Fastest recorded apparent motion of a comet or asteroid seen from Earth (degrees/day)? If I've done my maths correctly in this answer comet 2I/Borisov (C/2019 Q4) was seen to move about 0.2 degrees in 7 hours, which is almost 0.7 degrees per day.

I'm assuming this is pretty rapid motion for a comet, but I'm wondering if comets (or asteroids) have been observed to move even faster than this relative to the celestial sphere.

Yes I know the speed of apparent motion depends on incidentals like how close the comet happens to pass to the Earth. I'm just wondering if any particular comet stands out as been particularly fast, possibly even difficult to track because perhaps some software didn't allow such a large offset in tracking relative to the stars.

Question: What is the fastest recorded apparent motion of a comet or asteroid, seen from Earth (degrees/day) relative to the celestial sphere?

note: @JamesK's comment reminds me that I would like to exclude meteor tracks. For the purposes of this question the recorded tracks should be produced reflected light, either of the Sun or of radar illumination, and not be something burning up in our atmosphere.

This is more of a recent example than a record.
On 2019-07-25, asteroid 2019 OK passed about 65000 km from the Earth at a relative speed of 24.5 km/s.
The Minor Planet Center lists multiple
observations from sites in Italy and Armenia an hour before closest approach.
Using pairs of these (streak endpoints?) by the ISON-Castelgrande observatory, I compute motions of 29.8° per hour at 00:23 UT and 31.7° per hour at 00:26 UT.
The MPC ephemeris service
gives sky motions of 30.6°/hr and 32.5°/hr from site L28 at those times.
If someone had observed the asteroid from (25°S, 85°E) in the Indian Ocean at 01:22 UT, they would have seen it move 76° per hour.

Update: On 2020-08-16, tiny (H=29.8) asteroid 2020 QG passed by at 12.3 km/s, about 9320 km from the Earth's center or 2950 km from the surface.
A NASA news article said this was a new record for close approach distance.
An observer in the right place at 04:09 UT would have seen a maximum sky motion of
$$\mathrm{\frac{12.3~km/s}{2950~km} = 0.0042~radian/s = 860^\circ/hr}$$
From pairs of actual observations reported to MPC,
ZTF Palomar (I41) saw a sky motion of  566°/hr at 10:23 UT,
and ATLAS Mauna Loa (T08) recorded 439°/hr around 11:46 UT.


How does one determine the effective temperature of a star from its spectrum? Determining effective temperature of a star is in general a non-trivial task. Simple reason for this is that we can only study the electromagnetic radiation from a star, but not the temperature directly. The complexity is due to the fact that the radiation is produced in stratified stellar atmospheres, which are only partially characterised by stellar temperature, but also by many other factors, such as stellar mass, elemental abundances, stellar rotation, etc. What is more, the temperature of atmospheres varies with depth, whereas effective temperature is just a number.

From the other hand, temperatures and magnitudes are the most important quantities, characterising stars.

So, the question: How exactly does one use the spectrum to extract the information about the temperature of a star? By temperature here I mean effective temperature, or even the temperature profile of the atmosphere.

Note: This a rather textbook question. I created it because I have encountered a good existing answer by @Carl, previously posted in a bit less textbook discussion How well can we in principle determine $T_{\textrm{eff}}$ of a star? . This question seems to be a much better place for the answer.

Temperature ($T_{eff}$) can be quite tricky to determine accurately as it interrelates to a number of other fundamental measurements.
Firstly, remember that the spectrum we observe from stars are pin-point, they give us the entire overall result and not a specific location or part of the star. We need to dissect the various parts to arrive at the fundamental parameters. We arrive at our results by iterating the values of the fundamental parameters until a model spectrum matches the true spectrum we observe. The issue is, like you say, the existence of a whole lot of uncertainties.
The first of these (although it doesn't have a large effect) is the Uncertainty Principle itself. This creates natural line broadening due to the emitted photon having a range of frequencies. The width of the line is determined by;
$$\Delta E \approx \frac{h}{T_{\text{decay}}}$$
where $\Delta E$ is the uncertainty in the energy,
$h$ is the Planck constant, and
$T_{\text{decay}}$ is the amount of time the electron stays in a high energy state before decaying.
Fundamental parameters
The rotation of the star causes a Doppler shift effect on the line spectra making it broaden. The faster the rotation, the broader (yet smaller) the line. Like the Uncertainty Principle, this is natural broadening as it doesn't impact the abundance of any particular element in the star.
Measuring the rotational velocity ($V_{\text{proj}}$) depends on both its axis of rotation and our line of sight to the star. Therefore, we use a combination of both velocity about the equator ($v_e$) and the star’s polar inclination ($i$) to determine the projected radial velocity;
$$V_{\text{proj}} = v_e \sin i$$
Temperature ($T_{eff}$)impacts wavelength in such a way that higher temperatures impart higher random motions on the atoms. When these photons collide with an atom, they can cause the atom to become ionised, i.e. lose an electron. Different energy levels (and therefore temperature) will create different abundances at the various ionisation stages of atoms.
The temperature of the stellar photosphere decreases as we move away from the core. Therefore the line profile represents a range of temperatures. The wings of the line arise from deeper, hotter gas which displays a larger range of wavelengths due to increased motion. The higher the temperature, the broader the wings of the line profile ([Robinson 2007, pg 58][1]).
Here you can see the effect of various temperature values on the synthetic spectral line of FE I 6593 A. Red: $T_{eff}$ = 4000K; Black: $T_{eff}$ = 5217K; Blue: $T_{eff}$ = 6000K;
$T_{eff}$ on spectral lines" />
Microturbulence ($v_{\text{mic}}$) is the non-thermal localised random motion of the stellar atmosphere. It works in a similar way to temperature - an increase in the motion of atoms creates a wider range of wavelengths observed and therefore broader line profiles.
In strong lines, saturation can occur when there aren’t any more photons to be absorbed. As microturbulence in these areas increases, it presents more opportunities for photons to be absorbed. This broadens the wings of the line profile increasing the overall strength of the line. We can use this fact to determine $v_{\text{mic}}$, by ensuring that the strength of the lines (equivalent width) have no correlation with their abundances.
Finally, surface gravity which is a function of the star’s mass and size:
$$\log g = \log M - 2 \log R + 4.437$$
with $M, R$ being in solar units and $g$ in cgs.
A star with a higher mass but smaller radius will invariably be denser and under greater pressure. By definition, denser gas has a higher number of atoms per unit of area (abundance), leading to stronger spectral lines.
A gas under pressure provides more opportunities for free electrons to recombine with ionised atoms. For a given temperature, ionisation is expected to decrease with an increase of surface gravity, in turn increasing the abundance of atoms in the neutral or low ionisation states.
The measuring of $T_{eff}$
As we've seen, there are a number of ways in which a spectrum of a star can be altered. The one you're interested in is temperature. As temperature is interrelated to all the other fundamental parameters, we need to treat them together as a whole and tease out the value of $T_{eff}$.
We begin with a synthetic spectrum and modify its properties iteratively until it matches the shape of the star’s spectrum. Adjustments of one parameter will invariably affect the others. The spectra will match when the temperature, surface gravity, and microturbulence values (amongst others) are correct. This is obviously very time consuming although programs exist to help.
Atmospheric properties can also be determined by other less time consuming means. Photometric colours can be used as a proxy for temperature, and absolute magnitudes for surface gravity. However, these determinations can suffer from inaccuracies due to interstellar extinction and are at best a close approximation.
[1] Robinson, K. 2007, Spectroscopy: The Key to the Stars (Springer)


Astronomy & machine learning. Where to start? I am a Computer Engineering major and would like to learn more about machine learning methods in Astronomy. What books or other resources could I use?

This recent KISSCaltech lecture by Prof. Rob Fergus at the NYU might be of interest. "An Overview of Machine Learning Approaches: Applications to Exoplanet Detection".


How massive does a moon have to be to have a stable magnetic field? Is there a minimum mass or other minimum properties necessary for a body to have a strong, stable dynamo to create a magnetic field conducive for life?
For example, would it be possible for Titan to have a magnetic field if it had different core characteristics?

Scientists (and science fiction writers) have speculated about the possbilities of life under the surfaces or on the surfaces of large moons in our solar system or large exomoons of large exoplanets in other star systems.
So a good place to find any limits on the possible properties of moons that can have magnetic fields is a scientific discussion of the possibilities of life on exomoons.
Like, for example, "Exomoon Habitability Constrained by Illumination and Tidal Heating" by Rene Heller and Roy Barnes, Astrobiology, 2013.
https://arxiv.org/ftp/arxiv/papers/1209/1209.5323.pdf[1]
In section 2. Habitablity of exomoons, they discuss the qualities, including mass, necessary for exomoon habitability.

A minimum mass of an exomoon is required to drive a magnetic shield on a billion-year timescale (Ms ≳ 0.1M⊕,
Tachinami et al. 2011); to sustain a substantial, long-lived atmosphere (Ms ≳ 0.12M⊕, Williams et al. 1997; Kaltenegger
2000); and to drive tectonic activity (Ms ≳ 0.23M⊕, Williams et al. 1997), which is necessary to maintain plate tectonics and
to support the carbon-silicate cycle. Weak internal dynamos have been detected in Mercury and Ganymede (Kivelson et al.
1996; Gurnett et al. 1996), suggesting that satellite masses > 0.25M⊕ will be adequate for considerations of exomoon
habitability. This lower limit, however, is not a fixed number. Further sources of energy – such as radiogenic and tidal heating, and the effect of a moon’s composition and structure – can alter our limit in either direction. An upper mass limit is
given by the fact that increasing mass leads to high pressures in the moon’s interior, which will increase the mantle viscosity
and depress heat transfer throughout the mantle as well as in the core. Above a critical mass, the dynamo is strongly
suppressed and becomes too weak to generate a magnetic field or sustain plate tectonics. This maximum mass can be placed
around 2M⊕ (Gaidos et al. 2010; Noack & Breuer 2011; Stamenković et al. 2011). Summing up these conditions, we expect
approximately Earth-mass moons to be habitable, and these objects could be detectable with the newly started Hunt for
Exomoons with Kepler (HEK) project (Kipping et al. 2012).

Their sources for those mass limits are:
Tachinami, C., Senshu, H., Ida, S. 2011, ApJ, 726, 70
Williams, D. M., Kasting, J. F., Wade, R. A. 1997, Nature, 385, 234
Kaltenegger, L. 2000, ESA Special Publication, 462, 199
Gaidos, E., Conrad, C. P., Manga, M., Hernlund, J. 2010, ApJ, 718, 596
Noack, L., Breuer, D. 2011, EPSC-DPS Joint Meeting 2011, http://meetings.copernicus.org/epsc-dps2011, 890
Stamenković, V., Breuer, D., Spohn, T. 2011, Icarus, 216, 572
So Heller and Barnes beleive that the approximate lower mass limit for a habitable planet or moon or other type of world would be about 0.25 times the mass of Earth.  Note that they suggest that a world with of mass of only 0.12 Earth mass could retain a substantial atmosphere for a long enough time.
This is a somewhat lower mass than the 0.195 Earth mass suggested by Stephen H. Dole in Habitable Planets for Man, 1964), as the lowest possible mass for a planet to retain a dense atmosphere for geologic eras.  I don't know the reasons for that discrepancy.
https://www.rand.org/content/dam/rand/pubs/commercial_books/2007/RAND_CB179-1.pdf[2]
Returning to the subject of the lowest possible mass world to generate a magnetic field, it may be noted that the smallest worlds in our solar system to have detectable magnetic fields are Mercury and Ganymede.
The planet Mercury has a mass of only 0.06 Earth mass, but has an extremely large iron core relative to its size.  It rotates with a period of 58.65 Earth days, much slower than any other planet except for Venus.
The planet Mars has a mass of 0.11 Earth, almost twice that of Mercury, and rotates much faster than Mercury.

Mars lost its magnetosphere 4 billion years ago,[175] possibly because of numerous asteroid strikes,[176] so the solar wind interacts directly with the Martian ionosphere, lowering the atmospheric density by stripping away atoms from the outer layer.

https://en.wikipedia.org/wiki/Mars#Atmosphere[3]
The planet Venus has a mass of 0.72 Earth, about 12 times the mass of Mercury, but a much longer rotation period of 243.02 Earth DAys.

In 1967, Venera 4 found Venus' magnetic field to be much weaker than that of Earth. This magnetic field is induced by an interaction between the ionosphere and the solar wind,[104][105] rather than by an internal dynamo as in the Earth's core. Venus' small induced magnetosphere provides negligible protection to the atmosphere against cosmic radiation.

https://en.wikipedia.org/wiki/Venus#Magnetic_field_and_core[4]
Ganymede, the largest moon in the solar system, has a mass of 0.025 Earth, and has a weak magnetic field.
Callisto has a mass of 0.018 Earth, and Titan, largest moon of Saturn, has a mass of 0.0225 Earth.  They don't have detectable magnetic fields.
Their lack of magnetic fields may be due to their lesser mass compared to Ganymede.  Since all three, and all moons of Jupiter and Saturn which orbit closer than them, are tidally locked to their planets, their rotation periods are equal to their orbital periods.
So the rotation period of Ganymede is 7.154 Earth days, that of Titan is 15.945 Earth days, and that of Callisto is 16.689 Earth days.
So perhaps the slower rotation periods of Titan and Callisto prevent them from having magnetic fields.  All of moons of Jupiter closer than Ganymede, and all the moons of saturn closer than Titan, have rotation periods shorter than 7.154 Earth days, but even the most massive, Io, has only 0.6 the mass of Ganymede.
And of course the internal composition of a world, and other factors, will also influence whether it has a magnetic field and how strong it is.
If it is desired that an exomoon have a magnetic field to shield it from cosmic rays and its star's stellar wind, and so to keep it habitable, it is possible that an exomoon might not need its own magnetic field.  It could orbit within the magnetic field of its planet and be protected by the magnetic field of the planet.
Such a possibiity is discussed in "Magnetic Shielding of Exomoons Beyond the Circumplanetary Habitable Edge". Rene Heller & Joge Zuluaga.
https://www.researchgate.net/publication/256422936_Magnetic_shielding_of_exomoons_beyond_the_circumplanetary_habitable_edge[5]
And anyone interested in the potential habitability of exomoons should check my answer at:
https://worldbuilding.stackexchange.com/questions/198117/what-if-a-moon-orbits-far-away-and-extremely-fast/198135#198135[6]


Cosmological redshift vs doppler redshift I'm reading Harrison's "Cosmology: Science of the universe" because Harrison focuses on the distinction between cosmological redshift (he calls it expansion redshift) and the Doppler redshift.

He states that "they [Doppler redshifts] are produced by peculiar and not by recession velocities"
and
"[expansion redshifts] are produced by recession and not
peculiar velocities".

I understand the concepts of both kinds of redshifts but have a hard time understanding this strict separation. Please correct me, where/if I am wrong:

Suppose a galaxy has no peculiar movement. This means that its position will stay (approximately) the same in comoving coordinates. In fact though ( proper distance), it IS moving away from Earth with recession velocity V, caused by the expansion of the universe. So it should have a Doppler effect based on this recession velocity AND a cosmological (expansion) redshift should also take effect because the light gets stretched with the expansion of the universe. Even though the recession velocity is not due to a peculiar movement, it means that the source of light is moving away from the observer and hence the light should be redshifted and on top the light gets redshifted on its way through expanding space.

Please correct me or tell me if I am right or wrong, I have spent a lot of time reading but still don't fully understand this. Thanks.

After considering @benrg's comments, I realize that my first answer contained too strong statements about the relation between the two redshifts. I try here to moderate my answer, but you might want to accept their answer instead.

It is common to think of the two redshifts as having nothing to do with each other. Doppler shifts arise when the observer and/or the emitter moves through space, whereas the cosmological redshift can be derived considering stationary emitters and stationary observers in an expanding space.
Because the cosmological redshift doesn't involve movement through space, it is often considered completely different from the Doppler. However, it is also possible to derive the cosmological redshift by considering it as infinitely many infinitesimally small Doppler shift (e.g. Lewis 2016). I admit that I'm not well enough versed in general relativity to be certain about my statements, but just because an infinitesimally small patch of spacetime is flat doesn't necessarily mean that infinitely many such patches add up to be flat. However, as @benrg says, in GR there is only one redshift.
Different or the same?
The reason I think it makes sense to view the Doppler shift and the cosmological redshift as two separate mechanisms is the following:
In principle you could have a universe (non-capitalized, since it's not our universe, the Universe) that were static when a distant galaxy emitted a photon, then at some point expanded quickly by a factor of 2, and then again is static. In this hypothetical case, the observer would still measure the photon to have been redshifted by a factor of 2 (i.e. $z=1$).
That this is true can bee seen from considering the mathematical derivation of the cosmological redshift (see e.g. here) which involves an integral, the result of which only depends on the initial and the final state, not on the expansion history.
In contrast, if you and your friend stand still with respect to each other while your friend shines her flashlight at you, then run away from each other with a relative velocity of $0.6c$, then stand still before you receive the light (i.e. analogously to the hypothetical universe above), then you would measure no redshift; you wouldn't measure the special relativistic Doppler shift of $z+1 = \sqrt{\frac{1+v/c}{1-v/c}} = 2$ that you would if you were receding from each other while either emitting or observing.
In the real Universe, galaxies move through space (i.e. they change their comoving coordinate $\chi$), and space expands (i.e. the scale factor $a$ evolves). If the physical distance to a galaxy is
$$
d = a \chi,
$$
then the change in this distance gives their total velocity wrt. us, and is obtained through differentiation:

$$
\begin{array}{rcl}
v_\mathrm{tot} & =      & \dot{a}\chi + a\dot{\chi} \\
               & \equiv & v_\mathrm{rec} + v_\mathrm{pec},
\end{array}
$$
where dots denote differentiation with respect to time, and the two terms have been identified as the cosmological recession velocity, and the peculiar, "normal" velocity. Each of these terms give rise to a redshift, but through two very different mechanisms. Only the latter term is called a Doppler shift.


Is it coincidence that Einstein predicted exactly twice the outcome of Newton for the gravitational deflection of light? During a solar eclipse Sir Arthur Eddington measured the deflection of star light by the Sun. Einstein predicted a deflection of 1.74 arc seconds, but according to Newton the deflection should have been 0.86 arc seconds. So is it a coincidence that the prediction of Einstein is twice as large or is there a kind of law involved which makes Einstein's outcome of his GR twice as big as that according to Newton?

As far as I know Newtonian prediction is based on Soldner's work in 1801 and Einstein 1911 also predicted around $0.87$arcsec (he was also accused for plagiarizing Soldner's work the article:Historical Note on the Problem of Light Deflection in the Sun's Gravitational Field explains why the accusations were wrong).

The Newtonian approach accounted for the space curvature, as did special relativity which Einstein first used, but it missed the part that not only space is curved but space time is curved. So the factor of two was because of the spatial curvature, that's because by Minkowski metric we consider that light travels flat so that $\rm{d}s^2=0$.The correct answer would be to use the Schwarzschild metric and get the correct answer.

This paper gives more insight on the formulas and the correct approach. Gravitational Bending of Light


Is the Jupiter-Sun system considered a binary system of some type? Since Jupiter is very massive, it is the only planet (in our solar system) that has a center of mass with the Sun that lies outside the volume of the Sun. (Source)

If Jupiter was a star, they would form a « binary star ».

If the Sun was a planet, they would form a « double planet ».

Since the Sun is a star and Jupiter is a planet, does this have a particular name?

Does Jupiter have a special status or a particular effect in our solar system because of its heavy mass?

Since Jupiter-Sun's center of mass lies outside the volume of the Sun, that means that the Sun moves around that center of mass. Does this have an effect on Mercury, Venus, Earth and Mars orbits?

I'm not sure I understand your question entirely, but i'll do my best to offer a decent answer. It's true that the composition of Jupiter is very similar to that of the Sun (very similar approx. $H$ and $He$ abundance and pretty similar in density). The problem is that Jupiter is not nearly massive enough to have the internal pressure and temperature to undergo nuclear fusion. Jupiter doesn't have any particular special status aside from being the King of Planets in our solar system.

As for the last part of your question, all objects orbit around a center of mass. Though because the Sun is much more massive, the center of mass lies very close to the center of the star (except in the case of Jupiter, where the CoM lies outside of the sun and is approx the length of its radius). This is why all planets in our solar system orbit around the sun. This will indeed cause a slight perturbation of orbital alignment, but I don't believe it's significant.


What Right Ascension and Declinations are observable by a telescope from a particular location? I'm new to Observational Astronomy, and I just started observations with a telescope and read about RA and Declination(celestial co-ordinates). Although I think I understood the concept, I'm confused about what RA/Dec is visible from the location at present(Summer Sky).

From what I have understood, since I'm in Sendai, Japan(~38.26N, 140.86E), the +38 degree declination is at my zenith at midnight. So almost -30 Declination will be at my horizion. That should mean that I can see stars from -30 Declination to 90 Declination at my location. Is that correct?

What about RA though, what range of RAs will it be possible to view while its dark(2000 hours to 0300 hours)

The declination limit, looking south, is about -52 degrees.  This is basically your latitude minus 90.

At 8pm, due south will be about RA 12hrs, and at 3 am it will be about 19hrs.  The eastern horizon will be about 6 hrs ahead, and that increases as you look further north, until you hit the circumpolar region, when all RAs are visible.


How common are exoplanets in distant orbits? Are there many single transits in the Kepler data, which might be exoplanets with a longer orbital period than the time they have been observed for transits? Can any conclusions be drawn about how common they are? Is there reason to believe that the outer solar system is a common configuration? I suppose that slow moving planets are hard to detect also with doppler measurements. Will the upcoming telescopes aimed at nearby stars be able to find 10+ AU orbital radius planets?

There are many planets known which have orbits longer than the longest exoplanet orbital periods found by Kepler. These planets were discovered using the "doppler wobble" or radial velocity technique. The plot below (a few months out of date now) shows many planets orbiting with similar periods to Mars and Jupiter. The red points we discovered by transits (including Kepler planets), whilst the green were discovered by the doppler technique.

The sensitivity of current observational doppler techniques (shown by the blue line) means that so long as we observe for a sufficiently long time (decades, in order to see an oscillation in the radial velocity), we will be able to detect Jupiter-like (or even a bit lower-mass) planets out to much longer periods.

The frequency of planets in wider orbits is also probed well by surveys for gravitational microlensing.

Single transit events cannot be securely identified as being caused by planets. I think there are a number of other possible causes, both astronomical and instrumental, that might also be to blame. Nevertheless I'm sure somebody will analyse these in some statistical way at some point to draw some inferences, but at the moment, the securest data are from the radial velocity technique. These show that the solar system is not necessarily unusual at all - there are plenty of stars that have Jupiter-sized objects in Jupiter-sized orbits.

According to the review by Howard et al. (2013), from radial velocity surveys, about 10 per cent of G/K stars have a 0.3-10 Jupiter-mass planet in orbit between 0.03 and 3 au from the parent star and around 7% have a Jupiter-sized planet orbiting between 3 and 20 au. This latter number probably has a significant error bar at the moment, but it is consistent with the information from microlensing surveys  (e.g. Gould et al. 2010). This latter paper has an interesting statistic in the abstract. they say that if all stars had a solar system like ours, they would have detected 18.2 "events" - 11.4 due to Jupiters,  6.4 due to Saturns and 0.5 Uranus/Neptunes; they would also have detected 6 "two-planet" events. In actual fact they detected 6 events, one of which was a two-planet event, and thus conclude that a broad-brush figure for the frequency of "solar system occurrence" is 1/6.


Longitude of Perihelion at J2000? My textbook (and lots of other sources) shows the earth's longitude of perihelion at the epoch around 103 degrees. I understand that this means (in the neighborhood of J2000) the earth will reach perihelion about 103 degrees after the vernal equinox, however, this would be very close to aphelion (July 6ish) not perihelion (January 3ish).

Put another way: If the vernal equinox is March 20th & perihelion is January 3rd, shouldn't the longitude of perihelion be about -76 degrees (or 284 degrees)?

Help! What am I missing?

ANSWER--Credit to @Mike_G below, restated here because the final answer is in the comments: "At the March equinox, the alignment is Earth - Sun - first point of Aries." Beginner mistake: I was trying to find the earth's position in time but I thought the March equinox was when the earth was between the Sun & first point of Aries, which is the wrong arrangement.

The Earth's heliocentric ecliptic longitude is opposite to the Sun's geocentric ecliptic longitude. If the Earth's perihelion is at 103 degrees heliocentric longitude, then the geocentric longitude of the Sun at that time is 283 degrees, or 13 degrees past the December solstice.


Which telescope should I buy? I'm wondering which telescope I should buy, by the way, I'm new to astronomy and if you know any others that are under $100, It'd be appreciated. You don't need to take the price difference into consideration when replying, thank you!
1 : http://www.dshop.com.au/buy/astronomical-telescope-675x-magnification/F90060M?gclid=CjwKCAjw-qbLBRB7EiwAftBCI7h_7lH3E9goJG4Top0N-OHaraxB0cM8IiZgfygRkHD8DZ1ThARCdRoCzA8QAvD_BwE
Features:
Maximum magnification: 675x
Large aperture offers clearly illuminated views
Stable construction provides solid support and control
Metal tripod with slow motion control rod
Strong performance in both high and low magnification
Easy sky focusing with finderscope
Specifications:
Aperture: 60mm (2.36inch)
Focal Length: 900mm.f/15
Finderscope:  5x24
Diagonal Prisms: 90degree
Slow Motion Control Rod For Easy Vertial Micro Adjustment
Height: 128cm
Eyepiece: SR4mm H12.5mm H20mm
3X Barlow Lens, 1.5X Erector
Telescopic Aluminum Tripod
2 : https://www.bhphotovideo.com/c/product/617033-REG/Celestron_21035_Travel_Scope_2_7_70mm_Refractor.html
Specifications:
70mm Achromatic Refractor
400mm Focal Length, f/5.7 Focal Ratio
Anti-Reflection Fully Coated Optics
1.25" Rack-and-Pinion Focuser
Manual Alt-Azimuth Mount with Lock
10mm and 20mm Eyepieces, Erecting Prism
5x24 Finderscope
Aluminum Photo-Style Tripod
Backpack Holds OTA, Tripod, Accessories

Of the two, I would go with the Celestron - it's from a named and respected manufacturer with sensible specifications and relevant additional equipment. The eyepiece selection is thoughtful.

However, you pay the price because the Celestron's focal length at 400mm is much more limited. The best magnification you'll be able to get is 40x without a Barlow (good for the Moon but rather anaemic for the planets).

The other scope promises much higher magnifications but I worry that the manufacturer's name isn't stated, that there seems to be a number of redundant extra items and about the overall build quality.

Personally, I would consider the https://www.bhphotovideo.com/c/product/485261-REG/Celestron_21061_AstroMaster_70_AZ_2_7_70mm_Refractor.html as an alternative, though it is a little more expensive.


Why did the Event Horizon Telescope take so long to take a photo of a black hole? The Event Horizon Telescope is in the news now because they are about to release the first ever picture of a black hole. They have been working on this since 2006. Why did it take 13 years to produce the first picture? What was the critical mass that was recently hit that made the picture now complete?

A single telescope does not have the right angle size to picture the black hole that far away. The scientists had to use 8 telescopes to take a picture of the same point in the sky and merge the data from all sites using super computer. The data retrieval must point to the same object at the same time. Given the diverse condition of the 8 locations, there are only 5 days in a year when all of them have clear enough skies to point to the same location in a year.

Event Horizon Telescope


Implementing the VSOP-2000 Ephemeris For a project I'm doing, I need to implement the VSOP-2000 ephemeris for the Moon. However, unlike VSOP-87, there is little to no documentation online on how to use the data. For example, here is the first line of the x-term for the Moon in VSOP-2000:

  1   0  0  1  0   0  0  0  0   0  1  0  0   0  0  0  0   0  0  0  0     -.1053243070325313D-09     -.2560241101399878D-02

This is very different from VSOP-87, which I know how to use. Are there any resources online that could inform me on how to use this?

Probably the most reliable guide to vsop2000-p11.dat is the reference implementation vsop2000.for.
The input line in question is the first of 5399 k=0 records contributing to $x$.
Those are followed by 5842 k=1 records, 610 k=2 records, and so on up to k=8.
Then $y$ (k=0..8) and $z$ (k=0..7) follow the same pattern.

If I read the Fortran code correctly, the fixed-length (not blank-delimited) record format is:

Columns 1-7 are an index $i$
Columns 9-63 are 17 per-body coefficients $b_{kij}$
Columns 64-72 are three zeroes, ignored
Columns 73-126 are coefficients $s_{ki}, c_{ki}$ in astronomical units

To compute a position for time $t$ in Julian years since J2000:

For each body $j$, compute $\theta_j(t) = \dot{\theta}_j t + \theta_j(0)$, where $\dot{\theta}_j$ is the mean motion and $\theta_j(0)$ is the J2000 mean longitude from vsop2000-exp.txt
For each record $x_{ki}$, compute $\phi_{ki} = \sum\limits_{j=1}^{17} b_{kij} \theta_j(t)$
$x = \sum\limits_{k=0}^8 t^k \sum\limits_{i=1}^{n_k} (s_{ki} \sin \phi_{ki} + c_{ki} \cos \phi_{ki})$
For $y$ and $z$, repeat steps 2 and 3 using the respective sets of records.

This should give you a geocentric Moon position in astronomical units.


Earth's gravitational pull on ISS How do they make International space station to orbit the earth beyond Earth's gravity acting on it? We all know that ISS is rotating at an altitude of just 350km away. How could the ISS escape Earth's gravitational pull?

Your question presumes that the ISS is beyond Earth's gravity, that it has escaped earth's gravitational pull. This is not correct. All objects with mass in the universe affect all other things with mass in the universe, the effect just gets weaker with distance. So the ISS is feeling the effect of gravity from Earth significantly more than the moon is.

The reason the ISS doesn't just fall to the earth, either directly or gradually spiralling towards the earth, is that it is travelling fast enough around the earth that it is continually "missing" earth. It is sometimes described as 'falling' constantly around earth.

If I am to be properly correct though, the reality is that ISS is in fact falling towards the earth, getting closer and closer to Earth all the time. It needs occasional boosts to push it further back out in it's orbit.

Just to blow your mind a little bit: The ISS is pulling on the Earth with the same force that the Earth is pulling on ISS.


Can the Dark Energy be Pauli pressure? Some neutrinos are fermions. There are quite of few of them out there. Could their degeneracy pressure give rise to an expansion force to explain Dark Energy?

There are several reasons dark energy cannot be pressure due to the Pauli exclusion principle.  First of all, pressure does not cause expansion of the universe, because pressure is not a force-- pressure gradients are a force (per unit volume), and the cosmological principle precludes them.  Indeed, the pressure that exists everywhere only appears as a force when it is put into gravity, and there it is an attraction if it is positive, as it is for what you mean by Pauli pressure.  So what you'd need for dark energy is negative pressure, appearing in the gravity.

The second reason is that not only is what you mean by Pauli pressure positive, not negative, it is not any different from normal gas pressure, in the sense that it appears as a consequence of kinetic energy density in exactly the same way as any gas pressure does.  This is widely misunderstood, many people think there is some kind of extra pressure that comes from the Pauli exclusion principle, but this is not the case if you simply track the kinetic energy involved.  If you are tracking the kinetic energy, then gas pressure is 2/3 the kinetic energy density for nonrelativistic particles, and 1/3 for relativistic, and that's all-- it's just the same for "Pauli pressure," there's nothing at all different about it when you track kinetic energy density-- which is generally exactly what you want to do when you have an adiabatic expansion such as for the universe.  Hence, in order for us to understand the pressure from neutrinos, we only need to understand their kinetic energy density, which we think we do because it's pretty easy to track given some assumptions about how neutrinos couple to matter and what their mass is.

You might be wondering, if there's nothing different or extra about "Pauli pressure," then when can it be treated as an extra source of pressure?  Only when you are talking about the temperature, rather than the energy.   If for some reason it made sense to think of temperature as the fundamental constraint, rather than energy, then it would make sense to think of the Pauli exclusion principle as adding pressure.  But even then it's kind of a strange way to think about it, because all the Pauli principle does is constrain how the energy gets partitioned, which in turn affects the temperature.  The chain of causality in any gas is energy + partition -> temperature  and energy -> pressure independently of temperature, so it's kind of weird to do what is often done, and pretend that temperature + partition -> pressure.  It's only that last way that makes degeneracy look like an added pressure, and it would only make sense in a system whose temperature is, for some reason, being controlled.  That is rarely the case in an expanding universe, where the temperature is just whatever it needs to be given the energy and the contraints on its partitioning.


What does a pulsar look like if viewed from above? Have we ever seen a pulsar directly from above or below it's spin axis?

I am wondering if a pulsar seen from this angle would appear to be constantly spinning or if it would just appear like a normal star.

My understanding is that we "see" pulsars (also called "supernova remnants") by the radio and/or other electromagnetic radiations which they emit, which come to us in regular pulses, like the light-beam on the conning-tower at an airport.

This is because the beam of radiation shoots out along the magnetic-axis, while the object spins around its spin-axis, which might be considerably different from the magnetic-axis.

So if we are directly above, it seems that we might not receive any of the radiation, so we might not be able to "see" the object.


How would I define mass in a space I always heard that mass is a measure of inertia during my school days. Is this  applicable to bodies which are in space ? Suppose there is space craft in a space; how would I define mass ?
I read some article and consulted with my teachers; they said mass is similar to conservation of momentum but I do not know how. And then he said mass is independent of frame it is constant, whereas I see momentum is itself a frame dependent quantity and velocity is frame dependent. I am confused how I would actually define a mass in space, what is the correct definition for a mass ?

I am high school boy, I will be thankful if anybody answer my question.

To quote from wikipedia

  There are several distinct phenomena which can be used to measure mass. Although some theorists have speculated that some of these phenomena could be independent of each other,[2] current experiments have found no difference in results regardless of how it is measured:

  Inertial mass measures an object's resistance to being accelerated by a force (represented by the relationship F = ma).
  Active gravitational mass measures the gravitational force exerted by an object.
  Passive gravitational mass measures the gravitational force exerted on an object in a known gravitational field.

  The mass of an object determines its acceleration in the presence of an applied force. The inertia and the inertial mass describe the same properties of physical bodies at the qualitative and quantitative level respectively, by other words, the mass quantitatively describes the inertia. According to Newton's second law of motion, if a body of fixed mass m is subjected to a single force F, its acceleration a is given by F/m. A body's mass also determines the degree to which it generates or is affected by a gravitational field. If a first body of mass $m_A$ is placed at a distance $r$ (center of mass to center of mass) from a second body of mass $m_B$, each body is subject to an attractive force $$F_g = Gm_Am_B/r^2$$ where $G = 6.67\times 10^{−11} N kg^{−2} m^2$ is the "universal gravitational constant". This is sometimes referred to as gravitational mass.[note 1] Repeated experiments since the 17th century have demonstrated that inertial and gravitational mass are identical; since 1915, this observation has been entailed a priori in the equivalence principle of general relativity.

In space the third phenomenon (passive gravitational mass) is not terribly useful, because we may not have a known gravitational field to hand, or both the object being measured, and the instruments doing the measuring may be accelerating under the influence of the same field (this is what happens in orbit).

The others are fine. For measurement purposes it can be helpful to avoid having to measure forces exactly, and this is where conservation of momentum comes in. Suppose you have an object of unknown mass $m$ and an object of known mass, which, for simplicity of calculation we'll assume to be 1kg. Initially they are at rest relative to one another. Now we let some interaction happen between them (it doesn't matter what, so long as no other masses are involved) as a result of which they move apart. We measure the speeds of movement (in a frame in which they were both initially at rest), say $u$ for the unknown mass and $v$ for the 1kg mass (these must be in opposite directions). So conservation of momentum tells us that $mu - 1.v = 0$ or $m = v/u$.


If you're near a black hole and your time is slowed down, would a supernova be observable to both you and someone outside of the blackhole's pull? Let's say that Person A is on a planet orbiting a black hole (like the one from Interstellar) and time is slowed for Person A. Person B is back home on Earth. Now let's say a supernova happens and it's possible for both parties to see it happen at the same time (ignoring the time it would take for the light to get there). Would Person A see the supernova at the same time as Person B? And if so, if Person A is in 1,000 B.C. and Person B is in 2,000 A.D, would the supernova then happen at two different times at the same time?

Time is relative therefore this is a bit of a nonsense question.

But yes, if the gravitational effect near the black hole is strong enough (depends on how close you are) it would affect 'when' the observer would notice it.  This is because the light must travel a 'further' distance through the spacetime to get to you since gravity is distorting it.

You may find this interesting: https://web.archive.org/web/20130424041627/http://www.thebigview.com/spacetime/spacetime.html


What is the most appropriate way to estimate the helium composition of a star? Say we have to estimate the helium content in Proxima Centauri.

We begin by calculating the content of helium in the Sun (source): $24.85$% of $2.10^{30}$ kg.

Mostly all the energy is generated due to the product of nuclear fusion of hydrogen into helium by way of the proton–proton (PP) chain mechanism and the luminosity of Proxima Centauri is $0.00005\ L_☉$.

Thus, would it be a good approximation to say that the helium content is $0.00005 * 0.2485 * 2 * 10^{30}$?

Would you suggest a better way of estimating the helium content?

Nearly all the helium in the photosphere of the sun comes from the helium in the interstellar gas that collapsed to form the sun. That helium was produced shortly after the Big Bang (in about the first 20 minutes) while the universe was hot and dense enough for hydrogen to fuse to helium.  That produces a universe in which ordinary matter is about 25% helium and 75% hydrogen. (by mass) Over the eons, the gas is enriched in helium somewhat by previous stars to about 27% helium.

The outer layers of the sun still have this 25%-75% composition. Some of the helium has settled under gravity, reducing the composition of the photosphere to your 24.85% helium. The core is enriched in helium by fusion reactions. It isn't constant. The outer core is about 30% helium. The inner core has as much as 65% helium.  The average composition is about 28% helium, only slightly more than what it started with.

Proxima Centauri is fully convective, which means that helium from the core gets mixed up through the whole star. But it is such a dim candle that it has hardly produced any more helium than when it formed: about 27%.

So for any star, no matter how bright or dim (with a few exceptions) the helium composition is about 25%, or a little more.


Who first observed another planet rotating? We knew the earth rotates, but the moon is tidally locked.

To tell that another planet is rotating, you have to be able to distinguish its surface (I guess?). Who first did this?

Mercury's surface was virtually unknown until Mariner 10 flew by and Venus is covered in clouds. Of the terrestrial planets, that leaves Mars.

It was likely Christiaan Huygens in November 1659 who first realised Mars was rotating when he tried to map it. His estimate of roughly 24 hours was fairly close.

I believe Galileo's earlier observations on Jupiter were with a telescope not powerful enough to resolve smaller 'surface' features (like the Great Red Spot), which may have either proven that Jupiter rotates, or at least that something on the surface is moving across it.


Expansion of the universe speeding up I know that the expansion of the universe is speeding up, but is there an equation to measure this? Also, what is the cause of this?

The fact that the distance between galaxies is increasing is found by Redshift of light, where, the electromagnetic radiation from galaxies, as they travel large distances, get their wavelengths increased, and turn towards the red side of the spectrum. Upon working on these measurements, Astronomers concluded that the rate of increase of separation between two galaxies is directly proportional to the distance between those two galaxies.

So, consider two galaxies A and B, the distance between them being r

Then,  dr/dt α r
i.e, the speed at which they are separating is proportional to r.

The cause of this is not completely and surely known as of today, so it has been named a mysterious force called dark energy.


Isn't it likely that most extra solar planets don't transit their star in line with the Earth? In which case, for every planet that we detect, there should be significantly more that we can't detect because they don't transit in line with the Earth and, thus, are invisible.  Now, I suppose that we can reasonably assume that most planets in a given solar system orbit in a fairly well defined plane, so if we detect one planet then there are probably others in the same plane.  But, if we have detected no planets in a system, isn't it simply more likely that the planets are orbiting in a plane that doesn't bisect the Earth?  If that is the case, since there is such a tiny degree of variation that would allow a transit to be observed, shouldn't there be a huge number of planets that are not visible because the inclination of their orbital plane is something other than 0 (with respect to the Earth)?

Isn't it likely that most extra solar planets don't transit their star in line with the Earth?

Yes this is certainly true. The chances that a planet's orbital plane is directly along our line of sight is pretty small. And you are also correct to assert that it is more likely that most planets will orbit in the same plane. So if we don't see a planet around a star, it doesn't mean the planets don't exist. Absence of evidence is not evidence of absence.

In the early days of exoplanet detection, we had only a handful of known planets to study. This hardly precludes itself to statistical analysis. However, as the years have progressed, the number of known and confirmed planets has piled up, allowing astronomers to start inferring information based on statistical arguments.

One can argue that if we've spotted X number of planets after observing Y number of stars under Z conditions for T time then with a little bit of statistical analysis and math, there should be A number of planets in our galaxy, many of which we can't observe with current techniques. In fact, that is precisely what many astronomers have done.

Cassan et al. (2012) looked at microlensing detections and were able to infer, based on known detections and statistical analysis, that:

Planets are at least as numerous as the stars in the Milky Way.
17% of stars host a Jupiter-mass planets ($0.3$-$10\:M_J$).
52% of stars host "Cool Neptunes" ($10$-$30\:M_\oplus$).
62% of stars host an Earth-mass planet ($5$-$10\:M_\oplus$).

Swift et al. (2013) performed a similar type of analysis but focused mostly on M Dwarfs. They inferred that M Dwarf stars are expected to have, on average, $\sim2$ planets per star.

There are certainly many more such analyses and almost all of them come to the conclusion that there are hundreds of billions of planets in our galaxy. This can be confidently stated based on only the ~3500 exoplanets found so far.


How far can a planet with living beings get before we cannot see the living beings? Let there be a planet with Earth-like life. However, no trash or monuments or other visible structures, just organisms. How far can that planet be from Earth before our equipment just cannot observe the life anymore? Furthermore, let's say that our equipment is perfected and limited only by natural laws: how far can the planet be now?

The simple answer to this would be anywhere that is physically unreachable. If there happened to be Earth-like life out there in our solar system, we would likely be able to detect it, then land a probe and confirm.

When it comes to planets outside our solar system, then it becomes difficult. If there was no trash orbiting (and thus no way to notice slight dips in light etc) then we would likely be limited to studying the atmosphere using the transit spectroscopy method (more on this HERE) where we can detect the gasses and chemical makeup of an exoplanets atmosphere. There may possibly be other things to look for in xray or infrared spectrums, such as localised hot-spots if there is intelligent life, although depending on distance and size of the planet, this could be difficult.

Overall, no matter what we detected, if it was to far away for us to physically be there, then we would only be theorising. We could say 'This planet is a great candidate for life' and things along those lines and present papers on findings. We would, however never be able to confirm it.


A phonecall inside a blackhole I was wondering what a phone call would be like entering or falling t=into a black hole. Due to the massive density of the black hole and time would feel slower, how would that appear to the person on the other side of the call?

You and your phone will be destroyed before you get to the black hole, but
If we pretend that you and your phone can survive being in the vicinity of a black hole ... And let's suppose this is a regular black hole of about 5 to 10 solar masses, and you are falling in, you are not trying to orbit it.

As you fall towards the black hole, at first there is no problem. As you get closer to the event horizon you will accelerate, but until the last second or so the relativistic effects are quite small. Your phone signal will be redshifted, so the receiver will have to adjust the frequency that they are listening in on.

The last second before you fall in will last forever. Your signal will become redshifted further and further, your apparent rate of speaking will slow down and the last second before you pass the Event horizon will, to the outside observer, last forever.

You however are doomed. You pass the Event horizon and reach the singularity in a fraction of a second.


Why does earth's north pole roughly point in the same direction throughout the year? I know there is a gioscopic explanation but I was under the impression that Earth is traveling in a straight line through space that has been curved by our sun's gravity ,so my question is why doesn't one hemisphere always point away from the sun ? It just seems like a bit of a contradiction that gravity can curve the fabric of space but has no effect on earth's axis...

It is just conservation of angular momentum. In order to change the direction of the spin axis of the Earth (which would change the direction of its angular momentum) you would have to subject it to an external torque. The curvature of space has no significant bearing on the question.

In fact the direction in which the Earth's spin axis points does change, but on much longer timescales. The "precession of the equinoxes" (as it is called) is cyclical with a 26,000 year period.


Low precision GMST formula clarification According to the website http://aa.usno.navy.mil/faq/docs/GAST.php, a low-precision way of calculating the GMST is via the following formula:

GMST = 18.697374558 + 24.06570982441908 D

The website doesn't actually state where the numbers come from. Another website tries to do this:

"To compute of sidereal time with low precision use of the low precision formula: GMST = 18.697374558 + 24.06570982441908 D, where 18.697… is sidereal time of the reference time 2000-01-01 at 0 UT, 24.0657… is a ratio of synodic and sidereal periods of Earth and D is days (and its fractions) since the reference time."

but I don't understand what this 'ratio of synodic and sidereal periods' means. I know what the sidereal and synodic days are and their values, but how do we use those to end up with the number 24.0657...? If anyone could help me out that'd be much appreciated!

The number 18.697374558 was the value of GMST (in hours) at noon UT on 1 January, 2000. The number 24.06570982441908 reflects the fact that it takes slightly less than 24 hours (23.934469591898 hours to be precise) for a star to the appear to rotate by 360 degrees.

This is a consequence of the fact that the Earth orbits the Sun. Our 24 hour day is based on the time between two consecutive solar noons. Because the Earth moves a bit along its orbit, the Earth has to rotate by more than 360 degrees (with respect to the stars) to see the Sun rotate by 360 degrees.

Another way to look at it: The 23.934469591898 hour long sidereal day means that a solar day is 24.06570982441908 hours long expressed in sidereal hours.


How does this telescope work, and what is it for? Procrastinating, I clicked on a suggested video (How Large Can a Telescope Be?) and after about 01:20 I saw a clip of a telescope shown below.

I was confused as I saw the reflections in the mirror until I decided that there are actually two small objects in front of the primary mirror, one far from the primary where you'd expect to see a secondary, which I've labeled "A", and one very close to the primary, which I have labeled "B".

Both are supported by four vanes, and there does not seem to be a hole in the primary.

I've labeled the reflected image of the back side of B as B' (B prime), and I've indicated the reflections of an "A vane" and a "B vane" with arrows.

But "A" looks very small compared to what I'd expect, only perhaps a few percent of the primary's diameter, and it looks like the telescope is not in a conventional dome, but perhaps has only a sliding or otherwise openable roof (currently off screen) to protect it when not in use.

Question: How does this telescope work, and what is it for? Is it for imaging, or performing some other function?

https://www.youtube.com/watch?v=3wOFAkggSiU

I'm pretty sure this is one of the four auxiliary (or "outrigger") telescopes for the European Southern Observatory's Very Large Telescope Interferometer, part of the VLT complex at Paranal Observatory in Chile. It's used in conjunction with the four main 8 m ("unit") telescopes in an optical/infrared interferometric system, so its purpose is to send images of the sky in the optical or near-IR to be combined with images from the unit telescopes and the other auxiliary telescopes (via underground tunnels) in the interferometric system. (The octagonal gray frame is kind of a giveaway, and the fact that this video is ESO-based helps confirm this.)

https://www.eso.org/public/teles-instr/paranal-observatory/vlt/auxiliarytelescopes/

(Note that, per the link above, the primary mirror in each auxiliary telescope has a diameter of 1.8 m, while the secondary mirror is only 0.14 m -- very small, as you guessed.)

The object you identify as "B" is the tertiary mirror, which is flat, elliptical, and tilted at a 45-degree angle so that light from the secondary mirror is redirected off to the side. It sits in a mount (supported by the "B vanes") located just in front of the primary mirror, so there is no need for a hole in the primary mirror. You're probably used to versions of this (Nasmyth) design where the tertiary mirror is located behind the primary mirror, requiring a hole in the primary. Since the tertiary mirror in a Nasmyth design needs to sit along the elevation rotation axis of the telescope, whether it goes in front of or behind the primary depends on the overall telescope balance.

(Sometimes there is a hole in the primary, and the tertiary sits on top of a small "tower" that protrudes up through the hole from behind the primary; this would help disguise the existence of the hole. That's how the tertiary mirrors in the 8 m unit telescopes of the VLT work, though I don't think they have support vanes the way this one does. The existence of the support vanes and the nature of the B' reflection suggests there really isn't a hole in the primary here.)

The "dome" is an example of a "clamshell" design, in which two halves fold down out of the way; you can see one of the two hinges in the front, near the bottom right of your images. If you google for images of these telescopes, you can see them in the daytime, folded-up configuration.

Here's an image of one of the telescopes when it was still in the factory (before the mirrors were installed), which may help you understand the design better:


Is Darkness Faster Than Light? I've been looking around, and can't find a solid answer. Many websites say darkness is the speed of light due to the fact it is the absence of light, but other sources say otherwise, that in some cases darkness is actually faster in some cases. Could someone clarify this for me?

The darkness does not have a speed, as it's just the space the light travels through.  It's always there until the brief instant a photon whizzes through.

It would be analogous to asking what speed the road is travelling at, rather than the car.


What was Bradley's observation concerning abevation of light? As far as I understand, the effect of parallax exhibits itself in the shift of a star relative to the background stars on photographs.

As for the aberration of light, how is it reflected? It would not cause the relative shift of a star in a background, right? Because all the stars (the foreground one and the background ones) will be rotated as a whole, right?

So, what was Bradley's observation?

You are right that what you must use as a referent is different-- for parallax, your referent will be a much more distant object, because it shows little parallax.  In aberration, it happens to the whole visual field.  But what is shifted is the angle relative to the ground at which the light arrives, so if you set your telescope to see a given star, in principle you would need to correct for both parallax and aberration as the Earth orbits. But aberration is usually much larger than parallax because it scales with v/c rather than D/d (where D is 1 AU and d is the distance to the star).  For the Earth's orbit, v/c is about $10^{-4}$, but stars are much farther than $10^4$ AU.  As for Bradley's observation, a good version is given at http://cseligman.com/text/history/bradley.htm.  He was looking for the parallax effect, and stumbled into the much larger aberration effect.  He could tell it was something different because the sign was opposite and it did not depend on distance to the source.  This was 50 years after Romer, so the speed of light was already known.


Why do planets move in an elliptical orbit? I am a student of class 9 and recently came to know about kepler's law of planetary motion and got confused about the foci and elliptical orbits of planets and some stars.

Not sure if you're looking for a more mathematical answer or just the "why", but to answer the why, I'll start with some history on this.

Everyone who worked out a model for the Solar System, from Aristotle to Copernicus, liked circles.  Even though Copernicus correctly reasoned that the Earth moved around the Sun and not the Sun around the Earth, he continued to use circles in his models of the motion of the planets.

After Copernicus, Tycho Brahe, funded by the King of Denmark, had the best equipment at the time for observing the motion of the stars and planets and he was able to make star charts that were ten times as accurate as anyone before him. Brahe used equipment like this mural quadrant, and a large private observatory to take extremely accurate records.

Kepler, who was a better mathematician than Brahe, desperately wanted to get his hands on Brahe's star charts and the use of his observatory and equipment (so much so that when Brahe died, there were rumors that Kepler had poisoned him, though that probably didn't happen). When Kepler finally had everything at his disposal, he was able to work things out and study the Solar System more accurately. However, he still didn't know why the planets moved in ellipses; he'd only worked out that the ellipses fit the movement so well that it almost certainly had to be true, but he had no idea why.

Kepler, in fact, didn't care for ellipses.  He liked circles better, but he couldn't deny that ellipses worked.  Source.

Nobody knew why planets moved in ellipses until Isaac Newton was asked that question and had to invent calculus to answer it.  Calculus explains why planets orbit in ellipses, and that's the real answer.

If "calculus" isn't a satisfying answer, a way to kind of explain it would be to throw a penny out of a space shuttle (which isn't a good idea, but let's say you do). As the penny falls towards the Earth, it falls faster and faster (if we ignore air resistance) until it hits the ground.

Now, if you fling the penny from the space shuttle at a much faster speed and at a different angle, so that it only gets near Earth and misses the planet, it would actually start to orbit Earth. It would fall faster and faster until it passes the Earth, and then, like shooting a bullet into the air, the penny will slow down as it flies past the Earth.

According to Kepler's second law, the penny's fastest speed is at the point closest to the Earth (the perigee). That's essentially how objects in orbits work: as they move closer to the body they orbit, they accelerate faster and faster. Our penny will get so fast that, once it comes around the planet, it will be flung very far away, which will then slow it down. This is what creates an elliptical orbit.

Its motion is like a spring, falling toward the planet then flying away, but at the same time, orbiting in a circular motion with the spring motion, with 1 period per orbit. That motion of moving closer and then further in each orbit forms an ellipse.

It makes the most sense if you think of the velocity being greatest at the closest point and lowest at the furthest point. The low velocity moves it closer while the high velocity moves it back out further.  The total energy of the object in orbit (kinetic energy plus potential energy) remains constant.


Is there a possibility that a white dwarf can turn into a neutron star or a black hole? I know that a white dwarf is mainly balanced by electron degeneracy pressure and that if it gains more than 1.4 solar masses from any source (such as a companion star or a collision), it explodes as a type I supernova. But is there any possibility that a white dwarf can turn into a neutron star (or possibly a black hole)?

The answer is: to a neutron star - possibly; to a black hole, no.

The process whereby a neutron star is formed is known as an accretion induced collapse and is being seriously debated, especially in the case of white dwarfs that are born at the upper end of the "natural mass range" for white dwarfs and then accrete more mass as part of a binary system. An excellent read are the introductory sections of Taurus et al. (2013), who go through the motivation, process and (limited) observational evidence. See also Schwab et al. (2015); Ruiter et al. (2018).

Explosion vs Collapse

A white dwarf can respond to the accretion of material by exploding or collapsing. It depends on the competition between energy released in fusion reactions and the energy being locked away by endothermic electron capture (also known as neutronisation) reactions.

If thermonuclear reactions are initiated then the likely outcome is a runaway nuclear reaction - the pressure inside the star does not rise rapidly enough to prevent the entire star undergoing fusion. The energy released exceeds the gravitational binding energy and the likely outcome is a type Ia supernova.

On the other hand, the white dwarf is supported by electron degeneracy. If neutronisation begins to occur in the core, then protons (in nuclei) capture electrons to form neutrons. This destabilises the star causing it to collapse. The collapse would proceed (quickly) in a similar way to a core collapse supernova. The nuclei would dissociate, neutronisation would run to near-completion and the collapse would be halted by the formation of a neutron star.

There is little possibility that a black hole could be formed by such a collapse. The collapsing object would be of order 1.4 solar masses and comfortably smaller than the maximum mass of observed neutron stars (at least 2 solar masses). Therefore the collapse will be halted at the neutron star phase.

White dwarfs of moderate mass

Most white dwarfs of moderate mass have a C/O composition. They will need to accrete a lot of mass to get to a density (at about $4\times 10^{13}$ kg/m$^3$, reached at $1.38M_{\odot}$ in a non-rotating WD) where neutronisation becomes energetically feasible. Before this happens, it is probable that fusion reactions are ignited (due to high density, rather than temperature). The threshold density for ignition is *lower * for nuclei with lower atomic number (He < C < O) because of lower Coulomb repulsion, and the ignition threshold densities for He and C are also lower than the neutronisation threshold for C.

This means that in a C/O WD that has accreted a lot of matter, ignition could take place in C at the core, or it could be triggered in He (at even lower densities) at the base of a deep accreted shell of material. The outcome would likely be runaway thermonuclear fusion and the complete destruction of the star.

More massive white dwarfs

O/Ne/Mg WDs are made as the final stages of more massive stars ($8-10M_{\odot}$) and are probably born as remnants with much higher mass $>1.2M_{\odot}$ than typical C/O WDs. More massive WDs are smaller, with higher density. The neutronisation thresholds for O, Ne and Mg are only $1.9\times10^{13}$, $6\times 10^{12}$ and $3\times 10^{12}$ kg/m$^3$ respectively (all lower than for C, especially for Ne and Mg). This means that a O/Ne/Mg WD may have to accrete very little mass to reach this central density, begin neutronisation, which leads to collapse. In addition if such densities are insufficient to trigger C burning in a C/O WD, then they certainly won't be high enough to trigger burning in O/Ne/Mg because of stronger coulomb repulsion. Further, if little mass is accreted, then there won't be a deep envelope of accreted material in which to ignite burning off-centre.

For all these reasons, O/Ne/Mg WDs may be more likely (Liu et al. 2018; but see also Wang 2018) to collapse than explode (the collapse would cause a type of core-collapse supernova though).

Does accretion induced collapse occur?

At present there is only indirect evidence. When we look at recently formed neutron stars - identified as fast-spinning pulsars - we see they generally have very high velocities. These velocities are thought to result from an asymmetric "kick" delivered by a type II core-collapse supernova. This in turn suggests it might be quite difficult to retain a neutron star in a binary system, but many neutron stars are observed to be in binary systems, and many of these, particularly the millisecond pulsars, are thought to have undergone significant mass transfer in the past.

Further evidence comes from the retention of a significant population of neutron stars inside globular clusters. Again, the kicks might have been expected to expel most of these. In addition, there are a number of examples which appear to be "young", in that the ratio of their spin periods to rate of spin decay indicates they were formed recently. Since there are no high mass stars in globular clusters, and so no possible progenitors for these objects via the core-collapse of massive stars, then accretion induced collapse of a high mass white dwarf is a possibility.


What does the Reddening line represents in this U-B vs B-V plot? The plot is from this paper (pg-8).
They have also talked(in the same paper) about how central stars from galactic longitude range $20^\circ-80^\circ$ will have a very unusual Reddening line (pg-9). The current plot is for central stars from the galactic longitude range $200^\circ-300^\circ$

The group
of planetary nuclei plotted here are all those occurring in galactic longitudes 200o
-300°.
In the longitude range from 20° to 80° the points fall to the right of the reddening line in
Figure 3 (not plotted). If the latter energy-curves are well-behaved, a very unusual
reddening line would be needed to explain their position in the diagram. This is a puzzling result which is unexplained at present.

My interpretation of Reddening line: It is showing that if we plot real star data where the sample contains all the stars from bluer end to redder end on this U-B vs B-V plot, we will get  the Reddening
line on the plot.
Is this interpretation correct?

The reddening line is the locus that a star would move along in a colour-colour diagram as the interstellar reddening towards it increases.
i.e. It is an equation on the colour-colour plot of the form (for this colour-colour plot)
$$E(U-B) = f(E(B-V)) + c\ ,$$
where $E(U-B)$ is the colour excess in $U-B$ due to interstellar reddening, $E(B-V)$ is the colour excess in $B-V$ and $c$ is a constant chosen so that the locus would go through a star of the same spectral type but with no interstellar reddening.
Typically, for low reddening and a typical interstellar medium
$$ E(U-B) = 0.72 E(B-V) + 0.05 (E(B-V))^2\ ,$$
which I imagine is the line drawn on the plot you show.
To establish the reddening to a particular star you can follow it back from its observed position in the colour-colour plot, along a reddening line, until it intersects the main sequence locus (if it is a main sequence star). The intersection would give its unreddened, intrinsic colours and the x-axis increment through which you traced it back would give its $E(B-V)$.


Are Wikipedia's sun ecliptic-coordinate formulae accurate? I'm creating a C++ program that calculates the ecliptic coordinates based on the formulae from Wikipedia But, my calculations appear off. The mean anomaly for today, for example, should be 80.4-something; both my program and Google calculate approx. 79.2 - the result of (357.528 + 0.9856003*5927) % 360 (where 5927 is the number of days since January 1, 2000, GMT)

Assuming that you used the time of 170900 EDT(-4 UT), which is 230900 UT, the following is my math:

Let the Julian Date for 25Mar2016 at 230900, which is 2457473.464583, be set to JD;

Given the formula (on Wikipedia:n = JD - 2451545.0 ), your n is 5928.464583.
Then, taking the second formula ( g = 357.528° + 0.9856003° * n ) you get g=357.528°+5843.0964715441749 = 6200.3544715441749 mod 360 = +80.35447°.  Were you to round up this would give +80.4.

Without how you handled your date/time in c++ and your rounding, I would wager that is were the error occurred.


What's the farthest object as determined only by parallax? I'm wondering what the farthest object we know of, as measured only by parallax methods, from either earth or satellite based telescopes, and at either visible light or radio wavelengths.

Basically, if we had no other way of measuring distances,
 what would be the farthest object we knew about? I'm guessing this measurement would probably be one from Gaia, it being the latest telescope to measure such things TTBOMK.

The first data release from GAIA was in 2016 so fairly new. If you want to be certain you should take a look at it because as you say, GAIA will no doubt be the source of the most accurate parallax measurements in the near future. Speaking of accuracy, I believe that this is the essence of your question because it basically depends on how uncertain measurements you are willing to accept.
Basically, modern values of parallaxes are very small numbers, e.g. 0.1 milliarcsec which means a distance of

d(pc) = 1/(parallax in arc-seconds)

around 10,000 pc (one parsec is 3.26 light years). But the important point is the accuracy of these estimates. Clearly if your measurements is 0.1 milliarcsec, but the accuracy is 1 milliarcsec, it would be going out on a limb to postulate that you have measured a distance of 10,000 pc. So the real answer to your question is another question "to what accuracy?" So if you want something like 10 or 20 % accuracy you are more or less limited to our neighbourhood, say some 200 pc. If you check out ESA's website on stellar parallax, they will go up to 500 pc but it really depends on what level of accuracy your work requires.

If you check out the GAIA satellite then the aim is actually to go down to an accuracy of 24 micro-arc seconds so whatever object is measured to be farthest (with a given accuracy) is sure to replaced as the data from GAIA is analysed.

For fun, you might wanna look at this

also found at the GAIA website.. I think it provides a nice visual guideline to what distances we can expect (and it also nicely shows a ring at 500 pc - the limit of fairly accurate measurements from Hipparchos!)


Latest cosmological parameters I'm looking for the latest values (with uncertainties) of the four main cosmological density parameters $\Omega_i$ :
\begin{align}\tag{1}
\Omega_{\text{mat}} &={} ?,
&\Omega_{\text{rad}} &={} ?,
&\Omega_{\Lambda} &={} ?,
&\Omega_{k} &={} ?.
\end{align}
I know that $\Omega_{\text{mat}} \approx 0.30$, $\Omega_{\text{rad}} \approx 0.00$, $\Omega_{\Lambda} \approx 0.70$ and $\Omega_{k} \approx 0.00$, but I would like to have more precise values (with uncertainties, if possible).  Take note that these parameters are constrained by the following relation :
\begin{equation}\tag{2}
\Omega_{\text{mat}} + \Omega_{\text{rad}} + \Omega_{\Lambda} + \Omega_{k} \equiv 1.
\end{equation}

Of course, I checked Wikipedia but I don't trust it very much :

https://en.wikipedia.org/wiki/Lambda-CDM_model

I've also checked on arXiv.  For example :

https://arxiv.org/abs/1502.01589,

but I don't find clear final and consensual values in this paper.

Help would be appreciated.  Please state your sources.

Cosmological parameters are measured in a variety of ways, and their values will depend on which measurements you trust the most. The paper you link to (Planck Collaboration et al. 2016) with the 2015 results from the Planck observations of the cosmic microwave background is probably the one that most people will accept, but even in that paper you will find different values, depending on which observables you combine.

You will find the values in their Table 4. I think that most people use the values in the column called "TT+lowP+lensing" (e.g. Geil et al. 2016, Ricotti et al. 2016, and Liu et a. 2016), which is the "conservative" choice. However, you'll also find some (e.g. Chevallard & Charlot 2016 and Silk 2016) who use the values in the last column, called "TT,TE,EE+lowP+lensing+ext". These values takes into account external data (baryonic acoustic oscillations and supernovae data), which reduce the uncertainties, arguably to unnaturally small values. The TT, TE, TT, and lowP refer to the polarization maps used, and "lensing" to the weak gravitational lensing measurements by Planck.

Standard cosmological parameters

The table below is a modified version from the Planck paper where I show only the most used parameters:

Here, $n_s$ is the slope of the primordial power spectrum, $H_0$ is the Hubble constant in km s–1 Mpc–1, $\Omega_\Lambda$ and $\Omega_m$ are the density parameters of dark energy and total (dark+baryonic) matter, $\Omega_\mathrm{b}h^2$ and $\Omega_\mathrm{c}h^2$ are the density parameters of baryonic and dark matter, multiplied by the factor $h \equiv H_0/100$ (squared), $\sigma_8$ is the matter density fluctuations on scales of 8 (comoving) Mpc, $z_\mathrm{re}$ is the redshift at which the Universe was reionized (assuming instant reionization), and the last row shows the inferred age of the Universe in billion years.

Curvature density

The constraints on the curvature parameter $\Omega_K$ is given in Table 5, which has somewhat different combinations of data. All in all, Planck constrains the curvature to $|\Omega_K| < 0.005$, but you will rarely offend anyone by simply setting the curvature to zero.

Temperature and radiation density

The radiation density is a bit more convoluted. It has a contribution from both photons and neutrinos, and their densities are related as
$$
\rho_\nu = N_\mathrm{eff} \frac{7}{8} \left(\frac{4}{11}\right)^{4/3}\rho_\gamma,
$$
where $N_\mathrm{eff} = 3.046$ is the effective number of neutrino species. Following the procedure by Pulsar in this answer, but with updated parameters (i.e. the $N_\mathrm{eff}$ given above, and the average CMB temperature of $T_0 = 2.722\pm0.027$ (Eq. 83a)), I get that
$$
\begin{array}{rcl}
\Omega_\mathrm{rad}h^2 & = & \Omega_\nu h^2+ \Omega_\gamma h^2 \\
                       & = & (1.7018 + 4.6213) \times 10^{-5} \\
                       & = & 4.1620\times10^{-5},
\end{array}
$$
that is, with $h = 0.6781$,
$$
\Omega_\mathrm{rad} = 9.0513\times10^{-5}.
$$

Recap

So, answering your question is a little difficult, as there's no single answer, and since the curvature is given with 95% confidence ("2$\sigma$"), rather than 68% ("1$\sigma$"). For the Hubble constant and the matter and dark energy, I'd recommend $H_0 = 67.81\pm0.92 \,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$, $\Omega_\mathrm{m} = 0.308\pm0.012$ and $\Omega_\Lambda = 0.692\pm0.012$.

For curvature, I would use 0 (especially because your calculations probably would need to switch between ordinary and hyperbolic trigonometry depending on the sign),  but if you do want to include uncertainty, you can say $\Omega_K=0\pm0.005$ (95%). Or, you could simply use $\Omega_K = 1 - \Omega_\mathrm{m} - \Omega_\Lambda - \Omega_\mathrm{rad}$ and propagate the errors to get $\Omega_K = 0\pm0.017$, which will give you a more conservative value.

For radiation, to propagate uncertainties you would have to know the covariance matrix of the input parameters, but since the total error is dominated by that of the Hubble constant, standard error propagation $(\sigma_{\Omega_\mathrm{rad}}/\Omega_\mathrm{rad} \simeq 2 \sigma_h/h)$ yields a value of $\Omega_\mathrm{rad} = (9.0513\pm0.2456)\times10^{-5}$.

So, to be explicit my recommendation is:
$$
\{\Omega_\mathrm{m}, \Omega_\Lambda, \Omega_\mathrm{rad}, \Omega_K\}
= \{ 0.308, 0.692, 9.05\times10^{-5},0\}
\pm \{0.012, 0.012, 2.46\times10^{-6}, 0 \}.
$$
But I think the most important thing is to state where you take the parameters from. People rarely state why the choose a particular set of parameters, and although Planck gives very small error bars, other probes give error bars so small that they're basically mutually incompatible. That's why you can still easily get away with $\{\Omega_\mathrm{m}, \Omega_\Lambda, \Omega_\mathrm{rad}, \Omega_K\} = \{ 0.3,0.7,0,0\} \pm \{0,0,0,0\}$.


Can an SMBH recycle dark matter into energy? I just had a conceptual question, hopefully it makes for some really interesting answers.

I was wondering, as most galaxies have an SMBH at their centre (see Kormandy & Richstone 1995; Richstone et al. 1998), could dark matter accrete onto the SMBH over time (perhaps perturbed by a galaxy collision, interaction or merger) and then recycle this matter as energy? Can dark matter help an SMBH grow?

The accretion of dark matter by a black hole is possible, but difficult. Unless its trajectory takes it into the black hole, then it will simply approach the black hole, speed up and then escape again, since non-interacting dark matter cannot dissipate energy in any way.

Presuming a black hole did accrete the dark matter then the energy possessed by the dark matter would become part of the rest mass energy of the black hole. So yes, the black hole would grow. Not sure what you mean by recycle the energy. Energy is always recycled (conserved). If you mean can the black hole lose its rest mass energy? I suppose it may do via Hawking radiation, but this is vanishingly small for a SMBH. The dark matter will also carry angular momentum into the BH and there are ways that this rotational energy can be extracted (e.g. the Penrose process https://en.m.wikipedia.org/wiki/Penrose_process ).


If Venus had moon before and it fell back to Venus, why don't I see a unique, very big crater? Does lack of a very big crater support the idea that Venus never had big moons before?

If not, why can't I find a very big crater from the moon impact?

Venus has volcanism and weather, and those can cover or erode craters.  There are relatively few impact craters on Venus today, indicating that the surface of Venus is at most 600 million years old -- meaning that any older surface features have been erased.  The Earth's surface is even younger.

During the Late Heavy Bombardment, in the very early history of the Solar System, there was a period of time when many asteroids hit the Moon, creating many large impact craters.  It seems likely that the Earth and Venus got hit by many asteroids then, too.  Yet, no evidence of these craters is visible on Earth or Venus today. So, if Venus had a moon and that moon crashed into Venus thousands of millions of years ago, then the impact crater would likely have been erased by now.

And if Venus got hit by a sufficiently large object, then the energy of the impact might melt the entire surface of Venus, so that no lasting crater can form.

So, absence of very big craters from Venus is not evidence that Venus never had big moons.


Sun's apparent motion above the arctic circle during summer solstice I'm having difficulty with an apparent contradiction about the sun's motion during the summer solstice above the arctic circle.

The YouTube video The North Pole Entire Summer of never ending sun Summer Solstice High speed camera and solar motion models show the sun to move clockwise around the horizon for the full 360 degrees during the summer solstice.
However, when I walk in a circle counterclockwise, to represent the earth's rotation, and focus upon a fixed object outside the circle to represent the sun, I get a 180 degree motion, from "east to west", and then the sun's apparent motion returns again from west to east, not the full 360 degrees which everyone reports is the actual phenomenon.

Can someone help me to understand my dilemma?

Instead of walking in a circle, spin. Take a tiny step forward with your right foot, backward with your left. Then repeat. You are spinning counterclockwise when viewed from above.

Next, imagine holding a landscape photograph at arms length, and you initially facing a floor lamp. Now take that tiny step forward with your right foot, backward with your left. From your body fixed perspective, the landscape photograph doesn't move. It's the lamp that appears to be moving, and it moves to the right.

Next, imagine a helmet cam on your head that rotates to keep the floor lamp in the center of the camera image. From this perspective, it's the landscape photograph that appears to be moving, and it moves to the left.

That's what you are seeing in that youtube video.


Where do I find a better mirror? I recently bought a telescope, but it doesn't have a very clear image.  Where can I buy a better mirror?

Well, if you buy a car and then you go "this thing doesn't drive well, where do I find a better engine" - then folks are going to ask you questions such as: have you driven stick shift before, do you understand how the clutch operates, etc.

There are many reasons why a telescope does not provide a satisfactory image.

Perhaps the expectations were too high. An amateur telescope can provide quite amazing views, but it's never going to be equal to a large professional telescope in orbit around Earth.

Also in line with the expectations thing, all else being equal, a large telescope will always perform better than a small one. A 6" scope and a 24" scope will provide very different levels of performance. If you have the 6" scope, but you expect 24"-level performance, that's never going to happen.

Perhaps seeing (atmospheric turbulence) was bad during the time when you tested the instrument. For objects in the solar system, seeing is one of the major impediments - it reduces resolution and muddies up the image. It changes in time; sometimes it's good, other times it's bad.

Perhaps light pollution was the problem. DSOs (deep space objects, which have low brightness) will easily get masked by the regular light pollution in cities. It changes with the place; it's bad in cities, it's good far from cities.

Was the telescope collimated? Some telescopes, such as dobsonians (and most scopes with a newtonian OTA) require regular collimation. Many SCTs do too. Performing collimation on such a scope is like changing the oil on your car - you just have to do it periodically.

Was the telescope stored in a warm place (inside the house), then suddenly taken outside in the cold night? The temperature differential will be similar to bad seeing for up to one hour, until the scope and the air take the same temperature.

Was the scope properly focused? In many cases, finding perfect focus is a little tricky. A small turn of the knob could make a big difference.

Can you even reach focus? It's surprising how often people have weird setups where the eyepiece cannot even reach the focal plane of the primary mirror. In that case, no amount of twiddling the focuser will provide a good image.

Is the eyepiece bad? Sometimes mass-produced scopes come with eyepieces that are just not very good. It happens. Trying a known good eyepiece instead should provide proof.

Are the optics very dirty? A layer of dust on the primary will not affect performance in any way that you can perceive, so leave the dust alone. However, if there's a big fat oily smudge on the back side of the eyepiece, then of course performance will be degraded.

Are the optics seated properly in their mirror cells, etc? A primary mirror not sitting flat in its cell will mean the scope will never perform well; this could happen during transport with some cell designs.

Is the frame or the OTA intact? Has it been bumped or otherwise deformed?

Finally, yes, in some cases mass-produced scopes have bad optics. In that case there's nothing you can do except replace something - either the defective mirror or the whole instrument. Keep in mind that a custom mirror will cost more than your whole instrument, if it's high quality.

TLDR

You're not providing any details, so it's hard to say what's going on. Based on experience, I'd say the most likely candidates are:

If your scope is a dobsonian or an SCT, I would start with collimation. Learn how to collimate the scope, and do it regularly. It has a very, very large impact on observations of the Moon and the planets. Do this before you try anything else.

If you're observing DSOs (galaxies, nebulae), maybe light pollution is the reason. Load up the scope in your car and drive about 1 hour away from the city, and try again.

Learn how to focus the instrument. It could be trickier than it seems.

Finally, go to a local astronomy club meetup and try and use an instrument similar to yours. Maybe your expectations are not calibrated properly.


What happened to NS spectral types after OBAFGKM? What happened to NS after OBAFGKM? The full mnemonic was "O Be a Fine Girl Kiss Me Now Sweetheart" a few decades ago as I remember. I think it was that way in "The Larousse Encyclopaedia of Astronomy" which I once had. The Cambridge Encyclopaedia of Astronomy (which I still have, first published in 1977, Reprinted 1979) has "OBAFGKMRNS", this seems to stem from work done by Annie J. Cannon resulting in the classification of over 225,000 stars in catalogues published between 1918 and 1924, suggesting that "OBAFGKMRNS" is the accepted list "now" (1977?). So what happened to the "R" also? Of course you can't quote these mnemonics now, because of "Political Correctness", though you can always explain that G could be either Guy or Girl!

They still exist, but are perhaps less used because they don't indicate a particular temperature, or mass.
The stellar classification is based on the presence of particular lines in the spectrum:  A type stars have a strong hydrogen line. B type stars have Helium.  F and G  and K stars have various ionised or neutral metals. M type stars have compounds like TiO2 in their atmosphere.
But these are associated with temperature, and B type stars are hotter than A type (and O type stars are hotter still) hence OBAFGKM.
Asymptotic branch supergiants can have processes that dredge up carbon from their interior, producing carbon lines.  These are the R and N type stars, but R type stars are similar in temperature to K type stars. and N type are like M type.
S type stars have heavy elements produced by s-process neutron capture stars.  So if you are ordering the stellar classification by temperature, these don't have a natural place to go, and since the purpose of "Oh Be A Fine Girl Kiss Me (Right Now Sweetheart) is to remember the order of the letters then RN and S don't fit, since they are not ordered by temperature.
As for political correctness, many alternatives have been suggested, but none has caught on, perhaps because a good mnemonic should be a bit "naughty".


Why are there so many binary systems? Most star systems are binary, but why is that? Why would new stars form close to others, and not (relatively) evenly spread? And even if they are clustered close together, why are the majority binary? Or are binaries the ones that survive, and systems with 3 or more stars tend to be unstable and quickly collide?

If that is so, what makes binary systems stable? Why do the stars not simply spiral in and collide, or break away shortly after creation?

Collapsing gas clouds fragment into multiple cores because the Jeans mass, that determines the minimum mass that becomes gravitationally unstable to collapse, becomes smaller if the cloud is able to contract without heating up too much.  i.e.
$$M_J \propto T^{3/2} \rho^{-1/2},$$
where $\rho$ is the cloud density. Thus if the cloud density can increase but the temperature stays constant(ish), then the Jeans mass shrinks and the cloud becomes unstable to further fragmentation.

Multiple systems with $n>2$ are inherently unstable unless they are hierarchical. i.e. a star in a wide orbit around a close pair can be stable, as could two close binary systems orbiting each other. The condition for stability is roughly that the separation of the wider star must be 5-10 times that of the inner pair ([dependent on mass ratios and eccentricities - Eggleton & Kiseleva 1995). In most other cases, what happens is the ejection of other stars from a multiple system, leaving behind a binary system Durisen et al. (2001).

What makes a binary stable? We'll what would make it unstable? I'm not sure why you think they should spiral in towards each other. This can't happen unless there is some dissipative mechanism, like tidal interactions. Gravitational waves are ineffective in all but close binaries involving compact stellar remnants. They can't "break away" because they are gravitationally bound.

EDIT: Note that it isn't true that most systems are binaries. The majority of "systems" are in fact single stars. The binary frequency for solar-type stars is about 50% - i.e. as many singles as binaries; but the binary frequency for the much more numerous M-dwarfs is probably around 30% and so single stars outnumber binary systems, though it is not clear whether that is true at birth Duchene & Kraus (2013).


What is the minimum mass of a celestial object so that it can have a moon? I was wondering how massive something have to be so that it it can attract moons by pop culture standards (ellipsoid/round shape).
Could a planetoid have a moon? What is the relation between the mass of a moon and planet?

Planetoids can have moons and the minimum size is "pretty small". For example 2003 SS84is a small Near-Earth asteroid, with a diameter of 120m and a moon of about 60m in diameter, which orbits at a distance of 270m ever 24 hours.  It probably didn't form by "attracting the moon" but the moon probably formed as a result of impact splitting a "rubble pile" asteroid.  The moon, in this case, is similar in size to the main object.  There is a wide distribution of size ratios from close to 1:1 to very small objects orbiting much larger ones.
However one aspect of your question is different:  you ask for a "round" shape. That makes things harder, since a round shape requires a fairly large size to pull the object into a sphere.  There is probably only one object except for the major planets, with a spherical moon, and that is Pluto with its moon Charon (and even Charon isn't perfectly rounded, but it is close)
So it is quite easy for a planetoid to have a moon.  But it is hard for a planetoid's to be big enough to pull itself into ellipsoidal equilibrium


Coordinates in the Night Sky I've learned about Orbits recently, and there are 6 orbital elements. Now, I've heard of three coordinate systems whereas, they can help you measure positions of stars in the night sky. The Altazimuth, the Equatorial, and the Ecliptic coordinate systems. Now, I can't quite grasp what they mean, and how to use them. This is what I am asking for from an answer to this question.
From night to night, I know that star and planet positions may vary, because we're rotating, and the start move due to it, and even in the day, we're still spinning, and the star positions keep changing, whether or not we can see them, or they're blocked by the sun's brightness, so the next night, they'll be in a new place. That being said, I don't want an answer that says there can't be a coordinate system that doesn't relate to my position on Earth, because I want an explanation for those three coordinate systems that includes my position (preferably using latitude and longitude).
Assuming this can be feasibly explained in a post, thanks in advance!

The altazimuth system is directly linked to your position on the Earth. It is the “left-right/up-down” coordinates system; the azimuth being the “left-right” (with 0° for North, 90° East, etc.), and the altitude is the “up-down” (with 0° being the horizon, 90° being overhead, also called zenith). Of course, a star’s (or planet’s, comet’s, galaxy’s, etc.) position will change during the night and over the year.
The equatorial system is independent of any location on the Earth. It is akin to the longitude and latitude of the Earth, but for the sky. As you can guess from the name, its reference is the celestial equator, which is the projection of Earth’s equator on the celestial sphere. There’s an imaginary point along it, where the Sun crosses it from south to north (near March 20), that’s the reference point, to be more specific; it’s called the vernal point. Like longitude, right ascension is measured (in hours, minutes, and seconds of time) eastward from that point, from 0 h at this point to 12 h opposite it to 24 h back at this point. Like latitude, declination is measured north (+) or south (−) from the equator to the poles (+90° North, −90° South). Celestial bodies virtually keep the same right ascension and declination over a human lifetime, except for planets, asteroids, and comets, which keep moving.
Finally, the ecliptic system is used mostly for bodies of the Solar System. As you can guess by the name, it’s measured along the ecliptic instead of the equator. The ecliptic is the apparent path of the Sun in the sky over the year—it corresponds to the plane of the Earth’s orbit around the Sun. Coordinates in the ecliptic system are ecliptic longitude, measured eastward from the vernal point mentioned in the previous paragraph, from 0° at it to 180° opposite it to 360° back at it; and the ecliptic latitude, measured perpendicularly to the ecliptic, from 0° along it to +90° at the ecliptic north pole and −90° at the ecliptic south pole.
There are indeed formulas for converting from one system to the other. Between the equatorial and ecliptic system, it’s a fixed conversion, but between any of these and the altazimuth system, there’s a time factor, so you need to account for it. And as time zones vary from country to country (and sometimes within the same country!), we don’t rely on them but instead on true local sidereal time, which is basically the right ascension of the point in the sky that is due south (or north, in the southern hemisphere) of your location.
You also mentioned, in the beginning of your question, orbital elements, but these are completely independent of the coordinates system being used, although it is standard to consider them in the ecliptic reference system.