction on orbits is negligible.

Finally, none of this has anything to do with planets in orbit or not. Planetary orbits are due to gravitation.

Only another collision from a celestial body can interfer with the direction they are heading they outweigh most of the competition so distraction is rare.

Indeed, the only force massive enough to alter the orbit of a planet would be another planet, either hitting it full force, or passing closely thus altering its angular momentum. You'd need an awful lot of this to cause a planet to reach a solar system escape trajectory, however!

And it has nothing to do with colour or photon interaction. It's all about mass.

observation - How can apparent magnitude be negative?

Apparent magnitude is measure of how bright an object appears to an observer on Earth, meaning it's a function of both the object's intrinsic luminosity and its distance from us. The concept of magnitudes dates back to the Ancient Greeks, when stars in the sky were categorized into six magnitudes (the brightest being 1 and the faintest being 6). Each successively lower magnitude was twice as bright as the one before, meaning the scale was logarithmic. We still use magnitudes for historical reasons, though the scale was later standardized to use the formula

$m_x - m_{x,0} = -2.5log_{10}(frac{F_x}{F_{x,0}})$

where $m_x$ and $F_x$ are the magnitude and flux of the object of interest and $m_{x,0}$ and $F_{x,0}$ are the magnitude of flux of a reference object (where usually Vega is used to define the 0 point in magnitude). This means any object that appears brighter than Vega has a negative magnitude. There is no limit to how bright an object can appear, so there is no lower limit to magnitudes. The sun, for example, being the brightest object in our sky, has a magnitude of roughly -27.

botany - How deep in the soil can a seed be placed and still develop into a plant/tree?

To answer your specific questions...

Does nutrient availability limit emergence depth?
Yes, the size of the nutrient store in the seed does impose a theoretical limit on the maximum depth at which seed germination and emergence can take place.

What is the best depth for a given species?
The specific depth which gives best germination for any given species is very unlikely to be the maximum depth at which seeds of that species can germinate. This is because other factors also impose limits on germination depth.

Probably the most important other factor is light - most seeds require the perception of red light in order to trigger germination. For most species then, you are unlikely to find seeds germinating and successfully emerging from depths greater than the penetrance of light.

Another important factor in some latitudes is temperature - many seeds (from 26 families) use dormancy as a way of preventing germination in the wrong seasonal conditions. In species where temperature serves as a cue to alleviate dormancy, there will be a maximum depth at which a given seed can detect the temperature fluctuations at the surface. Beyond a certain depth, surface temperature changes will not perceptibly affect local temperature.

If you ever need to find out the best depth of germination for a given species, many can be found on google, or failing that it is trivial to perform a simple experiment.

Is there a mathematical relationship?
There is a general mathematical relationship, an allometric relationship (i.e. relating size to another trait), between seed size and maximum depth of emergence, described by Bond et al. (1999):

                                     
where:
dmax = maximum depth of emergence
w = seed weight
c = a constant which varies with phylogeny and environmental conditions.

Many species however will not submit to this equation. For example some large-seeded species (incl. coconut, coco-de-mer, avocado), have large seeds because they may need to enable long-distance growth along the ground in order to locate optimum growth conditions. Seeds of these species will therefore not conform to the depth-size relationship above.

Reference

Visibility of earth from moon during day-time of moon

Inspired by this question. I am curious whether earth, besides being nearly fixed on one place on the moon's sky, is it visible during the day-time on moon too?

My understanding is that earth should be visible as moon has no atmosphere. Also, if the NASA didn't edit the following photograph, it suggests that the day-time sky of moon is all black and earth should be visible in it.

I think, Stellarium don't take into account the atmosphere once you are on other planet.