the moon - Historical astronomical lunar tables

I hope this is the right place to ask this question, and I apologise if not.

I've come across a date written in an old book, but it is written using what seems to be a lunar/zodiac system.
Below is a picture of the date (ignore the text above it):

I'm trying to work out what this date would have been in our modern calendar, but I can't seem to find any tables or almanacs that go back that far.

Are there any publicly available tables that do go back to 1587? If not, how exactly could I go about calculating the answer for myself? I am a mathematician, so a bit of wacky maths should be within my capabilities.
This is the answer I'm probably expecting from this site: some kind of program/algorithm/formula to be able to work out on what date the moon would have been in Taurus in

If not, any ideas as to where best for me to ask/look next?

Why do astronomers use supernova to measure distance in space?

Type Ia supernovae are not common; they are rare events, happening maybe once per 100 years in a galaxy. Nevertheless they have two properties that make them fantastically useful for distance measurement.

1. They are "standard candles". The physics of the supernova detonation, thought to be when a white dwarf accretes matter and exceeds the Chandrasekhar limit, is very "standardised". The bomb goes off in exactly the same way with the same amount of identical explosive. That means to a good approximation, measuring the apparent brightness of a type Ia supernova and comparing it with nearby examples means that the distances to these events can be accurately estimated.




2. They are really luminous and last 2-3 weeks. This means that they can be seen at enormous distances, they almost outshine the galaxies that they are in, and they last long enough for astronomers to discover them in automated surveys and still have time to follow them up and measure their light curves and estimate their peak brightness.

The limitations are that you can't choose which galaxies you measure the distance to. You have to wait for a supernova to go off and that might take 100 years or more for a particular galaxy. There are also continuing debates about just how standard a candle these objects are. It is possible that more distant galaxies that we see in the very early universe made stars with a different composition (far less elements heavier than helium) and that this alters things a bit. Other problems are associated with "de-reddening" the supernova light curves to account for the possibility of obscuring dust in the host galaxy.

I am a bit puzzled by your question "Why can't the distances between our Sun and the celestial objects be measured directly?" - galaxies are much too far away to have a trigonometric parallax distance, so what were you thinking of? Methods like using Cepheids or RR Lyrae variables don't work for distant galaxies because we can't resolve the individual stars. You could estimate a distance using a measurement of redhift and a cosmological model for the expansion of the Universe, but really the point of using these supernovae was and is to test and improve the cosmological models.

gravity - LIGO gravitational wave chirp signal frequency

This question is about the LIGO gravitational wave signals' frequency. The signals start from about 35 Hz to peak at about 250 Hz, giving evidence of gravitational waves. The question is about the exact frequencies the waves' detection started to the exact peak of the wave.

It seems the LIGO graphs show the peak is a little
above 250 Hz. Are there tables published by LIGO that put the start and peak frequencies as 35 to 250 Hz? The graph in the paper by Abbott, et al shows the start and peak (the start is around 36-37 Hz, the peak seems to be at around 256 Hz) - are there exact numbers (tables) for the start and peak, that one can reference?

orbit - Does the orbital variation in planetary gravity affect the Sun's corona

Dimitris (see below) argues that the syzygies of the Earth and Venus and those of Mercury, Earth and Jupiter distort the Sun's corona, which in some way affects climate on the scale of hundreds of years. Is there any evidence to support the idea that the planets do have such an effect on the sun, other than the correlation noted by Dimitris?

Planetary orbits’ effect to the Northern Hemisphere climate, from solar corona
formation to the Earth climate.

Poulos Dimitris

Abstract
The four planets that influence the most the solar surface through tidal forcing seem to affect the Earth climate. A simple two cosine model with periods 251 years, of the seasonality of the Earth – Venus syzygies, and 265.4 years, of the combined syzygies of Jupiter and Mercury with Earth when Earth is in synod with Venus, fits well the Northern Hemisphere temperatures of the last 1000 years as reconstructed by Jones et al (1998). Later reconstructions that give too much emphasis on multy-centenial variation are due to increased error. The physical mechanism proposed is that planetary gravitational forces derange the Solar Corona that in turn deranges the planetary geomagnetic field causing
temperature variations.

http://www.itia.ntua.gr/getfile/1486/1/documents/PoulosPaper.pdf

imaging - Can I use an array of inexpensive cameras as an alternative to a telescope?

Yes it would. However, it also depends on your definition of "Detail". For some, it's a tight shot where the whole frame is only a few dozen arcseconds wide, and others think of detail as being wide field, but also "deep" - or having a very high signal to noise ratio.

If you want wide field, and a high signal to noise ratio, then yes, a multi-camera array will work well, and you'd be able to accumulate a large number of high integration time images in a relatively short amount of time.

So here's a breakdown of what you and others who may not know need to know about deep sky astrophotography:

The main goal for all photos is to produce as visually pleasing image as possible. A single image taken of any target will accumulate light from many sources. The first, and obvious source is the light from whatever target you're trying to photograph. The second, less obvious source is light pollution from man-made sources, the moon, and natural sky glow. The third source is the electronic and thermal noise of the camera itself.

Technically, all of these sources are "signal" - however two of the three are unwanted, so we will call the light pollution signal and the camera signal 'noise'. Longer exposures produce more signal from what you want, but also more noise from what you don't want.

Luckilly, noise can be reduced by using several techniques in conjunction with each other. First, you need to take long photos. The signal will increase linearly with time. A 60s photo with a signal of "x" will have a tenth of the signal as a 600s photo. Noise, however, increases with the square root of the exposure. Using the same example, the 600s photo will have a signal to noise ratio that is about 3.2 (sqrt of 10) times better than the single 60s photo.

Increasing the Signal to noise ratio by taking longer exposures is the first step to producing 'detailed' wide field images. Camera noise can be further reduced by taking many images. An array of cameras makes this trivial. Two cameras will produce twice the images as one camera, and 10 cameras will produce 10 times the images as a single camera. That's obvious. Why does it matter?

When we stack images, the more we have, the better the stacking algorithms can preserve the signal we want to keep, and reduce the camera noise we don't. Once again, the signal to noise ratio mathematics work out the same as above. 1 image will have a certain signal to noise ratio, but 10 images will have a signal to noise ratio that can be about 3.2 times higher - because the total integration time has increased ten fold.

Of course, if you're stacking photos, you should be calibrating them as well (dark frames, bias frames, etc) - which really reduce the amount of noise per sub-frame. Because of this, your actual SNR boost by stacking should in theory be a lot higher than 3.2.

If photographed a target across 60 degrees of the sky (30 before meridian and 30 after) then you'd be able to acquire photos for 4 hours. One camera taking 10 minute photos would get 24 photos in the 4 hour period of time. If you had 5 cameras, you'd get 96 photos. 72 additional photos in the same amount of time is a lot of extra data to help reduce the noise, and really let the faint details stand out above the background - especially if all of your photos were well calibrated.

telescope - What is a quaternary mirror and why does the E-ELT need one?

You may know that a standard Newtonian telescope has two mirrors, they are called the primary and secondary mirror.

The E-ELT has five mirrors: The quaternary mirror is simply "mirror number four", counting in the direction the light enters the scope.

It's complex because that's where the adaptive optics sits:

The quaternary mirror has an approximate diameter of 2.4-m (2380x2340mm). It is a flat adaptive mirror, with up to 8000 actuators, thereby allowing the surface to be readjusted at very high time frequencies. [...]

This mirror will correct in real time for high order wavefront errors (e.g. atmosphere, wind shake, low spatial frequency telescope errors) and small amplitude residual tip-tilt corrections.

Source: The European Southern Observatory's page on the E-ELT optics

How can the Kuiper belt coexist with the 9th planet?

The recent evidence for the 9th planet is an alignment in various parameters of known Kuiper belt objects.

These are not Kuiper belt objects. The Kuiper belt is taken to refer to the neighborhood of Neptune resonances. The scattered disc is another population, further out, and the objects relating to Planet 9 are further still.

Legend: grey = Kuiper belt, green = integer Neptune resonances, blue = scattered disc. The objects in the recent announcement are all off the chart, although their perihelions are <100 AU.

Source: Wikipedia.

The shape of the grey splotch at the left of this chart shows that the Kuiper belt is indeed a collection of populations dominated by Neptune. This isn't news, and it's still taken to be a belt. It goes all the way around the solar system and the orbital eccentricities are low.

This postulated planet might help explain the "Kuiper cliff", the otherwise unexpected cutoff before the 2:1 resonance. But, that's another question. Failing that, there's simply no relation between the Kuiper belt and Planet 9 except that they're both very far away.

general relativity - Is it ever possible that the graviation from the mass of kinetic rotational energy will overcome centrifugal force?

This is, I think, somewhat intuitively obvious, but this comment got me thinking about this:

what is gravitational force?

"Rotation speed can create centrifugal force opposing gravity and making things lighter," Rotation also stores energy, and energy is mass, per $E=mc^2$. A body spinning sufficiently fast will exert higher gravity, e.g. a slowly-spinning neutron star will have a weaker gravitational pull than equivalent neutron star that spins very fast."

Intuitively, my answer is "no way", added mass by rotational velocity I would think, could never be extensive enough to counteract the "flicking off" or centrifugal force of very fast rotation. Even with a Neutron Star I would think it's impossible, but I'm not 100% sure, so I thought I'd ask.

If we ignore relativity from motion but not $E=mc^2$, as that's the crux of the question

Centrifugal force = $m v^2/r$

Kinetic energy of rotation = $1/2 I w^2 = 1/5 m v^2$

Mass equivalent of kinetic energy = $1/5 m v^2/c^2$

gravitational force $g=frac{Gm}{r^2}$

so if we apply the gravitational force of kinetic energy

$g = G(1/5 m * v^2/c^2) / r^2$, or, simplified, $G*m*v^2 / r^2 c^2$

and we compare the two equations
Centrifugal $m v^2/r$
additional gravitational $G*m*v^2 / r^2 c^2$

we can remove $m*v^2$ from both on the top and 1 $r$ from the bottom

additional gravitation ratio to centrifugal force = $G/r * c^2$

$G$ and $c$ are numerical. $G$ is very small, $c$ is very big and the ratio grows smaller as the radius grows larger.

gravitational constant: $6.67408 × 10^{-11} m^3 kg^{-1} s^{-2}
c = 2.998 x 10^8 m s-2

the ratio, unless my math is broken, centrifugal force to additional gravity from added mass by kinetic energy of motion = $1/r * 1.35 * 10^{27}$, so you'd need a hugely small r, almost a plank length or a singularity where the added gravity from kinetic energy of rotation would overcome the "flicking off the surface" or centrifugal force.

When I work out the units, I get meters per kilogram, which I don't think is right. The units should cancel out with a ratio of two forces in opposite directions, so I think I made an error, but I don't see where I made it.

My question is two fold. 1) is my math broken? and if so, where? and 2) is the added mass from kinetic rotational energy ever relevant, say in a very rapidly rotating Neutron star? Could it ever assist the Neutron star in collapse or add gravity?, I can see how it could add to flattening, as rotation flattens objects naturally and perhaps, add a speck of gravity on the polls where centrifugal force is zero, but logically, I think, rotational energy would end up spinning any non black hole object apart, long before the added energy of rotation had enough mass to make a measurable difference on gravity. Is my sense right or is there a situation where added mass from energy of rotation could overcome the centrifugal force?

Why is it that we can"t feel how fast the Earth is moving?

Not only Earth is moving but also you with the rest like atmosphere, so you are in closed system. The edge of atmosphere does not fly off because it has nothing to rub against, it just stays in same speed. When you put your hand from the car and you feel the force against it, it is the force which you must make to push the air from your way. If the air would be moving at same velocity as you, your hand would not feel this and you would be able to hold outside of car even at 300km (do not do that, it is dangerous).

star - Has the sun become way brighter the last years?

Up to last year, I never got blinded by the sun in everyday situations (e.g. running towards it), at no time of the year I was. But since last christmas, it occured to me more often that I cannot watch in the direction where the sun is because it is so bright that it really hurts me.
My mother and my roommate have observed the same phenomenon independently of me.

This, of course, is only anectodal evidence, which is practically worthless. We all get older and we may have the same kind of "photophobia", or we just forgot about being blinded the last few years and just now consciously felt an remembered it.

Therefore I want to ask this question: is there any objectively observable increase of brightness over the last few months and if so, why?

galaxy - How many stars and galaxies can be seen by the naked eye?

How many of the luminous dots that we see naked are galaxies and not stars from our galaxy?

I imagine that the majority of the luminous points that we see naked eye during the night, are actually stars from our galaxy. But how many of them are other objects (other galaxies, nebula, etc.), excluding planets from our Solar System?

astrophysics - K-Pg asteroid impact

It was day and night. That obviously depends on your position on the earth, at any given moment there are places in each of the time zones on the earth.

If you are talking about the impact point, that would be easy to know from the direction it came from - for example if it came from the side opposite the sun, it was midnight.

space time - Could the Sun's bending of light be measured on photographic plates before Einstein's prediction?

I suggest reading the paper on the 1919 expedition to get a clearer picture of why they did it at that time and why it hadn't been done before. From reading chapter 2 I think the main reason was the astrophotography equipment required for the experiment and the alignment of bright enough stars close to the sun to observe the effect.

Of course before Einsteins prediction in 1911 no one had any specific reason to observe the stars close to a solar eclipse before and during totality, it might seem obvious now but that's hindsight for you.

exposure - How does one add false colour to a fits image

I have an exposure of a star field with a small galaxy located near the centre. It is of course a greyscale image. I would like to add colours (realistic) to this image for aesthetic purposes. What software can I use to play around with colours that is Linux compatible and relatively easy to get the hang of? I've been using DS9 which has some colour options but to my knowledge doesn't quite produce a realistic colour image.

orbit - Does the Sun's light travel fast enough to have a straight path to Earth?

Light travels in straight lines in spacetime, but not necessarily straight lines through space, and the same is true for free-falling orbits of test particles like satellites (when thrust-less). However, in relativity what constitutes 'space' means taking some 'slice of now' through spacetime, which is of course depends on the reference frame one chooses to use.

For weak, slowly changing gravitational fields appropriate to the Sun, a conventional choice is the static weak-field metric
$$mathrm{d}s^2 = -left(1+2frac{Phi}{c^2}right)c^2,mathrm{d}t^2 + left(1-2frac{Phi}{c^2}right)underbrace{left(mathrm{d}x^2+mathrm{d}y^2+mathrm{d}z^2right)}_{mathrm{d}S^2}text{,}$$
where $Phi$ is the Newtonian gravitational potential. Thus, we can compare to a hypothetical flat spacetime, with a flat Euclidean space given by $mathrm{d}S^2$. The trajectory of light follows null geodesics ($mathrm{d}s = 0$), so the light travels as if traveling through a medium of refractive index
$$n = cfrac{mathrm{d}t}{mathrm{d}S} = csqrt{frac{1-2Phi/c^2}{1+2Phi/c^2}}approx 1 - 2frac{Phi}{c^2}text{.}$$
A changing refractive index does make light bend; this is exemplified in the simplest case by Snell's law, although the continuously varying case found here needs the more general Fermat's principle.

You shouldn't take the analogy of a refractive medium too literally. Some differences between an actual medium is that there is no dispersion as the refractive index is independent of frequency, and that there is an additional effect of gravitational redshift not present in refractive media. But the latter is separate issue from the trajectories of light rays, so we can ignore it for our immediate purposes.

Light deflection is quantified by an impact parameter $b$, which can be thought of as the perpendicular distance between the Sun and a hypothetical flat-spacetime trajectory, and a scattering angle $theta$, the angle between that and its actual path in the limit of infinite distance. (Note that the illustration in wikipedia assumes a repulsive rather than attractive potential, but otherwise conceptually it's the same situation.)

Modeling the Sun as having a spherically symmetrical gravitational potential $Phi = -GM/r$, the leading-order light deflection is,
$$begin{eqnarray*}
theta = 4frac{GM}{c^2b} &quad&left(text{if};frac{GM}{c^2b}ll 1right)text{,}end{eqnarray*}$$
which is twice what one would get for a particle at light speed under Newtonian mechanics. For light ray just grazing the outer edge of the Sun (or being emitted from there), this angle is about $1.7$ seconds of arc. Completely radial light rays would not be bent in this spherically-symmetric case.

cosmology - Is there any other theory, apart from the Big Bang paradigm, which describes the birth of universe?

This is not an answer but a comment which is too long to post as such.

There is no such thing as a scientifically approved theory, there are theories consistent with the evidence some of which are simpler in some sense than others, and philosophy of science tells us that we should prefer those that are simpler.

Also I doubt that the big-bang theory is what you think it is, which is simply that at a certain time in the past the Universe was in a hot dense state which then expanded, forming the lightest chemical elements... The first part of this is a simple extrapolation of the observation of the way that galactic Doppler shift increases with distance and the observed cosmic microwave background. The second part is the consequence of the application of well know nuclear properties of matter.

Now there are plenty of more-or-less widely accepted additions to this, mainly to explain the uniformity and fluctuations in the microwave background, and the emergence of first stars and galaxies etc. But all of these are to some extent speculative since we have no generally accepted unification of Quantum Mechanics and General Relativity or require physics that has no predicted consequence (so far) beyond the phenomena it has been proposed in order to explain.

To make this approximate an answer I will point you to the Wikipedia page on Non-Standard Cosmologies

distances - What is the definition of "deep space"?

It's not in any sort of academic context, but in relation to space exploration NASA defines deep space as such:

Deep space is the vast region of space that extends beyond our Moon, to Mars and across our solar system.

Again, not exactly an academic context, but Govert Schiling (the author of Deep Space: Beyond the Solar System to the End of the Universe and the Beginning of Time) in relation to astronomy defines deep space differently:

In a sense, everything beyond Earth’s atmosphere is deep space. But more typically it’s a phrase that has been used by observational and amateur astronomers alike for extended objects beyond the Solar System. I decided to provide a brief introductory section on the Sun and the planets, but the emphasis is really on the objects in our Galaxy and the wider Universe.

So this more or less echoes what some others have contributed, but provides additional context.

the moon - Is there a threshold on distance/size for a tidal locking?

You can see an example of tidal locking and atmosphere simulation for a planet closely orbiting a dim star. They show a simulation of the atmosphere and some interesting theories about the movement of gasses due to tidal locking (convection) that occurs between the bright and dark side of the planet. The link goes directly to the discussion of tidal locking:

KEPLER 186F - LIFE AFTER EARTH

A similar question was posted in the physics forum and it looks like that could be close to the answer you are looking for:

Tidal Lock Radius in Habitable Zones

I am also interested in tidal locking and hope someone may post a more concise answer.

cosmology - What accounts into calculating the Hubble constant?

From my understanding, the Hubble constant $H_0$ calculates from observed redshifts $z$ of distant galaxys against their proper distance $D$. The current value appears to be 67.80(77) $frac{km}{s}Mpc^{-1}$

Calculating the Hubble constant via the redshift, I assume one only wants those velocity contributions due to the expansion of the universe, and not those from the real movement of the galaxies within in cluster (peculiar motion) or so. This, I also read in various online resources (which I can't recall right now).

On the other Hand, if your galaxy cluster is far enough away (eq. ~ 1 Gpc), one can neglect peculiar motion, which is in the order of 1000 $frac{km}{s}$ (1000 $frac{km}{s}$ / (1 Gpc $times$ 67.80(77) $frac{km}{s}Mpc^{-1}$) $approx$ 1.4%)

Nonetheless, how would you try to correct for the peculiar motion, or is it really just neglected? Calculating all the gravitational components in each cluster? Another idea could be to assume that the galaxies within the cluster move randomly in respect to each other and the peculiar motion cancels out over the average? Any other possibilities?

Note that this question is partly a copy of another question of mine at physics.stackexchange.com, which wasn't fully answered but commented to ask it here.

gas - What is a typical value for core-to-star efficiency?

I was reading Unfolding the Laws of Star Formation: The Density Distribution of Molecular Clouds by Kainulainen et al., which discusses star formation rates and efficiencies.

One variable used is $varepsilon_{text{core}}$, the core-to-star efficiency, describing how much gas above the critical value of $s$, the logarithmic mean-normalized density, forms a star.

I'm using this parameter in some calculations, but I'd like to use an average value, as opposed to a value for a given molecular cloud.

What is a typical value for $varepsilon_{text{core}}$?

general relativity - Gravity assist: why is the velocity doubled?

Since the collision is perfectly elastic, the ball's velocity goes from -80
km/h from the train's reference point (negative being towards the train) to
+80 km/h from the train's reference point, a speed increase of 160 km/h.

For a stationery observer, therefore, the velocity goes from -30 km/h
(towards the train) to 130 km/h, an increase of 160 km/h.

The situation you describe only applies if the train isn't moving.

I made an error in units below, assuming the train is moving at 50 km per second (not hour) and that the ball is moving at 30 km per second (not hour), am I'm too lazy to correct it. The general principle, however, still applies.

The confusion may occur because we're ignoring the train's loss of momentum,
which means the train is going slower after the collision, and moving
backwards in its own frame of reference. A slightly more detailed
calculation:

• Suppose the the train has mass M kg and the ball has mass m kg.




• From the "fixed" observer's point of view, the initial momentum (in
  Newtons) is:

$rho =50 M-30 m$

and the initial kinetic energy (in Joules) is:

$e=450 m+1250 M$

• Let $v_t$ and $v_b$ be the train's and ball's velocities in
  meters/second after the collision. Since perfectly elastic
  collisions preserve both momentum and kinetic energy, we have:

$m v_b+M v_t=50 M-30 m$

$frac{m v_b^2}{2}+frac{M v_t^2}{2}=450 m+1250 M$

There are only two solutions to the equation above, one of which is
the initial conditions. The other is:

$
left{{v_b}to -frac{10 (3 m-13 M)}{m+M},{v_t}to -frac{10 (11 m-5
M)}{m+M}right}
$

Plugging in 0.0585 kg for the mass of a tennis ball and 640000 kg for
the train, this becomes:

${{v_b}to 129.9999854,{v_t}to 49.99998538}$

effectively confirming the calculation.

I'm not convinced this is a good analogy, however. Gravitational boost
occurs when a planet's gravity almost captures a spacecraft, thus
nearly making it a satellite, and giving it the same revolution
velocity around the Sun as the planet itself. The wikipedia analogy
bears only a passing resemblance to this.

planet - North Pole Right Ascension/Declination to axial tilt conversion

This is wrong, but may help someone find the right answer:

You can convert right ascension and declination to a 3 dimensional
unit vector in the J2000.0 ICRF reference frame using the standard
formula for converting spherical coordinates to rectangular
coordinates and using a radius of 1:

{Cos[dec] Cos[ra], Cos[dec] Sin[ra], Sin[dec]}

The ra and dec of the north ecliptic pole is ra,dec of
{270, 66.5607083333} per Wikipedia, so the unit vector representing
the north ecliptic pole is:

{0, -0.3977771648286046, 0.9174820582119942}

As it turns out, there is conflicting data for Ceres, perhaps because
of the semi-recent DAWN flyby.

Instead, I'll use Vesta, where wikipedia explicitly states (https://en.wikipedia.org/wiki/4_Vesta#Rotation):

north pole pointing in the direction of right ascension 20 h 32 min,
declination +48 [... which] gives an axial tilt of 29 [degrees]

Converting to degrees, Vesta's north pole's ra,dec is
{308,48}. Using the formula above to find the unit vector, we get:

{0.411957936296447, -0.527282113378167, 0.743144825477394}

If we take the dot product of two vectors and divide by the product of
their lengths, we get the cosine of the angle between them. In this
case, both vectors have length 1, and the dot product is 0.891563
whose arc-cosine is right around 27 degrees.

So, if Vesta's orbit were the same plane as Earth's orbit, the axial
tilt would be 27 degrees.

However, since Vesta's orbit is inclined 7.14043 degrees to the
ecliptic, this answer is incorrect.

I don't think you can find the correct answer without using Vesta's
inclination and the longitude of its ascending node.

Both of these values are known, but I can't figure out how to use them
to get the correct asnwer.