Frequency analysis on Semiperiodic object(s) using Lomb-Scargle So I already did a frequency analysis of the SRv-type's light curve  using Period04. I already have the 'periodicities' of the object/star of choice which is quite hard since the latter/other periods are tucked away by yearly aliases (thank goodness that's a rough week).
The problem, however, is that the mentioned-software doesn't allow the user to acquire the Power Spectrum but only the Fourier data (as of the moment) which is important for visualizing the non-coherent signals. May I kindly ask if it's possible to deduce/extract the power spectrum from the fourier data using Lomb-Scargle periodogram? And if there's an existing python tool/code to it?
The current code, which is below bog standard I'm afraid, just visualizes the Fourier spectrum;
import pandas as pd
import matplotlib.pyplot as plt
with pd.option_context('display.float_format', '{:0.20f}'.format):
    data = pd.read_csv('FLSP.csv')
    print(data) #shows Frequencies/Amplitudes

x = data['Frequency']
y = data['Amplitude']

plt.xlabel('Frequency')
plt.ylabel('Amplitude')

plt.plot(x, y)

Which should give you something (as useless) like this;

It's the Long Secondary Period (LSP) of the variable star at ~2350 days. Attach here is the fourier data if you want to check out.
Help at this point is sincerely appreciated. Thnx a lot and clear skies.
Note: I tried using Astropy's and Scipy's Lomb-Scargle, and their implementations require the observations; light curve, so I'm stuck on 'looking' at the fourier spectrum...suggesting to me that's its back to square one (hoping not lol).

This is a partial answer based on the discussion in comments below the question.
If one already has an "amplitude spectrum" and one wants to convert to a power spectrum, all you have to do is take the absolute value of the amplitude and square it. In Python (introduced in the question) that's just np.abs(amplitude)**2.
When you take the Discrete Fourier transform of a time series it is returned as a complex array, you need both real and imaginary components to retrieve both the absolute value of the amplitude spectrum and its phase spectrum.
Sometimes you can see "amplitude" used for the absolute value of amplitude, but it doesn't usually hurt if you take the absolute value again to be sure.
I've combined the signals of random noise with $\sigma$ = 1, a sinusoidal 50 Hz electronic interference noise signal, and a slightly phase-jittery square wave with an amplitude of 0.5 and a period of 12 milliseconds or a frequency of about 83 Hz.
I've kept the average near zero so we don't get distracted by a giant peak at 0 Hz, and I've shifted zero frequency to the center of the spectrum using np.fft.fftshift so we don't forget that negative frequencies exist in Fourier space.
We can see that the FT has both real and imaginary parts and squaring the absolute value gives us power.
For the time series plots I only show the first 200 milliseconds for clarity, the series has 1,000 time intervals of 1 millisecond each.
I've also used np.fft.fftfreq to automatically generate the frequency assigned to each bin in the FT.

import numpy as np
import matplotlib.pyplot as plt

np.random.seed(42) # this way you can compare your output to mine directly

N = 2000
nt = np.arange(N)
dt = 0.001 # one millisecond per bin
time = nt * dt

twopi = 2 * np.pi

noise = np.random.randn(N) # normal distribution with sigma = 1
frequency_1 = 50
signal_1 = 0.2 * np.cos(twopi * time * frequency_1)

# jitter = (7 * np.random.random(N)).astype(int) - 3
interval_2 = 12
q_int = int(interval_2 / 4)
ints = np.arange(0, N, interval_2)
jittered = np.random.randint(-2, 2, len(ints)) + ints
signal_2 = np.zeros_like(nt, dtype='float')
for j in jittered:
    signal_2[j-q_int:j+q_int+1] += 1 # I suppose one could convolute or sum over np.shift
signal_2 -= signal_2.mean()

combined = noise + signal_1 + signal_2

ft = np.fft.fft(combined)
freqs = np.fft.fftfreq(N, dt)

ft_shifted = np.fft.fftshift(ft)
power_shifted = np.abs(ft_shifted)**2
freq_shifted = np.fft.fftshift(freqs)

things, names = ((noise, signal_1, signal_2, combined),
                 ('noise', 'signal_1', 'signal_2', 'combined'))
m = len(things)
fig, axes = plt.subplots(m+2)
for i, (thing, name, ax) in enumerate(zip(things, names, axes[:-2])):
    ax.plot(time, thing)
    ax.plot(time, np.zeros_like(time), '-k', linewidth=0.5)
    ax.set_xlim(0, 0.2)
    ax.set_ylabel(name)
axes[-2].plot(freq_shifted, ft_shifted.real)
axes[-2].plot(freq_shifted, ft_shifted.imag)
axes[-1].plot(freq_shifted, power_shifted)
axes[-1].set_xlabel('frequency (Hz)')
for ax in axes[-2:]:
    ax.set_xlim(-150, 150)
plt.show()


What is the Age Difference Between Earth & Jupiter? What is the age difference between Jupiter & Earth?

Which planet was formed first, or were they formed at the same time?

It is generally thought that Jupiter "formed" first via the core accretion process. An icy core forms first and then sweeps up the gas around it. This process is generally thought to take from a few million to ten million years. After that, the solar system gas was dissipated and Jupiter was basically as it is today, but larger and hotter.

Earth formed on a longer timescale in the maelstrom of rocky planetesimals that made up the inner solar system. An essential part of the process was the collision that formed the moon, which probably occurred between 30 and 60 million years after the protosun began to form. Further heavy bombardment of the Earth that may have significantly changed its surface and atmosphere continued for some tens of millions of years after that.

The wikipedia page on planet formation in the solar system looks ok to me
https://en.wikipedia.org/wiki/Formation_and_evolution_of_the_Solar_System
I may add some more academic references subsequently.


Traversing a proton within the Planck time If I could travel a super small distance (proton) in an even smaller time (Planck time), how long would it take me to cross the observable universe?

If you don't like math: A fraction of a second.
For math lovers:
One proton is $8.414 \cdot 10^{-16}$ meters wide, and one Planck time is $5.391247 \cdot 10^{-44}$ seconds. This means that we are traveling at $\dfrac{8.414 \cdot 10^{28}}{5.391247} = 1.5606779 \cdot 10^{28}$ meters per second. Dividing by $299792458$ m/s, or $c$, we get $5.205861 \cdot 10^{19}c$. The observable universe is 93 billion light years in size, so it would take $\dfrac{93 \cdot 10^9 \cdot 86400 \cdot 365c}{5.205861 \cdot 10^{19}c} \approx 0.0564$ seconds to traverse the observable universe. This obviously is impossible, as faster-than-light travel (username checks out) violates the rules of general relativity (as we know it).


What is this rapidly twinkling red, blue, and white star I saw? Last night, I was on my balcony at 1AM (PST) and I looked up and saw two stars near the horizon (I'd guess ~30 degrees above the horizon), and they were "twinkling" about twice as fast as other stars higher in the sky, and I could clearly see them changing from red to white to blue repeatedly. Other stars in the sky only appeared white to me, and didn't seem to "twinkle" as rapidly as these two stars did. The red and blue make me think of red-shift and blue-shift, but I don't know how I would see both from the same object.

What was I seeing?

I don't know if it helps, but I am in the Los Angeles area, and I was looking in a roughly north direction. almost exactly to the east, according to google maps.

Edit: I tried taking a picture, but light pollution from the nearby street lights wouldn't permit me taking a decent picture. However, I noticed a group of three stars close together in nearly a perfect almost vertical line, and managed to find that in Stellarium. I think I found the two stars I am seeing: Procyon and Sirius

Is there anything about either of these stars that would make them show as red/blue?

It's most probably Sirius.  At this time of year (at 1 am local time) it's low in the sky in the East, so there is a lot of atmosphere in the way, and as Sirius is a bright bluish star, it will show all the colours described as it twinkles.


What is the difference between Terrestrial Time (TT) and International Atomic Time (TAI)? The equation which relates Terrestrial [Dynamical] Time (TT, or TDT) and International Atomic Time (TAI) is the following, according to the U.S. Naval Observatory:

$TDT = TAI + 32.184$

As I have understood, both TT and TAI are continuous time systems (differently from UTC, which has leap-second gaps) and described by the same unit (the SI second), but counted from a different epoch (the origin T0 of one is 32.184 seconds after the other).

It appears that both time systems are essentially the same. So what is the actual difference between them? If the difference between them is the nature of how they are computed (e.g., TT is a theoretical timescale; and TAI is a statistical timescale based on several clocks), why the 32.184 shift?

It's historical. TT is a successor to The International Astronomy Union's ephemeris time, and the offset between TT and TAI is a result of getting TT to match ET when they switched.

Excerpts from https://en.wikipedia.org/wiki/Terrestrial_Time:

  Terrestrial Time (TT) is a modern astronomical time standard defined by the International Astronomical Union, primarily for time-measurements of astronomical observations made from the surface of Earth.[1] For example, the Astronomical Almanac uses TT for its tables of positions (ephemerides) of the Sun, Moon and planets as seen from Earth. In this role, TT continues Terrestrial Dynamical Time (TDT or TD),[2] which in turn succeeded ephemeris time (ET). TT shares the original purpose for which ET was designed, to be free of the irregularities in the rotation of Earth.

  The present definition of TT is a linear scaling of Geocentric Coordinate Time (TCG), which is the proper time of a notional observer who is infinitely far away (so not affected by gravitational time dilation) and at rest relative to Earth. TCG is used so far mainly for theoretical purposes in astronomy. From the point of view of an observer on Earth's surface the second of TCG passes in slightly less than the observer's SI second. The comparison of the observer's clock against TT depends on the observer's altitude: they will match on the geoid, and clocks at higher altitude tick slightly faster.

  Time coordinates on the TT and TCG scales are conventionally specified using traditional means of specifying days, carried over from non-uniform time standards based on the rotation of Earth. Specifically, both Julian Dates and the Gregorian calendar are used. For continuity with their predecessor Ephemeris Time (ET), TT and TCG were set to match ET at around Julian Date 2443144.5 (1977-01-01T00Z). More precisely, it was defined that TT instant 1977-01-01T00:00:32.184 exactly and TCG instant 1977-01-01T00:00:32.184 exactly correspond to the International Atomic Time (TAI) instant 1977-01-01T00:00:00.000 exactly. This is also the instant at which TAI introduced corrections for gravitational time dilation.

(Bold emphasis mine).

For more on International Atomic time see https://en.wikipedia.org/wiki/International_Atomic_Time . Basically, it's a weighted average of multiple atomic clocks - and the altitude effect mentioned in the middle paragraph above is why they added the gravitational time dilation corrections. From the Wikipedia link above:

  n the 1970s, it became clear that the clocks participating in TAI were ticking at different rates due to gravitational time dilation, and the combined TAI scale therefore corresponded to an average of the altitudes of the various clocks. Starting from Julian Date 2443144.5 (1 January 1977 00:00:00), corrections were applied to the output of all participating clocks, so that TAI would correspond to proper time at mean sea level (the geoid). Because the clocks had been on average well above sea level, this meant that TAI slowed down, by about one part in a trillion. The former uncorrected time scale continues to be published, under the name EAL (Echelle Atomique Libre, meaning Free Atomic Scale).[7]:215


Sun constantly converts mass into energy, will this cause its gravity to decrease? If the sun is constantly converting the mass into energy, then will its gravitational field continue decreasing?

If the sun is constantly converting the mass into energy, then will its gravitational field go on decreasing?

It's a very interesting question and the answer is yes!

The solar constant indicates the mean solar radiation of electromagnetic waves (mostly in visible and near infrared light and I'll answer based on that.

While the conversion of mass matter† to energy in the Sun's core now represents a loss of mass proper matter, it turns out that that energy (trapped in the Sun and slowly diffusing towards the surface) will have the same gravitational attraction as the matter it came from until it actually escapes the Sun!

There is some prompt mass and energy loss via neutrinos and it's significant, perhaps several hundred keV per neutrino I simply don't know the number yet. I'll ask a separate question about it. I'm guessing that losses due to the stellar wind are small, but I'll update here as soon as the following is answered:

How much mass does the Sun lose as light, neutrinos, and solar wind?

update: the answer there is that loss via neutrinos is only about 2.3% of the radiative loss, and on average loss via solar wind and coronal mass ejections is about 4E+16 kg/year, or about another 30% relative to the radiative loss described below.

The value $I$ is about 1360 Watts per square meter at $R$ = 1 AU which is about 150 million kilometers or 150 billion meters. So the total energy lost per second $P$ is

$$P = 4 \pi R^2 I$$

Taking the time derivative of $E = m c^2$ we get

$$\frac{dE}{dt} = P = \frac{dm}{dt} c^2$$

so

$$  \frac{dm}{dt} = \frac{1}{c^2} \ 4 \pi R^2 I$$

That means that the value of the mass that we use to calculate the Sun's gravitational attraction changes by about 4.3E+09 kilograms per second, or 1.3E+17 kilograms per year.

The Sun's current mass is about 2.00E+30 kilograms, so this effect changes by a very tiny fraction per year, about 6.7E-14.  Over the age of the Earth of 4.5 billion years, that's 3E-04, or about 0.03% if the Sun's output were constant. It has probably changed over this time of course, so this is just a rough estimate.

†Thanks to @S.Melted's answer for clarifying this.

added to my answer by @Tosic:

  The Earth feels no torque from any force during this (the force from radiation is radial), which means its angular momentum is conserved. This means
  $$R_1 v_1 = R v$$
  $$R_1 \sqrt{\frac{GM_1}{R_1}} = R \sqrt{\frac{GM}{R}}$$
  $$M_1 R_1 = M R$$
  We can see the Earth's orbital radius would change by a factor of 0.03% as well (M1 and M are solar masses).


Relative brightness of the sun's corona In preperation for the upcoming solar eclipse I'd like to practice my photography to be able to take some (hopefully) good photos of the corona during totallity.

What is the brightness of the sun's corona? How bright is it compared to a full moon (or other celestial phenomena)?

At what point during the "diamond ring" does it become too bright to look at safely without safety equipment, and what is the relative brightness of the diamond?

The corona varies dramatically in brightness from the inner portion near the sun to the outer portion a few solar radii away. The innermost portion is the brightest; at ISO 100 and f5.6, an exposure of ~1/125 second will capture it. So it's roughly the same brightness as the full moon. The outer corona will require much longer exposures of perhaps a half second, but note that during this time the sun will move noticeably with a long lens if you're not using a tracking mount, so some blurring will likely occur. Overall, practicing on the crescent moon will give you a pretty good idea of the dynamic range of the corona.

The diamond ring is dangerous to look at directly, as is any unobscured part of the sun's disk; you should look away at first glimpse, or earlier.

There's lots more detail scattered around the web, which you've probably found by now. A good overview is available at http://www.mreclipse.com/Totality2/TotalityCh12-3.html#SEExpoTab


Did we ever actually see the earth revolving around the sun? Is the geocentric model completely disproved? Did we ever actually see the Earth revolving around the Sun?

Also, is the geocentric model completely disproved or was it sidelined because the heliocentric model made things easier to understand?

(BTW I know Earth revolves around the Sun and am just asking out of curiosity.)

Thanks!

What you're asking, basically, is whether there are any proofs for the heliocentric model of the Solar System.

A literal naked-eye observation of the Earth revolving around the Sun would be rather difficult, since human beings have never gone to another planet yet, and have only been to the Moon briefly, decades ago.

Here are several proofs; some of them are historically relevant also.

Kepler's laws of planetary motion

This became one of the earliest proofs, as soon as Newton figured out the law of universal gravitation, and the "fluxions" (what we would call today differential equations). When you assume a heliocentric model, and the inverse square law for gravity, then Kepler's laws in a heliocentric model come out of the equations naturally, as soon as you do the math.

This is like saying: "if it's heliocentric, and knowing that the law of gravity is correct, then Kepler's laws should be such-and-such". And then: "oh, but the theoretical calculations for Kepler's laws match the actual observations with great precision. Therefore, our hypothesis (heliocentric, inverse square law) must be correct."

It was the earliest strong indication that the heliocentric model is in natural accord with the basic laws of science, whereas the geocentric view was becoming more and more contrived as evidence accumulated.

Tycho Brahe in late 1500s provided the enormous mass of observations of planetary motion. Johannes Kepler in early 1600s used Brahe's observations to come up with his laws empyrically (and also arguing for the heliocentric model). Isaac Newton in late 1600s said "yes, Kepler is right, because of mathematics and the law of gravity, and here's the proof from calculus".

http://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion

Stellar parallax

An early argument against heliocentrism was that, if the Earth was really revolving around the Sun, then very distant objects, such as the stars, would appear to be ever-so-slightly bobbing back and forth around their average positions. Since that's not the case, it was argued, therefore the Earth must be fixed.

You can see this argument in historical archives, proposed by theologians in the late 1600s, in favor of the geocentric model, and against the growing consensus then of "natural philosophers" (what we would call today scientists) that the heliocentric model was correct.

In reality, parallax does exist, it's just very small. It was measured experimentally in the 1800s, and was then quickly used to determine, for the first time, the distance to the nearest stars.

http://en.wikipedia.org/wiki/Stellar_parallax

Aberration of starlight

The direction where we see a distant star also changes when Earth's speed vector changes during its revolution around the Sun. This is different from parallax; it's more akin to the way raindrops on the side windows of a car leave diagonal traces when the car starts moving (even though the raindrops fall vertically).

It's essentially a relativistic phenomenon (when applied to light), but it can be partially explained in a classic framework. It was actually observed before parallax in the late 1600s (the heyday of Newton), but it went unexplained until early 1700s.

http://en.wikipedia.org/wiki/Aberration_of_light

Orbital mechanics of interplanetary probes

Landing a probe on Mars or Venus would simply not work if you assumed a geocentric model. A geocentric description of the Solar System might remain valid in a purely kinematic perspective (just the geometry of motion) as long as you remain on Earth. But the illusion breaks down quickly as soon as you start to consider dynamics (see Kepler's laws), and/or when you try to actually leave Earth (space probes).

Let me reinforce this point, since several answers and comments got it wrong: the geocentric and the heliocentric models are not completely interchangeable, or a matter of relativity. You could build an "explanatory" geocentric model, and it would be "correct", purely in a kinematic fashion (the geometry of motion), and only as seen from Earth. But the model breaks down as soon as you consider dynamics (forces and masses); it would also reveal itself as incorrect even from a kinematic perspective as soon as you leave Earth.

This is not just an artifice to simplify calculations. The dynamic calculations are wrong in a geocentric model.

In order to compute the very high precision trajectory of the space vehicle carrying the Curiosity rover and successfully place it on Mars, you must operate from a heliocentric perspective. The dynamics are all wrong otherwise. You would not miss the target only a little, in a geocentric approach, you would miss it by a lot - the vehicle would not even go in the general direction of Mars.

Miscellaneous

When observed in a telescope, Venus has phases like the Moon, and also grows and then shrinks in size, synchronized with its phases (it's large as a thin crescent, it's small when it's gibbous). In a geocentric model, the size changes could be explained by an elliptic orbit of Venus around Earth, but the phase changes synchronized with that are harder to explain. Both phenomena become trivial to explain in a heliocentric model.

It should be noted that this does not necessarily prove the heliocentric model, just the fact that Venus is orbiting the Sun, not the Earth. So it's an argument against pure (or strict) geocentrism.

Jupiter, when observed in a telescope, clearly has its own satellites. This was an early blow against a strict geocentric model, which assumed that everything must orbit the Earth. It opened the door to the idea that orbits could be centered on other celestial bodies too, and to the idea that things in orbit around larger objects could have their own smaller satellites (and therefore the Earth could orbit the Sun without losing the Moon).

The list could continue (and the full list is very long) but these arguments should suffice. You don't necessarily have to see something with your own eyes in order to know with certainty that it's there. In the case of Earth's revolution around the Sun, it was simply a matter of an overwhelming amount of evidence piling up in favor of it.

Geocentrism simply doesn't make any sense whatsoever in modern science and space exploration.


Where can I find a database on the masses of different galaxies? I want to get the data for masses of galaxies, more information about them can only be helpful. Can anyone point me to a site where I can get this sort of information?

As a starting point, I would suggest searching on VizieR for catalogues of galaxies. E.g. select "Galaxies" in the "Astronomy" menu (on the right-hand side of the initial form) and put "Masses" in the "Find catalogs" search box (the upper one, the lower one is for searching by position which is less useful here).
In the results section, the elliptical icons on the right-hand side allow you to see roughly what the sky coverage is for the catalogue. Once you select a catalogue, you can take a look at what fields are there.


How do higher numbers affect orbital resonance? When talking about orbital resonance, the most common resonances in our Solar System are 1:2 (think Jupiter's moons) and 2:3 (think Pluto and Neptune). However, the lack of larger numbers within these resonances confuses me. I accumulated some data about resonances and I figured out that the difference between the numbers gives more stabilized resonances. For example, a 1:29 resonance would not be stable due to the massive separation of the objects from each other and the gravitational meddling of other bodies. However, that also brings up a different question, and the one I will be asking. A 28:29 resonance seems somewhat stable; after all the objects are close together and will have high gravitational influence on each other. Space is big, so it is unlikely the objects would collide, so why do we not see any 7:8 resonances or other resonances involving high numbers in our Solar system?

General Explanation
In the history of our Solar System, the orbits of Solar System bodies have gradually evolved, passing through many "higher number" orbital resonance ratios. Only the stabilizing effects of the most stable "lower number" resonance ratios have been enough to overcome the forces causing orbital migration.
When we look at the orbital resonances in the current Solar System, we get a snapshot in time that just shows us the resonances that have been powerful enough to overcome other forces that cause orbital migration.
A rigorous mathematical explanation for why lower-number resonances have a greater effect than higher number resonances may be resonance-type specific and most have enough math to choke a cat.  Intuitively, if you push a child on a swing 28 times for every 29 times they swing, you aren't going to have much of an effect.  Orbital resonances are similar.
Specific Example (Mercury Spin-Orbit Resonance):
Common wisdom holds that orbiting bodies will eventually achieve a 1:1 orbital/spin resonance with a main body, like the Earth's Moon.  Though the Moon was spinning faster in the past, tidal torque caused it's spin to slow until the spin rate was equal to the orbital period.  As the Moon spun down, it passed through many other ratios of orbital periods.
The case for Mercury's orbital/spin resonance around the Sun is a lot more complicated.  Scientists and astronomers are still working out how and why Mercury was caught in a 3:2 spin/orbit resonance as it spun down, rather than passing through the resonance. From this excellent news article:

Tidal dissipation in the planet and the ensuing deceleration of its
spin inevitably carry the planet through a sequence of the spin-orbit
resonances. The question then becomes in which of these resonances the
planet should eventually get trapped...
the international team revisited the problem of tidal evolution of
Mercury's spin and found that the 3:2 resonance is indeed the most
probable end-state. The frequency-dependent tidal torque acts as an
efficient trap for the planet trying to traverse a resonance. The
efficiency of the trap strongly depends on the value of orbital
eccentricity, as well as on the temperature and viscosity of Mercury's
mantle.

Note:
Forces causing migration of planetary orbits include (but are not limited to):

Loss of planetary angular momentum due to gaseous drag in the early proto-planetary circumstellar accretion disk.

Tidal forces resulting in exchanges of angular momentum.

Collisions.

Various n-body gravitational interactions.

The Yarkovsky effect.

Resonances with other bodies.


Why has Moving Cluster Method been successful only for Hyades? At all places where I have read of Moving Cluster Method for measuring distance to a cluster and once applied myself on data, I have been told that the method has been successfully applied only to Hyades cluster in Taurus.

Question : What factor made the method unsuccessful for other known clusters?

Do you know this paper? Mamajek (2005). "A Moving Cluster Distance to the Exoplanet 2M1207b in the TW Hydrae Association".

I'll risk an opinion:
There is about a thousand clusters with kinematic measurements, but besides no more than half a dozen, all pm components are under 20 mas (milli-arcseconds) with errors more or less in the range 0.3-6 mas. Such errors are too high in relation to the pm values and as a result, the vectors will not converge.


Heliocentric distance from galactic x y and z coordinates How can I get heliocentric distance to a star when I have galactic x y and z coordinates known ?

The galactic coordinate system is centred on the sun with the xy plane in the plane of the galaxy and the x coordinate pointed towards the centre of the galaxy.

Since it is a heliocentric system the Euclidian distance is $\sqrt{x^2+y^2+z^2}$.


How does the stellar evolution for low and intermediate mass stars differ? I'm trying to understand stellar evolutionary tracks for different masses. So far I have the general understanding from a cloud of gas and dust to a white dwarf, neutron star or black hole.
What I don't understand is how the different evolutionary tracks differ from each other for low to intermediate mass stars. I have attached two figures that causes my confusion.
The first one shows many different evolutionary tracks and the second one shows the evolution of a low mass star of 1 Solar mass.
The first image shows that a star of 1 Solar mass stops at point 6, which corresponds to the Helium Core Flash. The second image shows that the evolution continues from this point, making the star bluer before it becomes a red giant again and collapses and explodes.
My question is this: Where does the evolution actually stop? Is the Helium Core Flash the last thing a Sun-like stars experiences or does it in fact follow the rest of the tractory?
All in all, I'm a bit puzzled by the difference between low- and intermediate mass stars evolution. Also it seems to have something to do with conective vs. radiative zones..?
Sorry for the large images!

Where does the evolution actually stop? Is the Helium Core Flash the last thing a Sun-like stars experiences or does it in fact follow the rest of the tractory?

Stars with masses below about 2 solar masses undergo a helium flash but their evolution does continue afterwards, as shown in the second diagram (Fig. 13.4). So no, the core flash is not the last thing a Sun-like star experiences.
The tracks are sometimes separated like this for the technical and practical reason that many programs that create stellar models struggle to model the helium flash. To get around this, the models sometimes stop at the tip of the red giant branch and are then restarted at stable core helium burning by "engineering" a model to the correct state, without going through the flash.


Confusing satellite-like object seen in night sky I was staring up at the stars around 11:30 PM and noticed an object moving. It had no lights, and was not bright whatsoever. It was moving smoothly and in a straight line, except extremely fast, and dead silent. It was fairly large too, compared to how big stars look. Never would have noticed it had I not been looking at that particular region of the sky at the exact moment I did. It had to have crossed the sky in less than a minute. In fact, I feel like I've seen this same kind of thing once or twice before.

My question is what did I see? The fact that it was extremely dim, extremely fast, and produced no sound really confuse me.

At first blush, this sounds like it was a large, low-Earth orbiting satellite, like the ISS for example. Depending upon the angle above the horizon the transit time could be very low s you described. Your position would determine whether the satellite was illuminated by the Sun or not at that local time.


Why does a planet rotate and revolve? Why do planets rotate and revolve in the universe?

As the planets evolve during their protoplanetary stage and accrete materials from the protoplanetary disks, which are gravitationally collapsing interstellar dust and gases, these accreted particles retain some of the angular momentum from the materials they form from and being in constant motion.

      Generated image (virtual fly-by) from a simulation of the accretion period of the protoplanetary disk, showing preservation of      angular momentum in the orbit around a Jupiter-size planet, as it clears its neighborhood. (Source: Frédéric Masset)

One nice description for this angular momentum preservation, and why the planets appear to rotate faster than their surrounding protoplanetary disk goes like this:

  Conservation of angular momentum explains why an ice skater spins more
  rapidly as she pulls her arms in. As her arms come closer to her axis
  of rotation, her [rotation] speed increases and her angular momentum remains the
  same. Similarly, her rotation slows when she extends her arms at the
  conclusion of the spin.

  Source: Scientific American article on Why and how do planets rotate? (George Spagna)

So it could be described as this axial rotation of planets resulting in conservation of the angular momentum of the materials in the protoplanetary disk, forming during the accretion period of the planetary system as the protoplanets gain in weight, and preserve this angular momentum due to inertia of their radial velocity.


What will be Supernova SN 2014J's peak brightness? What is the expected maximum apparent magnitude of supernova SN 2014J?
When will it attain peak brightness?

According to this website the peak visible magnitude will be about 10.5 around February 2nd.
Earlier estimates had been a little brighter around the same date.


How is the expansion rate of the universe obtained from Baryon Acoustic Oscillations? I understand that we can measure the sound horizon from the anisotropies in the CMB. Large galaxy surveys today can construct two-point correlation functions, which reveal peaks at distances equal to today's sound horizon, I think because the BAOs at decoupling essentially created dark matter gravitational wells for structure to begin forming in (correct me if I'm wrong). I'm under the impression that the sound horizon is a fixed physical distance, so its measured comoving distance at any redshift should always be the same. Am I right or wrong in this assumption?

I read some information from a few different sources on this topic which confused me. On Wikipedia: "Peaks have been found in the correlation function (the probability that two galaxies will be a certain distance apart) at 100 h−1 Mpc, indicating that this is the size of the sound horizon today, and by comparing this to the sound horizon at the time of decoupling (using the CMB), we can confirm the accelerated expansion of the universe."

But if the sound horizon is always at the same physical scale, how can we use measurements today to determine the rate of expansion? Do the correlation functions give you proper separations, and from that with different cosmological models we can determine what $H_0$ or $H(z)$ value gives a comoving separation equal to the sound horizon? And is this the same as the BAO signal?

I'm under the impression that the sound horizon is a fixed physical
  distance, so its measured comoving distance at any redshift should
  always be the same. Am I right or wrong in this assumption?

You are both right and wrong at the same time. Right in the sense that, yes, in comoving coordinates, the distance is the same at all redshifts. But when we use galaxy catalogues to measure the BAO distance scale, aka the BAO peak, it usually varies with redshift. This variation is very tiny, of the order of a 1-2 h$^{-1}$Mpc.

  But if the sound horizon is always at the same physical scale, how can
  we use measurements today to determine the rate of expansion? Do the
  correlation functions give you proper separations, and from that with
  different cosmological models we can determine what 0 or ()
  value gives a comoving separation equal to the sound horizon? And is
  this the same as the BAO signal?

Before redshift of galaxies were measured, people were just using the galaxy positions on the sky (Right ascension RA, Declination DEC) to calculate the angular correlation function. You can also see the BAO peak in the angular correlation function. Nowadays, since we have accurate redshift measurements, people calculate the 3d correlation function $\xi(r)$. To get constraints from BAO peaks in a very simple way, we can in brief, do the following.

To do so, we first need to convert RA, DEC and redshift to X,Y,Z cartesian coordinates. It is at this step, we assume a fiducial cosmological model, as in $\Omega_{M}=0.27, \Omega_{\Lambda}=0.7, h=0.67$ etc. Now, we have 3D information of our galaxies and we can count pairs and get the $\xi(r)$, from which we can calculate the BAO peak.

People can then use theoretical model predictions for $\xi(r)$, given a specific set of cosmological parameters. It is here that cosmologists vary the set of cosmological parameters, which will give us a specific $\xi(r)$.
Lets say, we have 10 variations of $\Omega_{M}$, $\Omega_{\Lambda}$ and $h$. This would give us 30 theoretical $\xi(r)$'s, and we can compute the BAO peak for all of them using an empirical fit model. These peaks can now be compared with the observed peak from the data and we can calculate the $\chi^{2} = \frac{(peak_{obs} - peak_{theory})^{2}}{err(peak_{obs})^{2}}$. This can give us constraints on the different cosmological parameters.

Of course, this is just a basic answer to your question and there are a lot of other processes involved! You can check out this paper which was recently published in arXiv which uses only the BAO peak information to get constraints on cosmological distance measures.


Studies on planets without parent star I am interested in studies and/or publications about planets that wander in space and are not orbiting around a star or its remnants. These dark and frozen worlds should've got ripped away from their respective system by a large passing body (another star? passing black hole) or any other astronomical event (which one?).

I am curious, whether there exist statistical estimates on the number of such planets and the rate at which they are produced. I am also interested in reading about the statistical models that can produce those estimates and encode researchers' beliefs. If there are no such models, why?

A provoking thought: According to Greek, such planets are really fitting their name πλάνητες ἀστέρες (planētes asteres, "wandering stars") or simply πλανῆται (planētai, "wanderers"). It is frightening to imagine such world that drifts deeper and deeper into empty space. Forever.

It would be more accurate to call them surveys than studies.   A study looks at an object in detail.  A survey counts and categorizes objects.

Rogue planets are enormously difficult to see, and only a few very large ones have actually been observed directly.  For the largest rogue planets, it's unclear if they actually are ejected planets as opposed to failed stars.

This article discusses rogue planets in pretty good detail.

In 2011 the MOA survey was published, estimating a high number of Jupiter sized Rogue planets, perhaps outnumbering stars.

More recently the OGLE Project estimated about 1/10th that many rogue Jupiters, which is more in line with statistical predictions.   Smaller planets are more likely to get ejected from their solar-systems than larger ones.    Rogue Juiters outnumbering stars seemed unlikely and my understanding is that most scientists are more in line with the OGLE estimate than MOA.

Both estimates rely on microlensing which amounts to very small and temporary changes when rogue planets drift between the Earth and a star.  There's a fairly large margin of error for both surveys.

NASA's WFIRST telescope should answer a lot of questions on rogue planets.   It's not expected to be launched till the mid 2020s though.


Is there any site with telescopes data? Is the data gathered from telescopes "open source"? If so, is there any website that group it all?

Most general purpose observatories release the data taken on their facilities after the expiration of the proprietary period (this is the time, typically 12-18 months, where the data is only available to the proposers for the telescope time so they can work on it without being "scooped"). "Focused telescopes" in the sense of those that are performing surveys or looking for specific types of objects, tend not to release the raw data, instead releasing processed data or catalogs.

Some examples of data archives that provide raw or processed data include:

the NOAO Science Archive: provides access to 40 telescope+instrument combination from the US National Optical Astronomy Observatory
the ESO Science Archive Facility: access to raw and processed data from the four 8-m VLT telescopes plus the VLTI and the smaller (1-3.6m) telescopes on La Silla, Chile
the MAST archive at StSci: contains data from Hubble, Kepler and K2, TESS, SwiftUVOT, XMM and many others
the ING archive: for the Isaac Newton Group of telescopes of La Palma (1.0m JKT, 2.5m INT, 4.2m WHT telescopes)

Querying multiple archives for data is the role of the Virtual Observatory; there is a page of links to tools at SAAO and the Data Discovery Tool


How can I see a star behind a mountain range? Do I move apart from it and up? Could some explain what I need to do to see a star that, from my point of view, is behind a mountain:

Obviously, if it's one mountain, and not a very high one, I can just climb it to see what's behind. But what if it's a mountain range or a very high mountain?

Intuitively, I know I can move away from it to change the angle, and see the star. But what's the math behind it? And how does going up (or down, maybe?) help?

Moving away will help. If the mountain is very steep, and you are standing right next to it, you can only see objects that are directly above or behind you. Example: if the mountain is 100 ft tall, and you move 100 ft away, then you can see objects in the sky that make a 45 degree angle off the horizon. Angle = inverse tangent (mountain height / distance from mountain)


Why don’t we feel the Earth rotating beneath us? I’m not sure all of the details in the experiment but basically it went as follows. Someone marked their location of take off in a hot air balloon and went straight up in the air. They stayed suspended for some time before coming straight back down. Apparently the result was that they had not drifted from the original take-off point underneath them. Maybe I got all of that incorrect, but I’m sure it’s close.

But if in that experiment the Earth had no effect on the landing point, why does the “rotation of the Earth” need to be factored into making the shot with a long range rifle for example? Now I know it isn’t the best comparison and that there are many different factors in both scenarios. Such as “the bullet is traveling and in the experiment the balloon wasn't”, but I’d just like to hear different theories and opinions as to why this is.

P.S. I know a popular answer is going to be that the Earth is too massive for us to feel its spin, but I’m thinking deeper than that!

There are two different things going on, and they aren't too related (from what I can see). For the hot air balloon situation, you would think that if you hovered above where you are for 6 hours, the Earth could spin under you, and you would land back in a completely different place. Unfortunately, because the hot air balloon was on the Earth to begin with, it was already moving with the Earth. There are many reference frames in play, even though we feel like we are standing still. Everyone on Earth is standing still on the surface. However, the surface is rotating around the Earth's axis. The Earth's axis (and the Earth itself) orbits the Sun. The Sun is goes around our galaxy and our galaxy is travelling through intergalactic space.

So how does this relate to the hot air balloon situation? Well because the balloon was on the surface, it was already moving with the surface of the Earth. Remember how I said that the surface of the Earth rotates around the Earth's axis? Well, since the balloon was on the surface to begin with, it will also rotate with Earth's axis, just like the surface! What if we wanted to achieve that hover effect? I said earlier that the Earth orbits the Sun. In order to achieve that hover effect, we would need our hot air balloon to orbit the Sun without rotating around the Earth. You can't do this with a hot air balloon, since the atmosphere also happens to move with the Earth, and hot air balloons can't go into space. To achieve that hover effect, we would need some sort of spaceship with lots of fuel. If I had a hovering spaceship that didn't orbit with the Earth (this would, again, cost lots and lots of fuel), then yes, I could hover in the same spot and have the Earth rotate under me.

Now, you're probably wondering how I could achieve this with a spacecraft that was originally on the surface of the Earth. There wouldn't be much point in this hovering effect since you may as well just fly to your destination (like the theoretical Big Falcon Rocket), but if you wanted to do it, you would need to go above the Earth's atmosphere, use your boosters to go opposite of the rotation of the Earth (to cancel out your velocity), hover for some time, then use your boosters to go along with the rotation of the Earth (to return to rotation velocity), and land on Earth. Of course, you could skip the 2nd part by using heat shields and ramming into Earth's atmosphere like every other spacecraft we have, and no rocket would ever do this as it's just much more practical to orbit around the Earth (hovering costs fuel, orbits don't).

Ok, what about the sniper situation? The Earth rotate on its axis. Since a day is 24 hours long, a location at the equator goes around once in the same time it takes for someone in the arctic to go around once. However, the further north or south you are from the equator, the slower that part of the Earth has to rotate to complete 1 rotation in 24 hours. Think about spinning a ball. The equator of the ball whips around, but the top and bottom move much slower. It's the same thing. Lets say my sniper is at the equator. When the sniper shoots east or west, they do not need to correct for the Earth's rotation because everywhere along that latitude, the Earth is rotating at the same speed. However, if the sniper shoots north, the bullet will go to the east. That's because when the bullet was shot at the latitude closer to the equator (shot from the south), that spot of the Earth was moving faster than the spot of the Earth that the target was standing on. It's like saying that the spot I'm standing rotates at a speed of say... 1000 mph, but my target's spot rotates at say...995 mph. Since my bullet came from the 1000 mph part, it will obviously go out of the barrel at high speeds because of the gun, but it will also be rotating around the Earth's axis at 1000 mph. However, once my bullet starts approaching my target's latitude, since the Earth rotates slower, it will appear to deviate in the direction of the rotation (which is east, or to the right). Why? Well since my target's speed is 995 mph from rotation, you do 1000 - 995 = 5. That means that my bullet will have a net velocity of 5 mph to the right relative to my target. That means my bullet will miss if the target is far away enough. If you shoot from the equator to the south, the bullet will also go east, so as a result, it will deviate to the left. It's the same effect, but "upside down". This effect is called the Coriolis effect, and it's what gives hurricanes their power.

Finally, you asked "why don't we feel the Earth rotating beneath us?" This is because we are moving with the Earth. When you are on a train that is going at a constant speed of 50 mph, you do not feel it move (you may feel some bumps due to bumpy tracks). It is only when you accelerate or decelerate do you feel things move. When you are on the train, your speed is constant, so you do not feel anything. It's the same for the Earth, except you are rotating around the Earth's axis at a constant 1000 mph. Nothing has changed, except the speed is faster.

I probably explained this horribly so feel free to ask anything.


Why is it so difficult to discover the earth Trojan? It is discovered only in 2011.

Why is it so late?

We know where to find it.

2010 TK7 has a diameter of about 300 meters (1,000 ft). Its movements do not bring it any closer to Earth than 20 million kilometers (12.4 million miles), which is more than 50 times the distance to the Moon. (from the wiki entry for it)

That should make it pretty clear? Can you see a fly from 200 km away?


Is the Cosmos a super set of the Universe and any hypothetical multiverse(s)? I'm quite lost on the definitions of words like omniverse, megaverse, multiverse, meta-universe, universe, and cosmos. I used to think the Universe and the Cosmos were essentially the same things, but now I'm reading the latter is an orderly system (not sure what's meant by that).

Can somebody help me understand what they each represent? In 1980, Carl Sagan said "The cosmos is all that is, or ever was, or ever will be." This sounds like the Cosmos would be the greatest of all and all else would fall into that. Would it be accurate, for example, to say Cosmos > Multiverse > Universe? What about the other things I mentioned earlier?

"Cosmos" is the English cognate of the ancient Greek word which refers to the entirety of existence. It is equivalent to the Latin-derived word "universe;" the only difference between the two is the language of origin.


If there was air between the Sun and Earth, how warm would we get? Based on this question: How loud would the Sun be?

If instead of a vacuum we had air between the sun and earth (let's say of the same composition we have here on earth) how warm would our planet get?

This is a really hard question to answer as Carl Witthoft alludes to. The main problem is that there is no way you can finagle putting at atmosphere around the Sun and have it extend to the Earth. It would be wildly unstable and no matter what happens, the general result will be an explosion of energy that kills the Earth. And you can't really say you're going to wave some magic wand and prevent all of that because you're then fundamentally altering physics and you won't end up answering the question you wanted to ask to begin with.

Your only way of answering this question is to look at the current atmosphere of the Sun and try to extrapolate. Take a look at the figure below. This represents both the temperature (solid line) and density (dashed line) of the Sun vs height from the surface (what the "surface" of the Sun means is a whole other question). The temperature at the surface of the Sun is a measly $5800\:\mathrm{Kelvin}$. As you increase in height, this temperature (i.e., the temperature of the plasma atmosphere above the "surface" of the Sun) doesn't change much. This region of the atmosphere is known as the chromosphere. But then, at a particular height known as the transition region, there's a drastic increase in the temperature. It goes from a few thousand degrees to a couple million degrees. This area of the atmosphere is the Corona. This graph cuts off at $10,000\:km$ but the Sun's atmosphere extends far out beyond Pluto.

Hopefully you can fully appreciate the point I'm getting at which is that Earth actually is inside the atmosphere of the Sun and that atmosphere is really hot! The reason we of course aren't boiled to death is that the atmosphere is incredibly sparse. Only a handful of particles per cubic centimeter. So what you're really proposing is to increase the density of this atmosphere by an insane amount. You want to make it approximately $10^{20}$ times more dense! If we just extrapolate properties, that's going to turn the already hot atmosphere of the Sun (around the Earth) into something incredibly hot. And the huge density means we'll really feel the effects. The Earth will likely be vaporized in a flash, before the cataclysmic explosion which I alluded to above can even happen.

So, long answer short: About a billion degrees.


If the Sun were bigger but colder, Earth would be hotter or colder? That is the question. I know the concepts of luminosity $(L=4\pi R^2F)$ and the flow $F=\sigma· T^4$, with $T$ the temperature in its surface. But how I use that to know if the Earth would get warmer or not if $R$ increases and $T$ decreases?

The equilibrium temperature of the Earth, $T_E$, scales roughly as $L^{1/4}$, which is proportional to $R^{1/2} T$, where $L$, $R$ and $T$ are the solar luminosity, radius and temperature.

The actual approximate relationship is derived by equating the power received by the Earth, which is proportional to the solar luminosity $L$, with the power radiated by the Earth, which is proportional to $T_E^4$ for a blackbody. Hence $T_E \propto L^{1/4}$.

So the answer to your question depends on by how much you increase the radius compared with the decrease in temperature.

There will be second order effects that do depend on the spectrum of radiation from the Sun (and therefore its temperature) compared with the wavelength dependence of the albedo and emissivity of the Earth. So I will post a better question...


Which would be safer: removing hydrogen or adding hydrogen to our sun? If we want to extend the lifespan of our habitable zone safely, would it be safer to institute a yearly 'tax' on our suns outer layer of hydrogen, or a yearly 'subsidy' to our suns outer layer of hydrogen?

(the feasibility of said tax/subsidy is beyond the scope of this question)

Another way to ask this question:
Assuming the core of the sun remains untouched,
but its outer layer of hydrogen were reduced by half,
would the suns main sequence remain intact for a longer time?

To avoid ending up with a white dwarf and its associated bad weather, what would be the ideal point to reverse the tax and re-subsidize?

My kneejerk reaction is that your only option is to remove a chunk of mass from the outer part of the Sun.

The Sun will respond (on a Kelvin-Helmholtz timescale), by contracting and becoming less luminous because the core temperature is lower in a less massive star. This will extend its main sequence lifetime, because only the central parts of the star are involved in fuelling the nuclear burning. The core is separated from the well-mixed outer convective envelope by a stable radiative zone.

Since the default solar evolution involves it getting slowly more luminous with time, it is possible that mass could be extracted at just the right rate to keep the Sun at constant luminosity. This sounds like an interesting calculation to do.

The alternative of adding "fuel" in the form of hydrogen won't work. The Sun's luminosity will increase because of its increased mass and increased central temperature. However, because the new fuel cannot be mixed into the core (only the outer part is mixed by convection), then the main sequence lifetime would be reduced.


Why are there optical differences between sunrise and sunset? I've noticed that sunrises are not reverse sunsets regarding the sky hues and lighting.

Sunrises always start pinkish and end with a glorious yellow sunshine.
On the other hand as the sun sets lighting gets orange and the sky gets a hellish red colour. I've noticed this many times regardless the weather or temperture.

Why is this contrast between colors (pastel-like in the morning vs vivid in the evening) in a fenomenon that should be same in both instances in a day ?

Am I the only one that perceives this difference ?

You are not the only one whom sees a difference but remember that the statistics are skewed. Fewer people are awake early to see, or photograph, the sunrise while more people are awake to see the sunset.
People and heating of the Earth are factors explaining the difference in the atmosphere through which the sunlight is filtered, that affects the appearance of the light.
It's well explained at our Photo.SE site:

Most of the physical reasons (I can think of) probably do favor sunsets as well. During the day, the sun is heating the air and the ground. This leads to air movement that tends to pick up particulates (e.g., dust, smoke, etc.) and keep it in the air. It also leads to water evaporating (relatively) quickly from anything like lakes and rivers in the area. In addition, quite a few things that people do (e.g., driving) tends to happen primarily during the day, so it "pushes" more pollution into the air during the day. Thus, at the end of the day, the light is shining through air that carries more particles, which almost inevitably does more to diffuse the light.
During the night, there's a lot less pollution being put into the air, and the lack of heating means the air tends to be calmer as well, so by morning many of those particulates will have simply fallen out of the air, and the humidity either formed into clouds or precipitated. The clearer air does less to diffuse the light, so the sunrise isn't as spectacular.

A more scientific, but not complicated, explanation is offered at Wikipedia's "Sky" webpage. The heat affects the Rayleigh scattering and the Lorenz–Mie–Debye solution or Mie scattering, that explains the differences between sunrise and the sunset - a difference in the number and movement of particles in the sky.
A somewhat related factor is explained in a Q&A over at Space.SE: "Where does space begin on planets without atmospheres?". The height of the atmosphere and its layers along with the distribution of its component gases is altered by the heating of the planet and atmospheric circulation.


Is the LSST (Vera C. Rubin) telescope going to change position every 20 or 35 seconds? This is a very particular question, I realize, but I am still curious...
The Vera Rubin telescope is supposed to take a 15-second exposure, then take 5 seconds to move to its next spot...
But I also read that it will take two consecutive 15-second exposures at each stopping point, then spend 5 seconds moving to its next position...
On Wikipedia, in the middle of the article on the Vera C. Rubin LSST Telescope, two consecutive paragraphs contradict each other, the first saying it will spend only 15 seconds at each spot before moving on, and the second saying it will take two consecutive 15- second exposures at each point.....

How the Vera C. Rubin telescope will physically carry out the actual survey of the sky (the Legacy Survey of Space and Time; LSST) is still subject to evolution and refinement. The broad survey strategy is set out in the LSST System Science Requirements Document which is (intentionally) relaxed on how it is actually carried out. The main requirements are:

The main survey, of at least 18,000 square degrees must be uniformly covered do that each field of the camera's 9.6 square degree field of view gets an average of 825 30s visits over the 10 years of the survey. (The 825 visits are totalled over all 6 filters)
Rapid revisits (anywhere from 40 seconds to 30 minutes) must be acquired over at least 2000 square degrees to discover fast transients. (This can mostly be done by overlapping the fields of view of adjacent fields slightly

Other science requirements such as needing to detect moving objects (one of the four main science goals) and being able to reach the stated parallax and proper motion goals also impose some constraints on how often and when you need to revisit a field.
These constraints apply to the main "wide-fast-deep" (WFD) survey which is what will cover the 18,000 $\deg^2$ area above. Within the main WFD survey area, the baseline plan is make 2 visits per camera field (9.6 $\deg^2$) in either the same or different filters each night (This allows the moving objects and rapid transients to be found and alerts sent out). This pair of visits is repeated every 3-4 nights during the time the field is visible. Each visit consists of 2 back-to-back 15s exposures in the same filter, separated by the readout time of 2 seconds. These two 15s exposures are then combined in the pipeline to make the final 30s image.
Exactly how the WFD field visits are strung together, how far the telescope moves between fields and in what filters it observes and when, and what the observatory spends the other ~10% of time on, is a very complicated problem and depends a huge amount on what objects you want to be sensitive to and what type of science you want to do. These various tradeoffs are still ongoing as part of the Survey Cadence and Optimization Committee. The report of the large number of survey cadence simulations has recently come out but a final survey strategy will not be decided on until April 2023 (timeline summary) after more meetings, reports and simulations. Also, while 1x30s exposure at each visit (rather than 2x15s which has higher overhead) and a shorter time spent on each field is preferred (it increases the survey throughput, letting you cover more fields/area per night), this will depend on the final LSST Camera performance. This is because the susceptibility of the cameras' CCDs to cosmic rays, and whether you need the 2x15s exposures to remove them, is still partly unknown until it arrives at the telescope and has been operating for some time.
The simulations are based on a telescope model, whose code is available on github. The time taken to go from field to field depends on a quite a few things such as how far the telescope has to move (as there is both acceleration and deceleration limits as well as optics settling time depending on how far it moved), how much the dome has to move and whether you are changing filters (a 120s operation).
In general, the total time spent on a field is going to be:
$$
t_{slew}+t_{settle}+t_{exp}\times num. exp. + t_{shutter}\times num. exp. + t_{readout }\times (num. exp.-1)
$$
where $t_{slew}$ is the slew time (depends how far you move), $t_{settle}$ is settling time (also depends how far you move but minimum of 3s), $t_{exp}$ is exposure time, $t_{shutter}$ is shutter open/close time (1s; it's really big...) and $t_{readout}$ is the readout time (2s; the $(num. exp.-1)$ comes about because you can slew to the next field while reading out the last exposure).
So for 2x15s exposures, we have minimum of $3s+(15s\times2)+(1s\times2)+(2s\times1)=37s$ between fields. For the 1x30s, we have minimum of $3s+(30s\times1)+(1s\times1)+(2s\times0)=34s$ between fields. Exactly how often the telescope will be moving about to a new field will depend on the survey cadence decided on as outlined above, but there will be a strong preference to keep the slew times short and the number of filter changes small to maximize the amount of fields and sky covered per night.


In System KIC 9246715, How Far Does One Star Orbit the Other? I was going to ask if it's physically possible for one red giant to orbit another, but fortunately, I have just found such a thing--a red giant binary system known only as KIC 9246715.  In this article, all I could find of the distance between the stars is that one orbits the other in an orbit of 171 days.
The only problem with that statement is that it doesn't tell how far one star orbits the other in miles.  In our solar system, a 171-day orbit would be midway between Mercury (orbital revolution:  88 days) and Venus (orbital revolution:  225 days).  Obviously, this couldn't be the case for two stars that are each bigger than our own sun.  But even binaries confer to Kepler's Third Law of Motion, that the farther one body is from its parent, the slower it'll complete one revolution.
So in this real system, have there been any estimations of how far one star orbits the others measured in miles or kilometers?

The AAS Nova article cites Rawls et al. 2016,
who analyze light curves and spectra of the eclipsing binary KIC 9246715 to estimate its physical properties.
Besides the stellar masses and radii in the article, their Table 2 lists orbital semimajor axis a = 211 R☉ = 147 million km = 0.98 au, and orbital eccentricity e = 0.356.
During the 171-day period, the distance between the two stars varies between 95 and 199 million km, or 0.63 and 1.33 au.


How can red supergiants be cooler than dwarfs of the same spectral type? The Stellar Classification Wikipedia article has a sentence:

Red supergiants are cooler and redder than dwarfs of the same spectral
type

Which seems to be true, my course has a similar statement. However, here is an HR diagram:

Spectral Class varies on the same axis as Temperature (see top horizontal axis). Does the claim that "red supergiants are cooler than dwarfs of the same spectral type" only says that since they are in the same spectral type, their temperature is roughly the same, but the red giant is a bit cooler, or is there more to the story and I am misusing the HR diagram?

The HR diagram has many forms. There is no one to one relationship between temperature, colour and spectral type that is true for stars of all surface gravities and metallicities.
Spectral type depends on temperature primarily but spectral features can also be weaker or stronger at lower surface gravity or with changes in metallicity.
For example, the Handbook of Space Astronomy and Astrophysics has two tables of spectral type Vs temperature relationships, one for dwarfs and one for giants. And indeed the M-type giants are cooler than M-type dwarfs of the same subclass by ~200K.


Where can I find historic rates of meteor impact events? I have a friend that watches a lot of TV news and they have the impression that impact events are becoming more common than in recent memory. Does the evidence support or refute this?

Thinking it would answer the question, I've tried to find data on the annual rate of impact events. But maybe there is another way to look at it.

The database of the meteoritical society might be what you are looking for. You can do a search by year.
Based on this data, I drew this quick plot of number of recorded meteorites per year

Now there are a couple things to see here. First is that we find as roughly as many meteorites as we did ten or twenty years ago. So these events are not more common now than they were in recent memory. This answers your friends' question. Second is that we have many more records of meteorites than we did 100 years ago. This is simply because these numbers are not the exact number of meteorites that occurred, but rather the number of meteorites that were recorded.

The International Meteor Organization also has some databases you can browse to find more information.


Gravitational waves: how can scientists derrive the sizes and spin and distance of two merging black holes? After recording the ripples of  passing gravity waves I would imagine that some high-power math would be required to derive information about the sizes, spin, and distance of the merging black holes. Is it possible to get some insight into how they figured out one was 31 solar masses and the other was 20? How do you "see" spin from a wave?

One element of the answer is that they do it in the other direction. They simulate many different possible black hole mergers and then look for the nearest matches to what they observe. I imagine there is some scope for interpolating between the simulations as well as running bespoke simulations to confirm the match. Spin affects the wave because the shape of space around a spinning black hole is different from that around a non-spinning one, which affects both the motion of the black holes and the processes by which the gravitational waves are radiated.


Calculate Day of the Year for a given date I am trying to make sense of the following formula that claims to calculate the day of the year given today's date represented by (day/month/year):

N1 = floor(275 * month / 9)
N2 = floor((month + 9) / 12)
N3 = (1 + floor((year - 4 * floor(year / 4) + 2) / 3))
N = N1 - (N2 * N3) + day - 30

Finally N supposedly refers to the day of the year.

What do the magic numbers 275,9 etc represent in the equation ?

This would really be better off in Stackoverflow, or another page.  Presumably, you assumed some of the calculations rely on concepts in Astronomy, which as far as I can tell having gone through them, they do not with the exception of the concept of leap years; which I think is general enough that it doesn't really mandate being in Astronomy.  Regardless:

The equation is a little abstract, so probably the easiest way to understand it would be through understanding each bit of data. Let's start with N2, I'll come back to N1:

N2 = floor((month + 9) / 12)

N2 will equal 0 if month is less than 3, and will equal 1 if it is greater.  This formula is to determine whether February has passed.

N3 = (1 + floor((year - 4 * floor(year / 4) + 2) / 3))

N3 will be equal to 2 if the year is not a leap year, and will be equal to 1 if it is a leap year. The math here is just to determine whether the current year happens to be a leap year.

N = N1 - (N2 * N3) + day - 30

Now we take N1 (which I haven't covered yet) and we subtract it by the product of N2 and N3, which will equal 0 if we haven't reached March, or if we have it will be 1 on a leap year, and 2 on a non-leap year. We then add the number of the current day of the month, and subtract 30.  So what does N1 do?

N1 = floor(275 * month / 9)

Well, lets look at the values returned compared to days in a non-leap year:

Month     N1         Day       Diff
1.        30         31          -1
2.        61         59          +2
3.        91         90          +1
4.        122        120         +2
5.        152        151         +1
6.        183        181         +2
7.        213        212         +1
8.        244        243         +1
9.        275        273         +2
10.       305        304         +1
11.       336        334         +2
12.       366        365         +1

A bit messy, right?  Now, remember that you're subtracting 30 from the total at the end to get N, and we're adding in the current date. This means that although we multiply our current month to get N1, we're actually using this to calculate the dates from the months prior to our current month! Thus if we take the value of N1, subtract it by 30, and compare it to the preceding month, the chart will come out like this:

Month     N1         Day       Diff
1.        31         31           0
2.        61         59          +2
3.        92         90          +2
4.        122        120         +2
5.        153        151         +2
6.        183        181         +2
7.        214        212         +2
8.        245        243         +2
9.        275        273         +2
10.       306        304         +2
11.       336        334         +2
12.       ---        365         ---

From this, you can see that the value of N1 will equal 2 greater than the actual date for any day in which it is March or later.  This is perfect, as N2 is already a formula determining this for catching leap days. Note, these would all equal +1 on a leap year, as in another day would have been added in February. Thus coming back to the final calculation:

N = N1 - (N2 * N3) + day - 30

We take N1 and subtract in by the value of (N2 * N3), which will be 2 on a non-leap year if February has passed, 1 on a leap year, or 0 if we haven't passed February.  We add in the days in the current month, and we subtract 30 to get the offset in N1 values.

This should consistently give you the day of the year.  As for special significance, they're being used only for mathematical properties. Dividing a number between 1 and 12 by 9, just so happens to equal less than 1 for numbers less than 3, and greater than or equal to 1 and less than 2 all the way up to 12, which is why it is we have started March. The 275 * (month / 9) formula I'm not sure where they came up with. AFAIK, they have no special meaning to time or the Sun-Earth system - rather, they were chosen because they calculate well in this particular usage.


which pulsar has the longest spin period so far? Vela X-1 has a 283s spin period, while this one may have a 2.7hr spin period.
However neither of them is included in ATNF catalog.
They are not included in $p-\dot{p}$ diagram either. The longest period is just about 10s in this diagram.

My question is since Vela X-1 is definitely a pulsar, why is it not accepted?
Which one has the longest spin period so far?

When this diagram is mentioned in textbooks, we do not say which kind of pulsars are included at all. Why?

$p-\dot{p}$ diagram only shows radio pulsars?  no!
you can see magnetars,radio quiet pulsars and pulsars which are only active in $\gamma-ray$

Except magnetars, there are no single pulsars only pulse in X-ray band?

If a pulsar pulses in X-ray band, that does not necessarily mean it is an X-ray pulsar?
Only pulsars in binaries and accretion-powered pulsars could be called X-ray pulsars? They should be radio-quiet?

In binaries, there are no pulsars which pulse in radio and x-ray band simultaneously?

Vela X-1 is an example of an accretion-powered pulsar. These emit pulsed emission because they are accreting material onto their magnetic poles. If the magnetic poles and rotation axis are misaligned this results in pulsed X-ray emission. The power for the pulsar comes from the infalling material.

The classic P-Pdot diagram you show is for radio pulsars, or rather rotationally-powered pulsars. These are objects that are pulsators by virtue of accelerating charged particles from their magnetic poles out along the field lines. These accelerated relativistic particles emit synchrotron and curvature radiation that is beamed and intensified. The energy ultimately comes from the rotational kinetic energy of the pulsar.

In other words, other than the fact that they both involve neutron stars, these are completely different phenomena.

In terms of radio pulsars, your diagram looks reasonably up to date. I think the longest period object on your diagram is PSR J2144-3933, which has a period of 8.51 seconds (Young, Manchester & Johnston 1999).

Your diagram has a line marked as "graveyard". I believe this is a locus defined by Bhattacharya et al. (1992) and has the form
$$P = 2.42\times 10^{-6} B^{1/2}\ s,$$
where the pulsar magnetic field $B$ is in units of Gauss (typical pulsar values would be $B=10^{10}-10^{13}$ Gauss (as also marked on your diagram). The theory behind this "death line" is discussed by Ruderman & Sutherland (1975). Briefly it arises from the requirement of a minimum potential difference to be generated such that accelerated particles produce energetic enough radiation to stimulate the production of further electron/positron pairs. If the magnetic field strength falls or the rotation period gets too long, then this mechanism fails and the pulsar is quenched.

PSR J2144-3933 appears to lie beyond this death line, but there are other ideas and models of how this death line may arise (of which I am not very familiar, but see for example Zhang et al. 2000).

Note that millisecond pulsars are thought to recycled accretion-powered pulsars. That is that they gain sufficient angular momentum from an accretion process that they spin up to become radio pulsars again.

Radio-quiet pulsars are thought to be rotationally powered but where the radio beam is narrower than say the beaming of gamma rays in which they are detected.

Magnetars are something different again. Their power comes from the decay of extraordinarily strong magnetic fields.
The soft gamma ray repeaters and anomalous X-ray pulsars fall in this category.

So you can see that the word "pulsar" might include all sorts of different types of objects and physics.


Do the neutrons in neutron stars emit the radio waves? Neutrons can, especially in extreme circumstances (and large concentrations) emit electromagnetic radiation. I specifically asked about this in Physics S.E. Has a free neutron ever been shown to absorb/emit/interact with a photon?, the answer has to do with the fact that although neutrons have no detectable electric moment or dipole, they DO have a small, 'negative' magnetic moment.
But little I have read about neutron stars says which particles in or around them emits the radio waves, except one brief article mentioning infalling electrons from a surrounding plasma.

The electromagnetic radiation comes from accelerated charged particles, mainly electrons and positrons.
The surface of a neutron star is not made of neutrons. It is a totally ionised gas of nuclei plus electrons, with a composition that could range from iron-peak elements to hydrogen and helium accreted from the interstellar medium.
As a result of their rotation and strong dipolar magnetic fields then there are also very strong electric fields at the surface of the neutron star. These fields  rip electrons and protons from the surface and accelerate them away -  electrons from the magnetic poles and protons at lower magnetic latitudes. The classic work on this is by Julian & Goldreich (1969).
The acceleration due to these electric fields is easily sufficient to overcome gravity. The particles are forced to spiral along the magnetic field lines by the Lorentz force, and produce synchrotron or curvature radiation directly. The photons from these processes are then capable of creating secondary electron-positron pairs that are accelerated in opposite directions, releasing more photons in a cascade.
No neutrons are harmed in the making of this radiation, other than the ones in nuclei at the surface of the neutron star that get bombarded by high energy accelerated electrons and positrons.


How to interpret error of absolute magnitude of a star given data from wikipedia? This Wikipedia page contains information about the star Spica in the rightmost table. A small snapshot of this table is below.

In this table, the absolute magnitude is given as −3.55 (−3.5/−1.5). Instead of seeing the value and error given as value ± error, I am thinking the error is given as it is because of absolute magnitude is given on a logarithmic scale (as opposed to linear scale). But I am not quite sure how to interpret this as an error reading. How does one interpret this reading?

Spica (α Vir) is a binary star.
The Wikipedia article cites Herbison-Evans et al. 1971, whose Table III says:

  Absolute magnitude of primary (Mv1)   -3.5±0.1
  Absolute magnitude of secondary (Mv2)   -1.5±0.2

So the confusing "-3.55 (-3.5 / -1.5)" can be interpreted as "combined (primary / secondary)."


Extraction of error spectrum in Sloan Digital Sky Survey I need to know does SDSS spectrum contains error spectrum as well?

http://www.sdss.org/dr12/spectro/spectro_basics/#AbouttheSpectra

The above link says, further HDU's contain error spectrum and than tells about the inverse variance. Does inverse variance contains error spectrum? If yes, how can I extract that, if not where can I find error spectrum?

The variance is the square of the standard deviation, so you can calculate 1$\sigma$ errors of the spectrum by taking the square root of one over the inverse variance of HDU1 (https://data.sdss.org/datamodel/files/SPECTRO_REDUX/RUN2D/PLATE4/spPlate.html#hdu1head).

(You will get infinity for bad pixels since their inverse variance is set to zero.)


Where is a Hollow Object's Center of Gravity? Let's imagine that the Earth's moon is hollow and the remaining mass is structurally strong enough to retain its spherical shape (like a really big tennis ball).

Given this scenario, where would its center of gravity be? What would the conditions be like inside? Also would one still be able to stand on the moon?

If by "a hollow object" you mean a spherical shell -- a theoretically perfect sphere Moon, with a perfect sphere of material removed from the center: The center of mass, and therefore the center of gravity remains at the geometric center. Ref Newton's Law; Bodies with spatial extent.

The body of the moon is plastic; Meaning the gravitational forces are strong enough to reshape the material... it basically flows into a sphere. So if you magically removed a small amount of the center, it would collapse into a smaller sphere. If you removed most of the center, then the "thin" shell may or may not be strong enough to support it's shape.

Inside a hollow spherical shell, there is no net gravitational force. So you would be in a zero-G environment everywhere within the shell. From outside the shell, the gravity is the same as if all the remaining mass was located at a point at the geometric center.


Quasars and Gamma Ray Bursts Has there been any study to suggest that either quasars of gamma ray bursts are more detectable from further distances in the universe?

I've read that quasars are the most powerful energy in the universe but gamma ray bursts are detectable from further distances. Logically one would think that if some entity's energy were detectable from further distances than another entity's energy, then it'd be the entity with more energy than the other.

I'm just looking to see if anyone has any definitive study references that suggest why either quasar or gamma ray bursts have more energy than the other and to explain why if true the one with less energy is visible from further distances than the one with more energy.

I'm not really an astrophysicist type guy but I do some reading and watch different scientific shows that interest my since a lot is still unknown, so I'm hopeful someone here more familiar with this can answer or clarify this for me or give me some references to anything that gives clue to any of this.

I apologize for not having direct reference to any readings I've read that bring about this question in my mind but I know I read something that said gamma ray burst are detectable from further distances than any other energy entity and I've also read quasars are the most powerful energy sources in the universe, so I'm just looking for some clarify on this topic whether right or wrong with anything further you can provide.

I think you may be mixing up "Energy" and "Power" a little here. Power is the rate of Energy output per time. If Power is the speed, then Energy is the distance travelled.

Quasars look like stars, from Earth, they give out a light, and while they do vary a little (or a lot) you would not expect to see new ones forming or existing ones fading away.

Gamma Ray Bursts are a short lived. While a quasar will emit more energy over it's long life. Gamma Ray Bursts are more powerful. They are formed mostly as a result of energy released in gravitational collapse during a hypernova. Moreover this energy is released in a narrow beam.

Imagine comparing a laser beam with a equivalent power light bulb.  If the laser is pointing right at you, it appears much brighter than the bulb.

So the answers:

GRB are more powerful than quasars, but only last a short time.
GRB are form a beam of radiation.
So if a GRB and a quasar are equally distant, the GRB would appear brighter.


What causes these asteroid trails to be so strangely shaped and repeated? The Phys.org news item Hubble is the ultimate multitasker: Discovering asteroids while it's doing other observations includes the image below.

What causes these asteroid trails to be so strangely shaped and repeated? Usually asteroids appear as short straight segments in long exposures from terrestrial telescopes.

  Some asteroids from within our Solar System have photobombed deep images of the Universe taken by the NASA/ESA Hubble Space Telescope. The asteroidal streaks in this image are created by our virtual neighbours; asteroids in the main asteroid belt between Mars and Jupiter. In the background are thousands of colourful galaxies, some of them billions of light years away. Credit: NASA, ESA, and B. Sunnquist and J. Mack (STScI); CC BY 4.0; Acknowledgment: NASA, ESA, and J. Lotz (STScI) and the HFF Team

What causes these asteroid trails to be so strangely shaped and repeated?

The linked image is a composite of several long duration exposures captured by the Hubble Advanced Camera for Surveys of the same region of the distant sky. Each individual exposure was about ~2200 seconds long. Nearby objects such as asteroids would have been subject to the proper motion of the asteroid with respect to the Earth and parallax due to the Hubble being in orbit about the Earth during those ~2200 second long exposures.

Each individual streak comprises both proper motion and parallax effects. The parallax effects are what make the streaks have a nonlinear shape. One orbit later (~5725 seconds), another ~2200 second long exposure showed a similar path for the same asteroid, but shifted a bit because of the asteroid's proper motion. This is why the composite image shows twenty streaks (per the linked article; I see sixteen) for only seven asteroids.


Conceptual Doubt regarding the calculation of the Solar Constant The Solar constant is the intensity of the solar radiation in the
  upper atmosphere. It's value is about $1400$ $Wm^{-2}$.

Now we begin by stating that the Power radiated by the sun is about $3.9*10^{26}$W. Then we using the Formula $I=\frac{P}{4\pi d^2}$, we compute the Intensity. (Here d is $1.5*10^{11}$m).

I understand the mathematics but have trouble in interpreting the meaning of the solar constant. The Intensity formula used above is generally depicted by authors as follows (as per my understanding): There is a sphere around the Sun of radius $d$ and if the power is radiated equally in all directions then each point on the Sphere receives an Intensity $I=\frac{P}{4\pi d^2}$. But the Earth is not a point object, so is this assumption valid?

Next, what do you mean by the upper atmosphere? I've come across another version:

  The Solar constant is the intensity of the solar radiation in when the sun is directly overhead. It's value is about $1400$ $Wm^{-2}$.

I guess my confusion is due to a lack of an intuitive understanding of this concept and I would be grateful if anyone could provide me with one.

Imagine you have a solar collector measuring one square meter, which can directly measure the power (energy per unit time) striking it. If there are layers of atmosphere above the collector, some of the incoming light is absorbed or reflected by particles in the atmosphere, so you won't measure as much power as if you place the collector in space at the distance of Earth's orbit, with no atmosphere between you and the Sun. But if you place the collector in the upper atmosphere of the Earth, the density of atmospheric particles above you is so small that they absorb/reflect only a negligible amount of power. Orientation is also an issue--you'll get the maximum power if you orient the plane of the collector so that it's perpendicular to the direction of the incoming rays. If it's not perpendicular, then if you approximate the incoming radiation as a set of parallel rays which each carry an equal amount of power (a standard approximation in optics) then fewer rays strike the collector when it's at some angle that's not perpendicular to the rays. All this is illustrated in the following diagram (from this page), with the two possible orientations of the solar collector illustrated in gray, and the light rays illustrated as black lines:

The solar constant is defined to be the power you get on such a surface of unit area if it's not being obstructed by atmospheric particles, and if it's oriented perpendicular to incoming rays so that it receives the maximum possible power at that location.

One aspect of the orientation issue that's important to understand is that if you orient your solar collector so that it's parallel to the ground (so that it would be lying flat if it was actually at ground level), then it will only receive the maximum power if the surface of the Earth is itself perpendicular to incoming rays from the Sun at that point, which is not true in the above diagram. It would only be true at the point on Earth that's the dead "center" of the Earth as seen by an observer looking at it from the Sun's position, or the point where the gray line connecting the centers of Earth and Sun intersects with surface of the Earth on this diagram:

That is also the only point on Earth where an observer on the surface would see the Sun "directly overhead", at the very top of the apparent dome of the sky. So even if you stripped away the Earth's atmosphere, only at this point on the Earth would the power reaching a square meter of the surface be equal to the solar constant, the average power per square meter over the entire illuminated half of the Earth's surface would be less than the solar constant since most of the surface is at an angle relative to incoming rays. Incidentally, the angle of the surface relative to incoming rays is also the main reason why at Northern latitudes, it's warmer in the summer (when the surface at a Northern latitude is closer to being perpendicular to oncoming rays) than in the winter (when the surface is at a more inclined angle relative to incoming rays), as illustrated in a diagram from this page:

It's true that since the Earth is not a point, an observer in the upper atmosphere directly above that point on the surface would not be at exactly the same radius from the Sun as the radius of Earth's orbit (defined in terms of the position of the center of the Earth), but if you actually do the calculation of the solar constant at the radius of Earth's center vs. the radius of the point on the surface that's closest to the Sun, the difference is very small so it can be ignored if you're only stating the solar constant to nearest Watt/meter^2 or even the nearest tenth of a Watt/meter^2. The nominal solar luminosity is defined to be $3.828 \times 10^{26}$ and 1 AU is defined as 149597870700 meters, and if you use the formula you mentioned to calculate the solar flux at 1 AU it works out to 1361.1664654 watts per square meter. The radius of the Earth at the equator is about 6378 meters, so at an equinox when the Sun is directly overhead at the equator, an observer seeing the Sun directly overhead should be at a distance of 149597870700 - 6378 = 149597864322 meters, if you use the same formula to calculate the solar flux at this radius you get 1361.1665815, so it only makes a different if you want precision beyond the second decimal place.


Calculating the Angular Momentum of a planet I want to calculate the Angular Momentum of any planet at any point on the orbit around the Sun.
Example: I want to calculate the Angular Momentum of Pluto today, on Jan 27, 2021 00:00:00 Hrs
Where can I get the actual data to calculate this?
I will need Mass, Velocity and Radius to calculate. Do I need the θ too?
Is it readily calculated by NASA?
http://hyperphysics.phy-astr.gsu.edu/hbase/amom.html

The big essential fact about momentum is that it is conserved (in the presence of a central force, as is the case here). So the Angular momentum of Pluto today is the same as it was yesterday, and the same as last year and (excepting perturbations) the same as it has ever been.
It is easiest to calculate for a body that is moving perpendicular to its position vector. This is always true for circular motion, about the centre of the circle.  It is not true for elliptical motion, except at apoapsis and periapsis.
At these times $L = mvr$.  For Pluto the periapsis speed is $v = 6.10\, km/s$ the distance is 4.44 billion km and the mass is $1.31 × 10^{22}$ kg.  To get the angular momentum you've got to multiply them together. If you want SI units, convert those km to m first.
The angular momentum today is the same.
Alternately you can use the relationship
$L= \sqrt{\mu \ell}$ where $\mu = GM =1.33×10^{20}$ and $\ell$ is the semi latus rectum or $\ell=a(1+e^2)$, and you have to plug in the semimajor axis for Pluto and eccentricity of its orbit.
Or you can look it up, since it is constant $L=3.6\times 10^{38}$


Energetics of Titans Tholin haze So, I've googled and wiki'd on this, but no clue, so I turn my question to the community:

Given Titan's nice orange haze that is supposedly composed of Tholins and strangely coincides with its stratosphere (taken from wikipedia)

I got to think whether the haze takes the role of clouds on earth when it comes to energetics. From this a bunch of questions arose:

Do we know whether the haze is responsible for heat trapping and thus, the formation of a stratosphere on Titan?
What about radiation transport modelling with those particles, is this successful?
Also I'd be interested in what we know about the stability of Tholins. Could they survive harsh irradiation that atmospheres on Hot Jupiters do encounter?

I would like to be able to make a statement such as "Dust" or "Haze-controlled" energetics for the stratosphere of Titan, if this is so. It would be nice if you helped to clear this up.

This is a part answer to your question, based on some recent research of the photochemical behaviour modelled and observed for Titan's tholin haze and modelling of Titan's stratosphere.

The process appears to begin, according to the paper Ice condensation layers in Titan’s Stratosphere (Barth, 2012) (Abstract only - paywalled), with

  Photochemical destruction of methane along with the destruction of nitrogen molecules from energetic electrons in Titan’s upper atmosphere result in the production of a number of hydrocarbon and nitrile compounds which may be capable of condensing at the colder temperatures of Titan’s lower stratosphere.

then, according to the paper Laboratory experiments of Titan tholin formed in cold plasma at various
pressures: implications for nitrogen-containing polycyclic aromatic
compounds in Titan haze (Imanaka et al. 2004), in particular in reference to Titan's stratosphere,

  . In the stratosphere (100–
  300 km), further chemical reactions are induced by the catalytic
  $CH_4$ dissociation by such molecules as $C_2H_2$ and
  $C_4H_2$ absorbing the long UV (> 155 nm) irradiation

The significance of these UV absorbing molecules is explained in the article Photochemical activity of Titan’s low-altitude condensed haze (Gudipati et al. 2013) (Abstract only - paywalled), they state that tholin haze could form on condensed aerosols in Titan's atmosphere, demonstrating that, at least part of Titan's atmosphere is photochemically active. Through modelling, they found that

  Detected in Titan’s atmosphere, dicyanoacetylene ($C_4N_2$) is used in our laboratory simulations as a model system for other larger unsaturated condensing compounds. We show that $C_4N_2$ ices undergo condensed-phase photopolymerization (tholin formation) at wavelengths as long as 355 nm pertinent to solar radiation reaching a large portion of Titan’s atmosphere, almost close to the surface.

and evidence of these ices is suggested in the article Titan’s aerosol and stratospheric ice opacities between 18 and 500 μm: Vertical and spectral characteristics from Cassini CIRS (Anderson and Samuelson, 2011) stating that the ices and aerosols

  appear to be located over a narrow altitude range in the stratosphere centered at ∼90 km. Although most abundant at high northern latitudes, these nitrile ice clouds extend down through low latitudes and into mid southern latitudes, at least as far as 58°S.


What's the oldest possible age of the universe? (LCDM parameter range) What would happen if you took today's six-parameter LCDM model down to the 5% confidence level?  Understanding that the number of permutations is high, I would be particularly interested just in the present age of the universe t0---- what are t0's higher and lower limits, assuming that all five of the other parameters are leaning all the way to the plausible floor at the 5% CL either to increase t0 together, or to decrease t0 together?

Q: What would happen if you took today's six-parameter LCDM model to the 95% confidence level? Understanding that the number of permutations is high ...

NASA's Graphical Parameter Comparisons webpage has this to say:

Sources in the literature for values of these selected parameters determined between the years 1996 and 2016 are shown in the Data References Table. It would be a formidable task to include all results from all experiments and every paper. Therefore, results presented here are representative, and there are multiple caveats associated with those choices. For example, we do not guarantee that the datapoints are statistically independent of each other, although the choice of data sources is motivated to some extent by a wish to include independent determinations. There also has been no attempt to estimate what fraction of the parameter uncertainties are due to cosmic variance, although this should clearly be a consideration.
We discuss the individual parameters in pages that follow, and include an accompanying graphical history for each. Vertical gray lines shown in the graphical histories are weighted averages of the WMAP and Planck data points rather than the entire sample. This is done as a visual aid, rather than a statement that this is the current best value: the WMAP and Planck data values serve as a common thread in each of the plots. In a few cases, liberties were taken in the quoting of asymmetric error bars from published results as symmetric error bars on the plots. For illustrative purposes, the differences are negligible. Quoted uncertainties are 68% confidence limits.
Data references Table - Contributed by the NASA LAMBDA Archive Team.

The published data isn't accurate enough to reach the confidence level you require.
Prof. Wright of UCLA has a webpage about the age of the universe using actual age measurements, not estimates from cosmological models, but a more authoritative source would be NASA's Lambda Archive Team which has a page on $t_0$.
There are many ways to tweak the values and make the calculation, the computed value is somewhat dependent of the exact cosmological model assumptions. Within the standard ΛCDM model framework, CMB measurements, combined with BAO and SNIa measurements of the expansion rate, are presently in agreement and tightly constrained. The gray vertical line, representing the weighted average of WMAP and Planck data points, is positioned at $t_0$ = 13.796 Gyr.
This graphic from their website sums it up: