What is the coordinate acceleration for pure radial motion? As seen in this answer and also in this book chapter 6, page 6-10 expression 22,  when you are falling in radially from being at rest at infinity your velocity in coordinate time, as seen from a distant observer, can be described by the expression:

$$\frac{dr}{dt} = -\left(1 - \frac{r_s}{r}\right)\left(\frac{r_s}{r}\right)^{1/2}c $$

If we assume ”$r$” to be the real radial distance and not just the ”r-parameter of Schwarzschild coordinates” this is the velocity $v$. I want to know the expression for the acceleration as a function of ”$r$” and ”$v$” for a small object of mass $m<<M$ moving purely radially inwards or outwards from a compact spherically symmetric mass distribution in the general case.

By taking the derivative with respect to "r" of the expression above I get:
$$\frac{dv}{dr}=\frac{0.5}{r}{\sqrt{\frac{r_s}{r}}(1-\frac{3r_s}{r})}c$$

From these two expressions we see that the maximum velocity is $v=\frac{2}{3\sqrt{3}}c$ at $r=3r_s$. From this finding, $\frac{dv}{dt}$ should not be so hard but this will not be a general expression but only the acceleration of something dropped from rest at infinity. In Schwarzschild coordinates the radial velocity of light goes as $v_{light}=c(1-r_s/r)$. At $r=3r_s$ this becomes $v_{light}=2c/3$ so when falling in from being at rest at infinity you reach maximal velocity when $v/v_{light}=1/\sqrt{3}$. Maybe you can say that gravitation becomes repulsive whenever you are travelling towards the center of the gravitational field faster than $1/\sqrt{3}$ times the local speed of light?

Question: What does the general expression for the acceleration of an object in coordinate time look like for an object moving purely radially inwards/outwards from the center of the gravitational field assuming a spherically symmetric non-spinning mass distribution and that the r-parameter of the Schwarzschild coordinates is the real physical distance?

I am looking for an expression for the instantaneous acceleration as a function of "r" and "v".

I know that in the weak fields of our solar system Nasa/JPL are using this expression to calculate Newtonian plus relativistic acceleration:

$$\frac{d\bar{v}}{dt}=-\frac{GM}{r^2}\left(1-\frac{4GM}{rc^2}+\frac{v^2}{c^2}\right)\hat{r} +\frac{4GM}{r^2}\left(\hat{r}\cdot \hat{v}\right)\frac{v^2}{c^2}\hat{v}$$

This is based upon a first order expansion of the Schwarzshild solution in isotropic coordinates. For pure radial infall you get:

$$\frac{d\bar{v}}{dt}=-\frac{GM}{r^2}\left(1-\frac{4GM}{rc^2}-3\frac{v^2}{c^2}\right)\hat{r} $$ which becomes repulsive whenever you are approaching the central mass faster than:

$$v=\frac{c}{\sqrt{3}}\sqrt{1-\frac{4GM}{rc^2}}$$

I am hoping for an exact solution for pure radial acceleration in "Schwarzschild coordinate space" and coordinate time similar to the approximate solution for pure radial acceleration you can get from the Nasa equation.

As per the answer by amateurAstro down below the expression should be:

$$\frac{dv}{dt}=-\frac{GM}{r^2}\left(1-2\frac{v^2}{c^2(1-\frac{2GM}{rc^2})}-\frac{v^2}{c^2(1-\frac{2GM}{rc^2})^2}\right)$$

This is consistent with stuffing:

$$\gamma=\frac{1}{\sqrt{1-\frac{2GM}{rc^2}-\frac{v^2}{c^2\left((1-\frac{2GM}{rc^2})(\hat{r}\cdot\hat{v})^2+|\hat{r}\times\hat{v}|^2\right)}}}\frac{1}{\sqrt{1-\frac{2GM}{rc^2}}}$$

into

$$\frac{d(m\gamma\bar{v})}{dt}=-\frac{GMm\gamma}{r^2}$$

to get (at least this is true for pure radial and pure non-radial motion):

$$\frac{d\bar{v}}{dt}=-\frac{GM}{r^2}\left(\hat{r}-2\frac{v^2(\hat{r}\cdot\hat{v})\hat{v}}{c^2(1-\frac{2GM}{rc^2})} -\frac{v^2(\hat{r}\cdot\hat{v})\hat{v}}{c^2(1-\frac{2GM}{rc^2})^2}\left((\hat{v}\cdot\hat{r})^2+(1-\frac{2GM}{rc^2}-\frac{v^2}{c^2})|\hat{v}\times\hat{r}|^2\right)\right)$$.

(I am trying to find a velocity-dependent term that together with the classical Newtonian acceleration should reproduce Schwarzshild orbits exactly)

As Rob Jeffries suggested in a comment, formulating a solution for arbitrary dr/dt requires another parameter, either the energy per unit mass, E/m, or the radius of the shell where the object was dropped.

A shell observer at $r_0$, sees an object passing by at proper radial velocity $v_0$, with corresponding energy/momentum as measured in the local Minkowski frame (using units with c = G = 1)
$$m^2=E_0^2-p_0^2=E_0^2-\gamma _0^2 m^2 v_0^2$$
$$\frac{E_0^2}{m^2}=\gamma _0^2 v_0^2+1=\frac{v_0^2}{1-v_0^2}+1=\frac{1}{1-v_0^2}=\gamma _0^2$$
where the subscripts indicate local shell coordinates. In Schwarzschild coordinates,  energy per unit mass is constant and defined as
$$\frac{E}{m}={ \left(1-\frac{2 M}{r}\right)}\frac{dt}{d\tau}$$
From equations 11 and 12 in section 6.3 of the Taylor, Wheeler, Bertschinger book (TWB) linked in the original question
$$\frac{E_0}{m}=\left(1-\frac{2M}{r_0}\right)^{-1/2}\frac{dt}{d\tau}$$
$$\frac{E}{m}=\sqrt{1-\frac{2M}{r_0}}\frac{E_0}{m}=\gamma _0 \sqrt{1-\frac{2 M}{r_0}}=\sqrt{\frac{1-\frac{2M}{r_0}}{1-v_0^2}}$$
The equation for direct radial inward motion is obtained from TWB section 8.4, equation 17, by setting the angular momentum to zero and taking the negative root .
$$ \left(\frac{dr}{d\tau}\right)^2=(E/m)^2-\left(1-\frac{2M}{r}\right)$$
$$\frac{dr}{dt}=\frac{dr}{d\tau}\frac{d\tau}{dt}=
\frac{1-\frac{2 M}{r}}{E/m}\frac{dr}{d\tau}$$
$$\frac{dr}{dt}=-\left(1-\frac{2M}{r}\right)\sqrt{1-\frac{1-\frac{2M}{r}}{(E/m)^2}}$$
Differentiating, and a bit of algebra yields the following result for the acceleration.
$$\frac{d^2 r}{dt^2}=-\frac{2M}{r^2}\left(1-\frac{2 M}{r}\right)
\left(1-\frac{3}{2}\frac{1-\frac{2 M}{r}}{(E/m)^2}\right)$$
These can be rewritten in terms of the Schwarzschild radius $r_s=2M$ and proper velocity, $v_0$, at $r_0$.
$$\frac{dr}{dt}=-\left(1-\frac{r_s}{r}\right)\sqrt{1-\frac{1-\frac{r_s}{r}}{(E/m)^2}}=-\left(1-\frac{r_s}{r}\right)\sqrt{1-(1-v_0^2)\frac{1-\frac{r_s}{r}}{1-\frac{r_s}{r_0}}}$$

$$\frac{d^2 r}{dt^2}=-\frac{r_s}{r^2}\left(1-\frac{r_s}{r}\right)
\left(1-\frac{3}{2}\frac{1-\frac{r_s}{r}}{(E/m)^2}\right)
=-\frac{r_s}{r^2}\left(1-\frac{r_s}{r}\right)
\left(1-\frac{3}{2}(1-v_0^2)\frac{1-\frac{r_s}{r}}{1-\frac{r_s}{r_0}}\right)$$

For known $dr/dt$ at a radius $r$, $E/m$ can be determined by inverting the previous expression for $dr/dt$. Then substitute $E/m$ in the equation for acceleration above.

$$(E/m)^2=\frac{(1-r_s/r)^3}{(1-r_s/r)^2-\left(\frac{dr}{dt}\right)^2}$$

The alternate approach, dropping an object from rest at radius $r_0$, uses equations 37 and 38 from section 6.7 of TWB, resulting in

$$\frac{dr}{dt}=-\left(1-\frac{r_s}{r}\right)
\sqrt{\frac{\frac{r_s}{r}-\frac{r_s}{r_0}}{1-\frac{r_s}{r_0}}}$$

$$\frac{d^2 r}{dt^2}=\frac{r_s\left(1-\frac{r_s}{r}\right)}
{2 r^2\left(1-\frac{r_s}{r_0}\right)}
\left(1+\frac{2r_s}{r_0}-\frac{3r_s}{r}\right)$$

The shell radius $r_0$ necessary to obtain a desired velocity at shell $r_1$ can be found using
$$\frac{E_0}{E_1}=\sqrt{\frac{1-\frac{r_s}{r_1}}{1-\frac{r_s}{r_0}}}
=\frac{\gamma_0}{\gamma_1}$$
where $\gamma_0=1$ for an object dropped from rest at $r_0$.

As an example, a plot is shown below using $v_1=1/2$ at $r_1=2r_s$. For these values $\gamma_1=2/\sqrt{3}$, $E/m=\sqrt{2/3}$ and $r_0=3r_s$. For reference, the green and red curves are Schwarzchild coordinate velocity and acceleration for an object dropped from rest at infinity, $E/m=1$. The dashed blue and brown curves are for an object dropped from rest at $r_0$, and the solid blue and brown curves for an object launched from $r_1$ with a Schwarzschild velocity of
$$\left(\frac{dr}{dt}\right)_{r_1}
=-\left(1-\frac{r_s}{r_1}\right)v_1
=\frac{1}{4}$$
Note that the dashed curves are overlaid by the solid ones from $r_1$ to $r_s$, since the initial conditions were chosen to result in the same velocity at $r_1$.


How can I collimate a dobsonian telescope with a laser collimator? I've had a dobsonian telescope for a while now, but I've never been able to collimate  it. I thought I had it collimated once, but when I looked through it the star didn't look right, it looked kind of like a round disc with a hole in the middle. I know the basic idea behind doing it with a laser collimator, you need to get the laser to line up with the center of the primary mirror as well as line up in the center of the target on the laser itself. However, I always have a lot of difficult accomplishing this.

For starters, I always have a lot of trouble getting the secondary mirror lined up. I'll get really close, but once I get close I have a lot of trouble doing the fine tuning to get the laser in the middle. I'll get it in, but when I got to tighten the other screws it will no longer be lined up. I'll try to account for the tightening as well, but it's always a guessing game for me when it comes to knowing which screws to loosen and tighten so that the laser winds up where I want.

I don't have that many problems with the primary mirror, and can normally get that lined up fairly easily. However, usually after lining up the primary mirror the secondary mirror no longer seemed to be lined up and I need to do the secondary mirror all over again. I'm assuming this is to be expected, but considering the difficulty I have lining up the secondary mirror it's very frustrating.

I've tried looking around at resources online, but none of them go into enough detail into aligning the secondary mirror. They just say "adjust the screws until it's lined up", which doesn't solve my problem, since I have trouble figuring out exactly which screws to adjust and when. I'm also confused as to what I should see through my telescope when it is properly aligned. Most of the sites just use a diagram, but I'm never really sure what to make of the diagram. The one time I did line it up it looked like a round disc with a hole in the middle, which I didn't think looked right.

So to summarize, the main problems I have are:

I have trouble fine tuning the secondary mirror, since I have trouble accounting for the movement of the mirror when tightening the two screws I didn't loosen
I'm not sure what a star should look like through my telescope when it is properly collimated

Can anyone offer some detailed instructions, and possibly a video, of how to collimate a dobsonian telescope with a laser collimator?

Note to future answerers: We have been working in the Astronomy.SE main chat room with OP to solve this problem and established it's actually a matter of focusing that resulted in problems described (telescope was either too intrafocal or extrafocal), not collimation itself. This is a video on YouTube that OP said was exactly how the problems looked like. Granted, focusing can be technically called collimation too, and indeed that demonstration video calls it as such (albeit misspelled), since it's still about adjusting positions of primary and secondary mirror in dobsonian or newtonian telescopes (dobsonian is a type of newtonian), but the former is a lot simpler to solve by turning focusing knobs and achieve optimal illumination.

Now, let's answer the actual question.

Preparing our laser collimation setup

First thing to know is that any laser collimator will likely have the beam strong enough to cause damage to your eyes if pointed directly in them, or damage strong light sensitive equipment, like for example digital cameras, so be careful with it and take precautions. Carefully read instructions on how to handle and use it, before you do so. This is how a laser collimator might look like:

Next thing to do is make it functional by inserting batteries or plugging its power adapter and testing it actually works by turning it on with the switch, if it has it (probably does). Once you'll establish your laser collimator is turned on and functioning by observing (usually ruby red) light beam spot it makes on any neutral surface, turn it back off and attach any adapters on it you might need to help you attach it to your telescope's focuser. This is how your telescope's OTA (Optical Tube Assembly) might look like:

Now remove the barlov and eyepiece from your telescope's focuser if you're using a direct focuser adapter for your telescope model, tighten the laser collimator's compression ring by rotating its rubber ring and attach the collimator into the focuser. Rotate collimator's compression rings to tighten it in place. If you're not using an adapter, your laser collimator might attach directly to your eyepiece. Consult your collimator's user manual, if unsure.

Rotate your collimator's targeting faceplate towards your telescope's primary mirror (on the telescope's bottom) and turn on the laser.

Angular alignment of the secondary mirror

By looking into the telescope top down from the secondary (top) mirror into the telescope, we will adjust the angular alignment of the secondary mirror by referencing the laser dot projecting on the primary (bottom) mirror. The secondary mirror's support has a center ring, the collimating cap, with three bolts on the top of the secondary mirror holder. Use Allen wrench or Phillip's screwdriver to rotate them individually to bring the laser dot into the center of the doughnut ring of the primary mirror, like so:

Try to bring the dot more directly into the doughnut center than on the photograph above, though. I intentionally didn't select a photograph from when it was in its dead center, because of the light reflection covering the doughnut ring completely, rendering it nearly invisible.

Primary mirror collimation

Unlock the primary mirror plate locking screws, position yourself towards the panel side of the primary mirror tube so you can observe the laser's center spot on the targeting faceplate on the collimator and start adjusting collimation screws by turning them gently until the laser dot is in the center of the targeting faceplate:

Another way to check your collimation progress when adjusting collimation screws is by checking the projection laser dots on the secondary (top) mirror and making sure both laser dots are overlapping into one single spot, but that might make your adjustments of the primary mirror somewhat harder, so using the targeting faceplate of the laser collimator would be my suggestion and perhaps check alignment of the secondary mirrors laser dots projection for the last finishing touches to collimate your mirrors as precise as possible.

Now you're done collimating your dobsonian or newtonian telescope, so it's time to remove the laser collimator by first switching it off, then untightening its compression ring and taking it out of the focuser. Once you've removed the laser collimator, insert in the focuser barlov and the eyepiece, and tighten them back in place.

Additionally, and as requested, here is the Youtube video of the Newtonian Collimation using SCA Laser Collimator that are also the source of the last two example photographs and inspiration to write the collimation procedure, courtesy of HoTechUSA YouTube channel. Make sure to browse through other similar videos on telescope collimation in this same channel.

Collimation of a dobsonian without a laser collimator

Standard collimation technique for the exact dobsonian telescope make OP mentioned has can be read in this Instruction Manual for Sky-Watcher dobsonians (PDF):

  Collimation is the process of aligning the mirrors of your  telescope
  so that they work in concert with each other to  deliver properly
  focused light to your eyepiece. By observing  out-of-focus star
  images, you can test whether your  telescope's optics are aligned.
  Place a star in the centre of  the field of view and move the focuser
  so that the image is  slightly out of focus. If the seeing conditions
  are good, you  will see a central circle of light (the Airy disc)
  surrounded by a  number of diffraction rings. If the rings are
  symmetrical about  the Airy disc, the telescope's optics are correctly
  collimated  (Fig.g).

  If you do not have a collimating tool, we suggest that you  make a
  "collimating cap" out of a plastic 35mm film canister  (black with
  gray lid). Drill or punch a small pinhole in the exact  center of the
  lid and cut off the bottom of the canister. This  device will keep
  your eye centered of the focuser tube. Insert  the collimating cap
  into the focuser in place of a regular  eyepiece.

  Collimation is a painless process and works like this:

  Pull off the lens cap which covers the front of the telescope  and
  look down the optical tube. At the bottom you will see the  primary
  mirror held in place by three clips 120º apart, and at  the top the
  small oval secondary mirror held in a support and  tilted 45º toward
  the focuser outside the tube wall (Fig.h). The secondary mirror is
  aligned by adjusting the central bolt  behind it, (which moves the
  mirror up and down the tube),  and the three smaller screws
  surrounding the bolt, (which  adjust the angle of the mirror). The
  primary mirror is adjusted  by the three adjusting screws at the back
  of your scope. The  three locking screws beside them serve to hold the
  mirror in  place after collimation. (Fig.i)

  Aligning the Secondary Mirror

  Point the telescope at a lit wall and insert the  collimating cap into
  the focuser in place of a regular  eyepiece. Look into the focuser
  through your collimating  cap. You may have to twist the focus knob a
  few turns  until the reflected image of the focuser is out of your
  view. Note: keep your eye against the back of the focus  tube if
  collimating without a collimating cap. Ignore the  reflected image of
  the collimating cap or your eye for  now, instead look for the three
  clips holding the primary  mirror in place. If you can't see them
  (Fig.j), it means  that you will have to adjust the three bolts on the
  top of  the secondary mirror holder, with possibly an Allen  wrench or
  Phillip's screwdriver. You will have to  alternately loosen one and
  then compensate for the  slack by tightening the other two. Stop when
  you see all  three mirror clips (Fig.k). Make sure that all three
  small  alignment screws are tightened to secure the  secondary mirror
  in place.

  Aligning the Primary Mirror

  There are 3 hex bolts and 3 Phillip's head screws  at the back of your
  telescope, the hex bolts are the  locking screws and the
  Phillip's-head screws are  the adjusting screws (Fig.l). Use an Allen
  wrench  to loosen the hex bolts by a few turns. Now run  your hand
  around the front of your telescope  keeping your eye to the focuser,
  you will see the  reflected image of your hand. The idea here being
  to see which way the primary mirror is defected,  you do this by
  stopping at the point where the  reflected image of the secondary
  mirror is closest  to the primary mirrors' edge (Fig.m).  When you get
  to that point, stop and keep your  hand there while looking at the
  back end of your  telescope, is there a adjusting screw there? If
  there  is you will want to loosen it (turn the screw to the  left) to
  bring the mirror away from that point. If  there isn't a adjusting
  screw there, then go across  to the other side and tighten the
  adjusting screw on  the other side. This will gradually bring the
  mirror  into line until it looks like Fig.n. (It helps to have a
  friend to help for primary mirror collimation. Have  your partner
  adjust the adjusting screws according  to your directions while you
  look in the focuser.) After dark go out and point your telescope at
  Polaris, the North Star. With an eyepiece in the  focuser, take the
  image out of focus. You will see  the same image only now, it will be
  illuminated by  starlight. If necessary, repeat the collimating
  process only keep the star centered while tweaking  the mirror.


What is the optimal escape trajectory from near a black hole? Consider a space ship that is being drawn closer to a black hole.  The crew begins to notice the effects, and discovers that they are nearing the black hole.  They then manage to halt their progression toward the black hole.  What would be the optimal escape trajectory, assuming they have not yet wandered too close?  I am thinking it would likely either be directly away from the black hole (figure 1), or parallel to the black hole (figure 2).  However, it is conceivable that there would be some other angle that would be optimal (e.g. 45 degrees in figure 3).

Figure 1: Go directly away from the black hole

Figure 2: Go "parallel" to the black hole

Figure 3: Go 45 degrees from the black hole

A follow up question is, would this answer be the same for other high gravity objects, such as neutron stars, or even regular stars?

Let's say for simplicity sake that they have already managed to halt their advance toward the black hole and they are now still (not moving) in relation to the black hole. If distance factors in, it would be interesting to see what the answer is at different distances (e.g. fairly close to the event horizon where I realize a tremendous amount of energy would be required, and for far away).

I'd like to know a little more about the geometry of the ship's trajectory. I would be asking for clarification in comments but I don't know how to put images in comments.

A good distance away the ship is moving nearly a straight line with regard to the large mass. As the ship gets closer the path gradually bends towards the large mass. If you're still a fair distance from the star, the path can be fairly well modeled with Newtonian mechanics and the curving path can be modeled as a hyperbola. The straight line the hyperbola is gradually deviating from is called an asymptote:

This illustration (page 36 of my coloring book) is a hyperbola about the earth, but it could also be a hyperbola about a larger mass.

Escape velocity climbs as you get closer to the mass. The hyperbola's speed is sqrt(Vescape^2 + Vinf^2). I use this right triangle and the Pythagorean theorem as a memory device:

If you did a burn in the opposite direction from your Vinf vector, it could reduce your Vinf and drop you even closer to the mass.

If your burn vector is at right angles to your velocity vector, it would increase your vinf (and thus raise point of closest approach aka periapsis). It would also change the direction of the hyperbola's asymptote.

To better answer your question, I'd need to know more about the geometry of this scenario, what Vesc and Vinf is and how much delta V the ship was capable of. It would be helpful to know at what distance our heroes discover they're in trouble.

If the ship has already fallen close enough that it's traveling an appreciable fraction of c, the above doesn't apply. Conics from Newtonian mechanics is a good approximation until you get too close to the black hole. Then you'd need general relativity to model the trajectory -- and that's above my pay grade.


Concerning fate of Milky Way Galaxy Andromeda Galaxy is the largest galaxy in the galaxy cluster[1] called Local Group[2] and our Milky Way Galaxy as well as some satellite/dwarf galaxies are orbiting around each other within Local Group. My question isn't Milky Way Galaxy supposed to orbit around Andromeda Galaxy[3] so how come the predicted merger[4]? Whose slowing down or do the gravity work differently as things are being scaled further up?

The Milky-Way does not orbit the Andromeda galaxy, they both move under the influence of all the members of the local group. Even if one were orbiting the other the orbit need not be near circular but could be a very eccentric (elongated) ellipse.

The projected merger is because the tangential component of Andromeda's velocity with respect to the Milky-Way is small compared to its radial component That is Andromeda galaxy appears to be moving almost directly towards the Milky-Way (which is what it says in the Wikipedia page you link to, but it is not too difficult to find primary sources using Google, here is an arXiv paper reporting a proper motion study of Andromeda and reporting such).


Will the detection of colliding neutron stars by LIGO help answer the question of where heavy elements came form? I have read about and heard about the theory that most of the heavy elements we have here on Earth came from colliding neutron stars rather than by way of supernovae.

I'm not sure how strong this theory is, but I wonder if a LIGO (and backup observational/ gamma ray) detection would help answer this question in any way.

I just found out the answer to my question from a live press release on You Tube that has been covered by blogs like this: http://news.nationalgeographic.com/2017/10/gravitational-waves-discovered-neutron-stars-pictures-science/

What LIGO does is tell you what is causing the gamma ray burst that can be studied different ways. The highly anticipated announcement I just watched explained that LIGO observed the neutron star merger (they're calling the event a kilonova). With the help of VIRGO, the area of the sky could be narrowed down better than previous LIGO detections. LIGO told us the masses of the bodies, the distance to the objects, how much mass was ejected and thus what the end product of the collision is expected to be.

We have observed many gamma ray bursts and this one was not particularly bright, but LIGO told astronomers to check this one out and so it was being observed within a few hours.

The electromagnetic spectrum of this event was collected and very broad lines were detected which are believed to be caused by heavy elements being ejected at near light speeds.

By our knowledge of how often these "kilonovae" should happen and the volume of elements heavier than iron thought now to have been emitted (16 Earth masses was the number mentioned in the press release), it can be said that a majority of many of the heavy elements we see around us may have come from these events.


How to use an app to help find deep sky objects in general, either attached to a telescope as a finder scope or held on hand? A friend with a young child asked me for a beginner scope recommendation of less than $300 to see the planets as well as some brighter deep sky objects in a relatively dark suburban location. My first inclination is to recommend the Starblast 4.5 using a more novice-friendly Orion 7-21mm zoom eyepiece, but I think a major hurdle for complete beginners like her is not knowing how to star hop to find deep sky objects, and with a child around, patience is extremely limited.
With the current technologies on the night sky phone apps, I thought it might be useful and much cheaper than buying GOTO scopes to attach the phone to the manual telescope, such as a Starblast, to use as a secondary finder. I found YouTube videos about apps:

             (Sky Safari Pro) or like this

             (advanced app), but I don't know if amateur astronomers actually use the phone as secondary finders now. (Since I was last active with an astronomy club over 15 years ago, before any night sky phone apps became popular!) I searched online, but most web pages are not recent. So I am writing here. Do people now attach a finder-app loaded phones/iPads to telescope bodies as a secondary finder? If so, what apps do they use and does it work well (does it need frequent realignment)? I'm thinking of using a bunch of masking tape to attach the phone to the Starblast, or are there piggyback mounts I can buy for the Starblast?

If the above is not an option, I did find the Sky Safari Pro tutorial,

            , where you hold the phone/pad with this app to help guide the manual telescopes to find the objects, where the biggest advantage in star hopping is that the app can flip the stars as they are shown in the upside-down view through a Newtonian scope like Starblast. For people who have tried the Sky Safari Pro tool holding the phone/pad on their hands to help with their manual scopes locate objects, is it easy to use and helpful in the field?
Are there other recommended apps helpful in finding deep sky objects?

You can do this. It's possible to mount a phone on a telescope and use it as an electronic finder with an app like Sky Safari or Stellarium Mobile.
However:

The accuracy of it very much depends on the accuracy and sensitivity of the phone's sensors, and this varies quite a lot between phones
If the telescope tube contains ferrous metal then it'll throw off the magnetic sensor
The light of the screen adversely affects your eye's night adaptation

These things may or may not be issues depending on each individual situation.
I've tried it a couple of times but never been happy with the results. The two phones I tested weren't accurate enough.
Some people do use their phones as finders, but they tend to do it with digital setting circles (DSCs) such as the Nexus 2. This has a pair of 'encoders' which you attach to your telescope, and these are used to track the movements of the telescope's axes, and update a mobile application accordingly.

Personally, I use a Telrad non-magnifying sight to get myself in the general vicinity, then move over to a magnifying finder scope, and finally to the eyepiece.
I star-hop using a combination of three, depending on my target. A lot of the time I can go straight from the Telrad to the eyepiece. Other times I'll star-hop with the finder telescope.

If your friend is short on available observing time, then it can help to plan sessions beforehand: a list of targets and directions to them from the nearest brightest recognisable star.
A program like Kstars can be used to assist with this: it has a star-hopping planning tool as well as a "what's up tonight?" tool.


What is an asteroid called if it is in deep space? Pretty much every definition I've read about asteroids is something like "Asteroids are rocky worlds revolving around the sun that are too small to be called planets."

What would an asteroid be called if it was in deep space, not orbiting any sun?

We have only discovered one such object so far: ʻOumuamua. (Although a second, 2I/Borisov, was discovered in 2019) The general practice has been to call it an "interstellar asteroid" or "interstellar comet", even though it is not in orbit.  You could just call it an "interstellar object".


What to look for in a tripod for binoculars? I have been looking at large astronomy binoculars in the 25x100 and 20x80 range. Clearly, these monsters cannot be held steady by hand, or even be held at all for any reasonable length of time using one's arms.

Somewhat related question What should I look for in an astronomy tripod? isn't helpful as a telescope tripod needs to hold a much bigger weight, so it stands to reason that one would get away with something lighter and cheaper for big binoculars.

What's recommended, tripod-wise? Any related pro tips from the binocular crowd?

Get the sturdiest tripod that you can!

My 20x80 binoculars are sitting on a ball and socket mount on a very heavy tripod. I use the Benro BH1 ball mount, which is an absolute joy to aim. This mount is rated for 6 KG, almost three times the actual weight of the binoculars. With high-powered binoculars the aim must be adjusted every minute or so, therefore you will want an easy to adjust mount that is far from the limits of its capacity. You really can't beat the ball and socket design for this purpose in my opinion. I need to photograph the setup for another question and when I do I'll check what model the tripod is.

Look also at the tripod of the user who asked the above-linked question. These large binoculars need the absolute heaviest duty tripod that you can reasonably afford. You will need all-metal construction and I would suggest a hook as well to hang a weight from, this will make the tripod steadier. You will have a hard time going overboard, so if you find two tripods with a similar price but one is heavier-duty, get that one.


Is there a cosmic, rather than technological, upper limit to what a telescope can resolve? Space radio interferometers could have a baseline of millions of kilometers, but is there a point where a larger baseline doesn't improve the resolution anymore because the photons observed are distorted before they arrive? This question deals with technological limits of resolution. I'm instead asking about cosmic limitations due to for example interstellar and extragalactic gas which scatters light.

This paper about results from the RadioAstron space/Earth interferometer is well above my pay grade, but it seems to be about this problem. The executive summary says:

  At longer baselines of up to 235,000 km, where no interferometric
  detection of the scattering disk would be expected, significant
  visibilities were observed with amplitudes scattered around a constant
  value. These detections result in a discovery of a substructure in the
  completely resolved scatter-broadened image of the pointlike source,
  PSR B0329+54. They fully attribute to properties of the interstellar
  medium.

Q: "Is there a cosmic, rather than technological, upper limit to what a telescope can resolve?

  Space radio interferometers could have a baseline of millions of kilometers, but is there a point where a larger baseline doesn't improve the resolution anymore because the photons observed are distorted before they arrive?'

Deviation of the wavefront from a blackbody spectrum and the processes that create them is reasonably well understood, the phases of the incoming wavefronts can be measured to enhance the directionality and reduce the distortion. Correlation of the different frequencies allows detection of even small amounts of distortion.

See: "The evolution of CMB spectral distortions in the early Universe" (29 Sept 2011), by J. Chluba, R.A. Sunyaev and the Video: "Spectral distortions of the CMB and what we might learn about the early universe", by Jens Chluba, and "Science with CMB spectral distortions" (27 May 2014), by Jens Chluba.

The method is explained in the paper "Cosmic Microwave Background constraint on residual annihilations of relic particles" (9 Aug 2000), by McDonald, Scherrer, and Walker, page 2:

  "In this paper we compute the energy injected into the CMB by annihilating particles as a function of their mass and annihilation rate (i.e., the product of cross section and abundance squared). We derive constraints on the particle properties by comparison with the observed limits on chemical potential ($\mu$) distortions, and Compton-$y$ distortions (§II). We compare these constraints to similar constraints obtained from the production of deuterium by photodissociation of primordial helium (§III.A), and from the diffuse photon background produced after recombination by extragalactic annihilations (§III.B), and annihilations in the Milky Way halo (§III.C).

  II. DISTORTIONS OF THE CMB ENERGY SPECTRUM

  We consider first the effect of annihilation products on the CMB energy spectrum. The distortion of the spectrum takes place in two steps: first the high energy annihilation products rapidly dissipate their energy into the background photons and electrons, and then the low energy background evolves more slowly in an effort to restore the Planck spectrum. The permanence of distortions produced after $z ≃ 10^6$ is simple to understand in the following way: A Planck spectrum with a given photon number density must have a specific energy density. For $z ≃ 10^6$, photon nonconserving processes (double Compton scattering and bremsstrahlung) are inefficient
  in the background plasma. Therefore, if energy is injected into the CMB but not the the correct number of photons, a Planck spectrum cannot be restored. We now discuss in more detail the form of the distortions pro-
  duced in different redshift intervals. ...'.

See: "New Technology Offers to Broaden Vision for Radio Astronomy". Even a tiny antenna is very capable, this tiny antenna can see in seven directions simultaneously:

[Click image to zoom]

Adaptive optics are used with some optical telescopes, no reason not to apply the same algorithm to VLBI.

"Adaptive Optics: ESO's Very Large Telescope sees four times first (laser) light" (07/11/2016):

  "... even at the best optical sites on Earth such as the 2600 m summit of Cerro Paranal or the 4205 m summit of Mauna Kea in Hawaii, refractive-index variations in the atmosphere lead to a seeing-limited resolution of 0.4 arcsec in contrast to the theoretical diffraction-limited resolution of less than 0.02 arcsec for the 8.2 m VLT telescopes.

"PKS 1954-388: RadioAstron Detection on 80,000 km Baselines and Multiwavelength Observations" (5 May 2017), by Edwards, Kovalev, Ojha, An, Bignall, Et al.:

  "1 INTRODUCTION

  A major challenge in astronomy is the struggle to observe objects with an angular resolution sucient to probe the underlying physical mechanisms. The longer wavelengths of radio-astronomy initially made the quest for high angular resolution more difficult, but the relative ease of preserving phase information enabled the technique of Very Long Baseline Interferometry (VLBI). Intercontinental VLBI routinely achieves milliarcsecond-scale angular resolutions, and extending the baselines between telescopes into space, with satellitebased telescopes, currently yields the highest angular resolution achieved in astronomy.".

There's a newer paper than the one you linked to, using a longer baseline, with better results. See: "PSR B0329+54: Substructure in the scatter-broadened image discovered with RadioAstron on baselines up to 330,000 km" (13 Sept 2016), by Popov, Bartel, Gwinn, Johnson, Andrianov, Fadeev, Et al., the conclusion on page 8:

  "7 SUMMARY AND CONCLUSIONS

  Here we summarize our observations and results and give our conclusions.

  (i) We made VLBI observations of PSR B0329+54 with RadioAstron at 324 MHz on projected baselines up to 330,000 km or 350 M$\lambda$. Our goal was to investigate scattering properties of the ISM which affect radio observations of all celestial sources. While the results of such observations are in general influenced by the convolution of source structure with the scattering processes, pulsars are virtually
  point-like sources and signatures in the observational results can be directly related to the ISM scattering properties.

  (ii) Visibility function at short ground-ground baselines manifests a single bright spike in delay-rate space that vanishes on long space-ground baselines. Thus, the scattering disk of PSR B0329+54 was completely resolved on ground-space baselines of 15,000 to 30,000 km. The FWHM of the
  angular diameter is 4.8 ± 0.8 mas at 324 Hz.

  (iii) The diffractive length scale or size of the diffraction spot near Earth is 17, 000 ± 3, 000 km.

  (iv) With the assumption of turbulent and large-scale irregularities in the plasma, the effective scattering screen is located at d/D = 0.6 ± 0.1 or somewhat more than half of the distance from Earth to the pulsar.

  (v) At longer projected baselines, up to 330,000 km, significant visibility amplitudes were detected, although none were expected from the scattering disk. They are scattered around a mean which stays approximately constant up to the longest baselines. This result indicates that substructure was discovered in the scatter-broadened image of PSR B0329+54.".

Your last question:

  Q: "I'm instead asking about cosmic limitations due to for example interstellar and extragalactic gas which scatters light."

Because you are looking at so many frequencies from so many angles for an extended period of time it's possible to use a supercomputer to see through (around) atoms, fog, dust.

Here are some layperson's explanations:

Ted Talks video: "See invisible motion, hear silent sounds"

CNN YouTube: "New technology makes troops invisible"

Ascendent Technology YouTube "Thermal Infrared & Visible CCTV PTZ Camera Fire Monitoring & Detecting See Through Smoke & Haze"


How to properly observe Solar Total Eclipse in 2019? I live in South America and I'm willing to travel to Chile in July 2019 to observe (in a curious way, not professional) the Solar Total Eclipse.

According to my research, one of the best cities to observe the eclipse is in the coastal city of La Serena at 3:20 PM Local Time. However, I don't know if it can be seen from "anywhere". I was thinking of observing it from the beach. Does that make sense?

I also know that I need some kind of special sunglasses for safety. I found some on Ebay but I don't know if they are authentic. I would appreciate if someone could point me to a website that sells authentic eclipse glasses for a reasonable price and international shipping.

In short, my question is, will I be able to amazingly see the eclipse by just traveling to La Serena and go to the beach on the specific day and time? Is there a better place to see it? Is it better to see it from a high ground? Any tips are welcome.

Thanks.

You can get some equipment especially designed for the eclipse.
But don't forget about a simple pinhole camera
Important to note that you shouldn't look through the pinhole. Instead, you are using the pinhole to project an image of the sun on the ground behind you.
The gaps between leaves in a tree act as natural pinholes so if there are any trees around you, you should be able to see lots of eclipse images

One of the easiest ways to safely watch a solar eclipse is to use 2 sheets of cardboard and make your own simple pinhole projector.
A partial solar eclipse projected on cardboard. Solar eclipses can look spectacular, but the Sun’s UV radiation can cause permanent eye damage or even blindness. Make your own pinhole projector to view a solar eclipse.
Projecting the Sun through a box projector, or projecting using binoculars or telescope (edit: shown below), or simply 2 pieces of card is a safe and easy way to view a solar eclipse.

Source

Light filtering through the leaves of trees during a solar eclipse creates images of the crescent sun, like the image created by a pinhole camera. (Bigstock)

Source


Problem about orbiting of a satelite Problem: There is an satellite that orbits the Earth. Its orbit is circular and its orbital plane is perpendicular to Earth's Equatorial Plane. It orbits is such that a person who is on equator see this satellite on Zenith every 12Hours.

a. Find Satellite's Period.

b. What is distance of satellite to person when it Sets on person's horizon?

My problem with this question is:

For the first part: somewhere said period is 12hours. Somewhere else said the person on equator also moves (Earth Rotation) so 12 hours is P/2 and P= 24hours. What is the right answer?

For the second part: Which of the following figures show the position of satellite when it sets? (Figure 1 or 2?)

I'm treating this as a homework question. You have made a clear attempt, so it is not off topic, but you should clarify the source of the problem.

You will first need to establish if the rotation of the Earth is to be considered (it makes a big difference, but in a homework, sometimes the Earth stops spinning!) You have then correctly found the two possible answers, either 24 or 12 hours.

You can also consider the motion of the Earth around the sun (as the Earth only takes 23hr, 56min to spin once, relative to the stars). What this means is that, after 12 hours, the Earth has turned more than 180 degrees, and there is no orbital plane that passes over the person, and through the origin, and is perpendicular to the equator. However the error is only about 1 degree, so probably can be neglected for the purposes of homework.

For the second part, the first picture is certainly wrong, as the line of sight (in green) passes through the Earth. The second shows the line of sight (now in red) being tangent to the Earth's surface. You will need to use orbital mechanics to find the distance from the Earth's centre to the satellite, and find out the distance from the Earth's centre to its surface (green lines), giving two sides of a right-angled triangle. And the rest is maths.

There are a few details that may affect the answer: you should consider if the motion of the person will affect the answer to part (b) (and if not, why not). You should know that the Earth is not spherical. You should consider whether the optical refraction of light by the Earth's atmosphere should be considered.  However, in a homework problem, such issues can often be ignored, as their effects are small.


Might the universe be really small yet seem infinite from spherical reflections? if a single source of light quite small, was placed somewhere inside a completely reflective sphere, quite large, what would the reflections look like from the light source's point of view within that sphere?

:)

extra-credit: what would it look like over the course of time with movement starting from the moment the light turned on?

Found this question has been asked over in MathOverflow, sum clue?

The simple answer to your headline question is: No.

That kind of correlation between different angles has been investigated, and no significant correlation beyond 1 degree (first acoustic peak), e.g. large-scale circular structures in the CMB, has been found.

Details about the investigation of non-trivial topologies within the limits of recombination, see this Planck paper.

Closer-by structures (like galaxy clusters) visible by telescopic observations are correlated to gravitational lensing of the CMB, as described in this Planck paper, or by the integrated Sachs-Wolfe effect, as described in this Planck paper, since galaxy clusters tend to be correlated with dark matter halos.

Repeating pattens (from mirroring) would be evident from 3d maps (like this one for the local 300 Mpc environment) of the mass distribution or telescopic observations.

Multiple images of distant objects are well-known, e.g. as Einstein crosses, or Einstein rings, caused by gravitational lensing.
This can include mirroring in extreme cases; it's well-known how this would look like.


Estimating the abundance of type-Ia supernovae using magnitude limited survey I would like to estimate the fraction of type-Ia supernovae in a magnitude limited survey (only including those supernovae that are brighter than a certain than a given fixed apparent magnitude).

I currently have the following information:

Type-Ia supernovae are usually around 2.5 magnitudes brighter than core-collapse supernovae.
Volume limited surveys found that 30% of supernovae within a fixed distance from us are type-Ia.

How can I solve this problem using the above information? The formula for apparent magnitude $m = -2.512\log_{10}(d/10\mathrm{pc})+c$ seems like an obvious starting point, but doesn't seem to lead anywhere. A worked solution would be very helpful.

The rudiments of a solution would be take your apparent magnitude limit and work out how far away you can see type Ia supernovae and how far away you can see core collapse supernovae.

Within the latter distance, 30% of the supernovae would be type Ia. Beyond that distance, all of the observed supernovae would be type Ia.

You then need to be assuming something about the density of supernova progenitors and thinking about any spread in their absolute magnitudes. And you didn't mention any observational uncertainties.


What does it mean for a planet to be just 300 kms across? The recently discovered Planet Goblin is just 300 kms across. While I perfectly understand what it literally means, I'm having trouble wrapping this in my head. Does it mean I can imagine the planet as (approximately) a sphere for 300 kms diameter? That gives a radius of 150 kms, and thus a circumference of 942 kms.

A circumference of 942 kms means that the longest distance between any two points on the surface of the planet is just 471 kms -- That's just over 1.5 times the distance between London to Paris or 100 kms more than the distance between New York and Washington DC or about half the distance between St. Petersburg and Moscow.

Is the planet that small, really? Or am I missing something?

The reports that Goblin is a planet are incorrect. It is a dwarf planet, at best. Dwarf planet and planet are mutually exclusive terms. Planets are objects whose masses are so large that they have collided with, ejected, or captured the vast majority of the smaller stuff that share the same orbit as the planet.

Moreover, Goblin might well not even be a dwarf planet. The distinguishing characteristics of a dwarf planet is an object that is large enough to have pulled itself into a more or less spherical shape via self gravitation but is not anywhere near large enough to have the potential to clear the neighborhood of its orbit about the Sun.

The threshold that distinguishes dwarf planets from small solar system bodies is the potato radius. Objects somewhat smaller than the potato radius look like lumpy potatoes, while objects larger than that have sufficient mass to more or less pull themselves into hydrostatic equilibrium. The potato radius depends a bit on composition, and probably also depends on distance from the Sun. (Water ice is quite rocky at trans-Neptunian distances from the Sun.) The potato radius for rocky objects is about 300 km (diameter = 600 km), while for icy objects it's about 200 km (diameter = 400 km).

With an estimated diameter of 300 km, Goblin is probably too small to qualify as a dwarf planet, let alone as a planet.


Is there a reason two interstellar objects were found in the solar system in such a short period of time? ʻOumuamua was discovered on October 19, 2017, and 2I/Borisov was discovered on August 30, 2019.  Has something changed about observation techniques or equipment that made it possible to find these objects in less than two years?

I find it hard to believe that there has been no interstellar objects passing through the inner solar system in the past ten or twenty years, then suddently these two did, quite close in time.

I recognize it is incredibly hard to find such small, dark objects moving so fast, especially if they are outside of the ecliptic, and there be many more passing through that just haven't been discovered. Should this be chalked up to pure luck?

The biggest contributor to these two observations happening so close to each other is capability and interest.

Our ability to observe such objects is ever increasing.
Once we saw one, our interest increased and we looked harder ie. lots more funding for the necessary observations to be made that would identify such objects.
We saw the next one soon after.

There is unlikely to be anything special here, just the politics of scientific funding.

It may be that these were two objects that coincidentally passed through the solar system at nearly the same time, but we are going to need to make many more observations of such events before making such a determination.


Why Earth does not have rings? Why does the Earth not have rings like Saturn or Uranus?

A moon is held together by its own gravity, and pulled apart by the tidal action of a planet. If a moon comes too close to a planet it will be ripped apart by the planet's gravity and become a ring. The closest a moon can come to a planet is known as the Roche limit, and it is dependent on the mass and density of the planet and moon.

A large planet, such as as Saturn, has a large Roche limit, and has collected many moons over its history, some of which have been ripped apart to make rings.

The Roche limit for the Earth-moon system is at a radius of about 10000 km, which is very close to the surface of the earth (about 6400km). And unlike the gas giants, terrestrial planets don't have many moons.

So with few moons, and small Roche limits, the opportunities for ring formation are rare. That said, Phobos, one of the satellites of Mars, is now so close to Mars, that it is held together as much by its rigidity as by gravity, and is slowly getting closer to Mars. Within 50 million years it will be ripped apart and form a (small) ring around Mars


What is precipitable water vapor in millimeter-wave radioastronomy and how is it measured? Looking through ALMA on-line documentation and articles, the quantity PWV (precipitable water vapor) is a central theme. I have two linked questions.

What (actually) is precipitable water vapor, and is there a "non-precipitable" component also? Why not just call it "water vapor"?
How is it actually measured? It seems to be monitored by radiometers, but I don't understand how that can be used.

I usually see numbers of the order of 5 millimeter to 10 millimeters for very high altitude locations like Atacama. I assume this is the column-integrated amount, expressed in units of liquid, is that right?

As defined quite clearly, with equations, here

The total atmospheric water vapor contained in a vertical column of
unit cross-sectional area extending between any two specified levels,
commonly expressed in terms of the height to which that water
substance would stand if completely condensed and collected in a
vessel of the same unit cross section.

edit
There's, for example, an ALMA paper which discusses measurements and comparison with meteorological stations.


What does "EW(Hα)" mean? I came across a website saying that classical T Tauris have an "EW(Hα) > 5 angstroms".

Does anybody know what "EW" means?

Thanks.

"EW" stands for Equivalent Width. The equivalent width of a spectral line is essentially the range of continuum one would integrate over to get the same flux as the spectral line.

In your case, this is the Hα line.


What planet is better than earth to infer solar system configuration? The mankind had to work some centuries to infer the real configuration of solar system, starting from greeks, Ptolemeus, until Copernicus, Galilei, Kepler, Newton etc. Is there any planet where we could better/faster determine the configuration?

(For example, having a moon is an advantage. Maybe having two moons(or none) or having a thinner atmosphere or being closer/farther to the sun may helped.)

The area of the question starts from inferring the form of the planet(round) until the discovery and orbits of all eight planets.

PS: The question ignores the fact that life is not possible on another planet. It is from pure astronomical viewpoint. On Neptune you have a different sky, so other questions/answers.

I will argue Mars.

A smaller diameter makes it easier to determine the round shape of the planet, and determine the diameter more exactly.
A thinner atmosphere and almost no clouds are better conditions for sky observation.
The small orbital radius of the Mars moons makes it easy to figure out the distance to them. (For example, Phobos is only in the sky about a third of the time seen from equator). Also, the fact that there are two of them makes two data points to discover Kepler's third law.
Closer vicinity of the asteroid belt calls for an earlier discovery of objects with an irregular orbit.
The Moon is a much closer example of something orbiting another object than Galileo’s discovery of the Jovian moons. "Mars is the centre of the universe" disproved.


What happens when a huge star is being consumed by a black hole What happens when a huge star in its later stage( iron core) is near a black hole that slowly consumes some of the matter from the star? Would this offset the imbalance caused by iron fusion prolonging/ deterring a supernova?

A large star near a black hole may lose some of its atmosphere by having it spilling over its Roche lobe and into an accretion disk around the black hole. That disk will also radiate significant amounts of UV and X-rays that can blow away more atmosphere. That could in principle lighten the star, lowering the pressure and temperature in the core, and extend its lifespan a bit. Usually the amount of mass lost is just a few percent, so this is not a big difference. However, as discussed in the first link, for some heavy stars significant amount of matter may be lost (especially since the star itself may radiate so much that it loses its outer envelope). It does not prevent the star from eventually going supernova, but can change what kind of object it turns into (neutron star or black hole).


How do/did we figure out that planets move in orbits? I've learned that planets move in orbits around the Sun, but I really don't know how I would come to this conclusion myself. I've only seen planets in the sky a couple times (knowingly), and I am curious how we know for certain today that planets really move in orbits around the Sun (i.e., rather than moving but not around the Sun or moving around the Sun in a shape that is not a regular path).

I've also heard that people in the past knew about orbits even when they thought that Earth was at the center of the solar system. How did they figure this out in their times with their technology?

I've also heard that people in the past knew about orbits even when they thought that Earth was at the center of the solar system. How did they figure this out in their times with their technology?

The same celestial objects (stars, planets, the Moon) could be seen every year. So, people figured out there was a pattern to it.
At first, geocentrism was prevalent due to prejudice and the  stars "orbiting" the Earth in a clean manner (no strange effects). If you stare up at night, stars move in a circle with the center approximately at the Pole Star. This can be mistaken for the Star moving around the Earth.
Objects like visible planets (with retrograde motions) were relatively small in number — we have the inner solar system plus Jupiter and Saturn, so people gave plausible explanations for these. Ptolemy theorized that the planet orbits around a "ghost point", for example:

(This is called the epicycle theory)

how we know for certain today that planets really move in orbits around the Sun (i.e., rather than moving but not around the Sun or moving around the Sun in a shape that is not a regular path).

The heliocentric model came up because of various smaller discoveries:

Jupiter has moons, so not all celestial bodies orbit the Earth
Venus has full phases, but in the geocentric model only a small subset of the phases should be visible
Theories like epicycles were not as accurate as they were expected to be, and people started realizing this over time.

With all this in mind, the simple solution of heliocentrism explains everything in a much cleaner fashion, without leaving much doubt in mind. Kepler's laws make much cleaner predictions.
Remember that people had catalogs of astronomical data to verify things with.
Later on, once Newton came up with his theory of gravitation, Kepler's laws made even more sense (though I believe that Newton formulated his theory with some inspiration from Kepler's laws). I'm pretty sure that there were on-Earth experiments to verify the inverse-square as well.
Nowadays, sending probes to space and getting a different perspective has confirmed this with 100% certainity for us. More or less.


How to photograph Saturn I have been given an old telescope (Newtonian, 910mm) and I got a mount for my DSLR and I'm trying to take some pictures of various objects.  In particular, I'm trying to photograph Saturn, but I'm having some difficulty getting a decent image.  This is about the best I could get:

This was taken with a 2x Barlow T-mounted to my DSLR via an extender, basically a metal tube about 2.5 inches long (without both of these my focuser doesn't have enough range unless I drop an eyepiece into the extender, then the Barlow isn't needed).  I shot on a one second delay to allow any vibration from the shutter opening to subside.  Focus was manual via the telescope itself, and was at the sharpest spot I could find.   I had been outside for some time and temperature difference between the air and the scope were minimal.

I'm wondering if there are any pointers or maybe an idea of what I might be doing wrong.  This above exposure was 1/10s, ISO 5000.  I thought maybe the ISO setting was too high, but the image below was at 1/15s, ISO 2000 and doesn't look any better:

There seems to be some chromatic effect, in the first image the upper-left side has a blue tint which works its way across the spectrum to red in the lower-left.

The scope was collimated using the star method as I don't have a tool, so it may not be perfect but it's pretty good.

The thing that I am really puzzled about is that through the lens the image did not look like either of these pictures.  Although a bit small, I could make out the variations in color and some of the texture of the planet and its rings.  The image was sharp and there were no chromatic aberations.  Aside from having the rough shape of Saturn, the pictures have almost no resemblance to what I saw right before I took the pictures.  (This is looking directly through the lens of the camera through the optical viewfinder, live preview also looks ok, but the screen resolution isn't as great.)

Hmm - good question, Michael; and thanks for sharing your images with us.

For me, it's about eliminating possible causes:

Collimation - OK (the star test should be fine for your purposes)
Focus - probably OK (suggest you try the same kit - scope, Barlow, DSLR - during the day and focus on a far distance object if possible to double-check)
Chromatic Aberration - Newtonians are less prone certainly, but it can happen. The most common cause is atmospheric dispersion caused by looking at an object very low in the sky. Your profile doesn't say where you're observing but Saturn is very low in the Northern Hemisphere sky at the moment
Magnification - one problem for you is that your kit is probably not optimal for small objects. I have a 1000mm FL Newtonian and need to use a x3 Barlow and a specialist CMOS/CCD planetary camera for Saturn as it looks too small on my DSLR (which gets used for larger objects)

You said "The thing that I am really puzzled about is that through the lens the image did not look like either of these pictures" - please could you clarify what you meant by lens here; were you talking about using the scope, Barlow and a separate eyepiece, or were you using scope, Barlow and DSLR with live preview lens?

It is much more common for people to image the planets using a video stacked approach. The idea behind this is that they are very bright (so exposure times can be very short) and the limiting factor is random blurring caused by atmospheric seeing.

If you take a video file of say 1000 frames at 20fps then there is software available that can combine these into good shots (e.g. AutoStakkert, Registax). What DSLR are you using? For what it's worth, the human eye is very good at picking details and resolving out the seeing on bright objects, so my first photo attempts were dreadful compared to what I could see in the eyepiece.

I'd recommend you try reading up on this approach for inspiration. Some (but not all) DSLRs do allow video recording modes, so it might be worth a try.

Saturn is definitely a tricky object to go for - at the time of writing Jupiter isn't in a good Northern Hemisphere space (I'd normally suggest trying it out as a somewhat easier target). Have you tried practising by imaging parts of the Moon - it's a much easier target to build confidence and practical experience?

Please don't be disheartened ... it took me a couple of years to get semi-decent pictures of Saturn


Is this object Mars? I decided to try the Astrophotography mode on my phone. Looking at the image I saw an object I didn't notice at the time.
What is the blue circle right of the Moon and just on the top right of the hole between the clouds?

I tried to look at a planetarium (time and location included, no idea about the azimuth). I couldn't find a planet looking at this direction.
According to this (if I understand it correctly) the only planet visible at that time and location should have been Mars.

Is this object Mars?

I'm pretty sure this is simple lens flare.
While lens surfaces in cameras have antireflection coatings that strongly reduce reflections, they are not perfect. A very bright small spot like the Sun or Moon or a bright streetlight will result in near-mirror images opposite the center of the image.
Analysis from this answer:

Here I used this effect on purpose to reduce the brightness of the partially eclipsed Sun before it became annular. Solar eclipse of December 14, 2020):

complete image:
 click for larger


What are the exact causes that differences in sunrises are shorter around june and december but long in march and september? The changes of the sunrise and sunset times not expire regularly in a straight line but according to a sinusoid. Around the solstice (summer solstice on June 21st and winter solstice on December 21), the day length changes the least. The difference in sunrise and sunset in the days around the solstice is only a few tens of seconds per day.

At equinox (March 20 and September 23), the length of day and night on earth is everywhere the same. The difference in day length from day to day around the equinox is changing rapidly. The difference in the length of day can rise around the time of equinox to 3 minutes per day.

This asymmetrical lengthening and shortening of the days is because the "middle" of the day, every day a little shifts. This has to do with the fact that the earth does not describe a exact circle orbit around the sun, but an elliptical orbit. Because the Earth's axis at an angle to the orbit around the sun takes the shorter and longer are the days asymmetrical.

As for the elliptical orbit, I understand that, for example,the sun is the farthest from  the earth in the summer (on northern hemispere). Because of the (second) law of areas of Kepler is the speed of the earth there  the slowest. In itself, I understand then that at that time the difference in day length also is shorter. That is during June 21 (summer solstice).

But so is apparently also on the winter solstice on 21 December. Precisely at that time the earth is closest to the Sun (perihelion) so the earth has a higher speed. But because of the higher rate would you expect the day lengthening/shortening would therefore be greater. However, it appears not to be so. How is that possible?

So the question in short terms:

sunrise in june on the 21 at 5:30 and 22 at 5:31 etc
sunrise in march on the 21 at 7:30 and 22 at 7:33
So the difference is in june one minute and in march it is 3 minutes. And the question is what causes that the difference is bigger in march then in june. See http://www.timeanddate.com/sun/france/paris for a bigger disquisition

If I understand you correctly, you ask why does the length of the day not change much around the winter solstice.

There is no mystery here: The rate of change of day length has (almost) nothing to do with the elliptical orbit of the Earth.

In autumn the days get shorter; the rate of change of day length is negative. In spring the days get longer; the rate of change is positive.
At midwinter solstice the rate of change of day length is zero: The length of the day doesn't change.  Exactly the same is true in midsummer. Mathematicians would say that the "day length function" is stationary on those dates.

This has nothing to do with the elliptical orbit. Every smooth function is stationary at its maxima and minima. The reason is not astronomical but mathematical.

Note that the Earth is furthest from the sun (and therefore moving slowest) on July 4th.


How do we know galaxy GN-z11 is as far away as it is? The news media recently reported that galaxy GN-z11, formed just 400 million years after the big bang, is much further away than originally thought at perhaps 13.2 billion light years.  They say,

  The key to the discovery was precisely measuring the shift of the
  galaxy's light into longer, redder wavelengths, which correspond to
  how far the photons had traveled before reaching Hubble's eye.
  (Reuters).

So, suppose I'm standing next to, but safely away from, some train tracks.  As the train approaches, sounding its whistle constantly, the pitch will appear to increase until it reaches me, upon which I will briefly hear the whistle's 'rest pitch' (the pitch of the whistle as if I were aboard the train), then it would decrease as the train moves away.  Thus, the lower the pitch, the further away the train is.  I hope that's right.  So, if we know the rest frequency of galaxy GN-z11, its measured frequency indicates how far away it is from earth. Trouble is, how do we know what the rest frequency of light coming from this distant galaxy is?

So GN z-11 is the latest "furthest away" galaxy. It has a claimed redshift measurement of $z=11.1$, meaning that we are seeing the light it emitted about 400 Myr after the big bang (dependent on an assumed set of cosmological parameters).

To get the redshift measurement, the discoverers used grism (relatively low resolution) spectroscopy in the near infrared. What they were looking for is the rest-frame Lyman alpha continuum break, which would be redshifted into this wavelength range.

The Lyman alpha continuum break is caused by the absorption of nearly all ultraviolet photons with wavelengths smaller than 121 nm. These high energy photons are capable of photoionising the neutral hydrogen present in the intergalactic medium at a range of lower redshifts. This neutral hydrogen is present in abundance at redshifts greater than 6, as it had not yet been re-ionised by quasars and starlight. The consequence is that no light is expected to reach us from rest-frame wavelengths shortwards of 121 nm, but a galaxy's light can reach us from rest-frame wavelengths that are longer than this. When one observes the spectrum, we see flux at long wavelengths which suddenly cuts off at shorter wavelengths. The wavelength of the break is $\lambda = 121 \times (1+z)$ nm, where $z$ is the redshift.

It is this Lyman alpha continuum break that has been identified at an observed wavelength of 1470 nm. This leads to the redshift estimate of $z= (1470/121)-1 = 11.1$.

The details are presented in Oesch et al. (2016). The continuum break is the only thing visible in the spectrum. The authors are confident that this is what it is because their previous broadband photometry had given them an estimated redshift of $>10$ (which is why they observed this candidate in the first place).


What will eLISA be trying to observe? eLisa - Evolved Laser Interferometer Space Antenna.

  A LISA-like mission is designed to directly observe gravitational waves, which are distortions of space-time travelling at the speed of light. Passing gravitational waves alternately squeeze and stretch objects by a tiny amount. Gravitational waves are caused by energetic events in the universe and, unlike any other radiation, can pass unhindered by intervening mass. Launching eLISA will add a new sense to scientists' perception of the universe and enable them to listen to a world that is invisible in normal light.

The "duh" answer to my question is "Gravity waves, of course."

Obviously this is not what I'm asking.

So, we've deployed eLISA and we spotted a gravitational wave passing. Well, there went the wave. What data do we extract from it? What discoveries, observations will we acquire? What else can we discover with it than "there was a gravity wave, and it came from this direction"?

The first observation is whether gravitation radiation exists as predicted by General Relativity. Evidence from observations of binary neutron stars says it does, but it remains a major unknown.

Now Gravitational waves have been discovered, and the predictions of GR have been confirmed

Gravitational astronomy will be more like listening than looking. Right now I can hear my kids playing upstairs. I can learn a lot about what they are doing just by noting that sound waves are passing, from a particular direction.

Since the first detection of gravitational waves, we have observed several black hole merges, of different sizes. We have also observed neutron star collisions and subsequently observed these at gamma ray and visible wavelengths. This confirms these events as a potential source of many heavy elements.

We would expect extreme gravitational events to produce particular wave forms, for example black hole mergers should make a "tone" that rises in pitch as the two event horizons merge at faster rates. Again we have lots of theory on this but if we can "hear" these events we can check if GR does correctly model gravity in these situation, or if there is something missing.

Most interesting would be if we do hear Black hole mergers, but they don't sound like what we have expected. That would mean that there is more to gravity than we understand, and would lead to new science.

Now after several observations, it is the combination of gravitational observations combined with light observations that are producing the most science. E-Lisa will allow much smaller to be detected.


Musk's Roadster is now Russell's Teapot? Space Exploration Stack Exchange were not very fond of this question and closed it, so I think it might fit better here.

Summary: Assuming no foreknowledge of its existence, can Elon Musk's Tesla Roadster be detected and categorised from Earth?

Bertrand Russell argued the following, an argument that is now known as Russell's Teapot (emphasis mine):

  Many orthodox people speak as though it were the business of sceptics to disprove received dogmas rather than of dogmatists to prove them. This is, of course, a mistake. If I were to suggest that between the Earth and Mars there is a china teapot revolving about the sun in an elliptical orbit, nobody would be able to disprove my assertion provided I were careful to add that the teapot is too small to be revealed even by our most powerful telescopes.

Well now we have a midnight cherry red automobile, between Earth and Mars, in an elliptical orbit around the Sun.

Now assume that this civilisation we have falls. The knowledge of this automobile is lost. Then in some centuries/millennia later humanity has managed to work its way back to a level of technology comparable to what we have today.

Could this automobile then be detected by "our most powerful telescopes"?

I guess that we could detect it as an orbiting object; it is large enough to be found. Can you confirm or refute this guess?

And if the automobile can be found as an orbiting object, could it then be determined that this is not an ordinary space rock, but that there is something very odd about it? That is to say: is it possible to do any kind of characterisation of the automobile that makes it stand out as an unusual space object. There is some uncertainty about the exact state of the automobile after such a long time, but there is no question that chemically and optically, this will be no space rock, even in several thousand years. The question is if we can detect that.

Digging some more I found about the (re)discovery of J002E3, which turned out to (probably) be the Apollo 12 Saturn-IVB upper stage. I quote from this paper (emphasis mine):

  Additional spectral observations were completed in May 2003 at the Air Force Maui Optical Supercomputing (AMOS) site. Through the modeling of common spacecraft materials, the observations of J002E3 show a strong correlation of absorption features to a combination of human-made materials including white paint, black paint, and aluminum. Absorption features in the near IR show a strong correlation with paint containing a titanium-oxide semiconductor.

So, in short: could we do the same with this automobile and find that this is not a plain old space rock? How close would it have to be for this?

Small objects in space are hard to detect and characterise. Compare the tesla with another small solar system body: 2015TC25, It is currently about 0.3AU (3/10 of the distance from Earth to the sun) and it has a magnitude of about 30. That is perhaps detectable if you know exactly where to look. It would not be "discoverable" at this distance, with current technology. We know about because it made a close pass of Earth in 2015, and it was discovered during that pass.

The Tesla could be discovered if its orbit takes it close to the Earth. Observations of its spectrum would indicate it is not a regular asteroid, The cherry red paint would mark it as being utterly different from natural asteroids.

How close depends on what technology you have. By comparision with 2015 TC25, it might be observable from 0.3AU, if you have enough time and money to spend.


Is it odd that our Sun has so many planets? The internet is abuzz with the new evidence of the possible ninth planet in our Solar System. With those nine objects, the asteroid belt, the Kuiper belt, and everything else in our Solar System - the number of objects which orbit our star is pretty massive. Is this odd for a star its size? It's known that our Sun is a comparatively small star, so does having so many planets make it an oddity? Or do larger stars tend to have dozens or planets, even hundreds?

From an exoplanet-finding point of view, the Sun has between one and three planets.

The major exoplanet-finding techniques in current use involve watching for either periodic Doppler shifts as the planet's gravitational pull causes the star to wobble, or periodic brightness shifts as the planet transits the star.  Both require that the planet is large enough and close enough to generate a measurable signal and that the orbital period is short enough to let astronomers distinguish periodic variations from one-off variations; the transit method additionally requires that the planet's orbit crosses the star from the point of view of Earth (which favors close-in orbits).  Looking at the Solar System with these techniques:

Mercury: too small
Venus: Maybe visible
Earth: Maybe visible
Mars: too small
Jupiter: Highly visible
Saturn: Orbital period too long
Uranus: Orbital period too long
Neptune: Orbital period too long
"Planet 9": Orbital period too long

If you look at this graph of exoplanet discoveries, Jupiter is solidly in the cluster of blue Doppler discoveries, Saturn is just past the "we've been watching for one full orbit" right-hand edge of that cluster, Earth and Venus are somewhat below the sloped minimum-period-mass line, and everything else is nowhere near the detection range.

The reason the Sun has far more known planets than any other star is simply because we've got a better look at it.


How uniformly is cometary debris distributed along the entire orbit of the comet? Meteor showers occur when Earth passes through debris shed by a comet (if I get that right). Does this mean that the entire orbit of the comet is strewn with debris? I suppose that the debris follows the same orbit as the comet core it was released from as it passed close enough to the Sun for ices of different kinds to sublimate. Is the debris that cause meteor showers somewhat evenly distributed along a huge ellipse crossing the orbits of Earth and (often) Jupiter?

Are such cometary debris streams, and not only those crossing Earth's orbit, somehow cataloged and characterized?

There are more or less two possibility to get debris in a sufficient amount: collisions and breakup/ejection.

Collisions

This is mostly the case for asteroids. At the point of impact you get a cloud of debris which basically moves as if the colliding bodies combined into a bigger one. However, for small objects, the fragments in the cloud are not gravitationally bound to each other and thus can separate over time. If larger objects survive the initial collision, they can undergo further impacts at different orbital positions leading to the seasonal meteor showers.

Breakup/Ejection

This is mostly the case for comets. Particles get ejected from the original body by heating of the surface, or other effects. This is only possible for small grains, which are massively influenced by stellar radiation pressure. The result is an effectively smaller gravity of the star and thus a different orbit. Being on different orbits, theses particles naturally have to separate.

Conclusion

In both cases you get a physical separation over time. Being that all of the quantities involved are on a continuum, you effectly have a smeared distribution of fragments over an orbit. Depending on the special case, you can get a distribution over the whole ellipse, but a part of it seems more likely.

EDIT: As you asked about catalogues, I would recommend you looking at https://www.ta3.sk/IAUC22DB/MDC2007/ plus at http://adsabs.harvard.edu/abs/2006mspc.book.....J for a scientific read.


Does the "photo of a black hole" taken by EHT actually depict a visible light-based scenario? I'm still at my bachelor of electronic and communications engineering, so I'm not an expert of radiotelescopes; however, I surely know that, typically, radio communications are characterized by a RX that has at least one bandpass filter.

As a result, I am led to think that the "photo" of the black hole, recently released to the public, is a reconstruction picture taken from various radiofrequency signal samples that got mapped to the frequencies of visible light spectrum (since the Event Horizon Telescope project is based on radio telescopes), hence it doesn't depict what we would actually see (neglecting any impossibilities in doing so), but rather, it shows the radio emission from whatever is close to the event horizon of the black hole.

Is this correct, or is it actually a photo capturing visible light?
I mean, scientifically, this is a huge step; however, I have the feeling that the news got "boosted" a little too much, in medias, as they just talk about "photos" and "light" surrounding a black hole - not specifying anything else for a inexperienced audience.

You are right that the image is not a visible light image. It was taken using very, very short wavelength microwaves -- the wavelength is 1.3 mm which is only a bit shorter than the Far IR!

I think you may be reading a bit too much into the word "photo" which, as you say, has the everyday meaning of "taken with a camera using lenses in visible light."  I think most astronomers would use the term "image" rather than "photo" -- if you look at the NSF's press release, for example, you'll see that it uses the term "image".  (Links to all six of the papers the EHT team has published are in the NSF press release if you want further details.)

(I'd bet that "image" was changed to "photo" somewhere in the journalistic process as being "more meaningful" to the average reader.)

The appearance to the naked eye would probably not be terribly different, since the photon orbits don't depends on frequency.

As far as whether it got "boosted" or not, you're right to be very skeptical -- university publicity offices are prone to exaggerate and sometimes pump relatively ordinary science into amazingly important-sounding "breakthroughs."  In this case it all seems pretty straightforward, though. And there's nothing special about a visible-light image over a microwave image -- both are useful and neither tell us everything. (It is not uncommon for images in the visible to be among the least useful -- with X-ray or radio actually telling us much more.)


When we say a galaxy is 200 million light years away, does this account for the expansion of space in the time it took it's light to reach us? When we see a distant galaxy, the light we are watching started it's journey millions of years ago. In all that time, space has been expanding, so if the initial distance was A, right now the distance would be A+B and, surely, light would take more time to travel A+B than it originally did A.

Usually the distance data doesn't take the expansion of the universe fully into account. That's not important for a distance of 200 million lightyears, but as we get close to the border of the visible universe, it makes a relevant difference. Therefore astronomers prefer to talk of redshift instead of distance.

To be a little more precise: The light travelling from a galaxy to Earth takes only that part of the expansion of the space between the galaxy and the Earth into account, which it still has to traverse.

For a proper distance measurement in the everyday's sense, we would have to send light to the galaxy and reflect it back to Earth, measure the time, divide it by the speed of light, and take the half of the resulting distance.

If this experiment would be performed, some of the now visible objects would never reflect light back to Earth, because they are already too far away.

For a more detailed discussion about the various notions of distance, read Wikipedia's article on distance measures in cosmology.
The usual distance data is provided in the (one-way) light travel distance.


Is there a single, most prominent helium line? There's a prominent spectral line at ~1420 MHz caused by hydrogen.  From what I can tell, this is the most notable electromagnetic emission from hydrogen in radio astronomy.

Is there an equivalent helium line -- a single, prominent radio emission of helium in space?

(And are there equivalents for other elements too?)

I am unaware of there being a single Helium line that is as prominent or useful as the 21cm line is for Hydrogen.

The 21 cm line is due to a hyperfine transition in the 1s ground state of hydrogen. It is notable due to its strength, and also its usefulness for mapping neutral hydrogen in the universe, which is impossible in the optical since it is absorbed by neutral Hydrogen. It was in fact the first such spectral line observed in the radio part of the spectrum, and is so fundamental that the Pioneer and Voyager probes used the length of the transition as a time and distance unit on the plaques they carry, so that potential extraterrestrials could decode the messages.

There are many more spectral lines in the radio from heavier elements, and a whole field dedicated to observing and understanding them: radio frequency spectroscopy.

For Helium in particular, hyperfine lines can also be observed; the hyperfine transition of the $^3$He isotope has a wavelength ~4mm. The same isotope also emits a recombination line, where an electron recombines with an ion and falls to lower energy levels, emitting a photon in the process. The wavelengths of such recombination lines for Hydrogen like atoms (singly ionised in the case of Helium, so that it has a single electron) can be estimated using the Rydberg Formula:

$$\nu = R_{m}c\left[\frac{1}{n^2}-\frac{1}{(n+\Delta n)^2}\right]$$

where $R_m$ is the reduced mass Rydberg constant, $n$ is the final state and $\Delta n$ is the change in state.

Estimating the strengths of such lines is a little more complicated, but there are details here.


The Sun and Regulus together at the summer solstice Viewed from Stonehenge, when did the Sun and Regulus rise together at the summer solstice? Also when did Regulus rise before the Sun and would have been seen before the dawn glare?

Using sky simulator software (in my case Guide 9.1), the sun and regulus rose at the same time in about 2500 bc.  I think you would need to go back another 1500 years to be able to see regulus in the pre dawn glow. At that time, the sun would be about 7 degrees below the horizon as regulus was rising.

However the surrounding woodland wasn't cleared until about 5500 years ago (per Wikipedia), so it may not have been possible to make this earlier observation.


What happens when star spins too fast? Some stars such as a neutron star can spin very fast around 600 times a second, wouldn't the star be ripped apart? Although the gravitational pull is very strong but gravity is the weakest of all known forces and any good theory why neutron stars spin that fast does it have anything to do with the magnetic field and the surrounding dust clouds?

The structure (mass versus radius and density profile) is influenced by its rotation rate, but not by as much as you might think.

Even in Newtonian physics you can think of a mass element $m$ at the surface of a star of mass $M$ and radius $R$, rotating with angular velocity $\omega$.

A condition for stability would be that the surface gravity is strong enough to provide the centripetal acceleration of the test mass.
$$ \frac{GMm}{R^2} > m R \omega^2$$
If this is not satisfied then the object might break up (it is more complicated than this because the object will not stay spherical and the radius at the equator will increase etc., but these are small numerical factors).

Thus
$$ \omega < \left(\frac{GM}{R^3}\right)^{1/2}$$
or in terms of rotation period $P = 2\pi/\omega$ and so
$$ P > 2\pi \left(\frac{GM}{R^3}\right)^{-1/2},$$
is the condition for stability.

For a typical $1.4M_{\odot}$ neutron star with radius 10 km, then $P>0.46$ milliseconds. A proper General Relativistic calculation of this limit would give a similar result, but depends to some extent on the equation of state of the neutron star.

Happily, this is easily satisfied for all observed neutron stars - they can spin extremely fast because of their enormous surface gravities and all are well below the instability limit. I believe the fastest known rotating pulsar has a period of 1.4 milliseconds.

You also ask how pulsars can attain these speeds. There are two classes of explanation for the two classes of pulsars.

Most pulsars are thought (at least initially) to be the product of a core-collapse supernova. The core collapses from something a little smaller than the radius of the Earth, to about 10km radius in a fraction of a second. Conservation of angular momentum demands that the rotation rate increases as the inverse of the radius squared. i.e. The spin rate increases by factors of a million or so.

Pulsars spin down with age because they turn their rotational kinetic energy into magnetic dipole radiation. However, the fastest rotating pulsars - the "millisecond pulsars" are "reborn", by accreting material from a binary companion. The accreted material has angular momentum and the accretion of this angular momentum is able to spin the neutron star up to very high rates because it has a relatively (for a stellar-mass object) small moment of inertia.


Could stars form outside of galaxies? Is it possible for there to be a dense enough nebula to form stars outside of any galaxy? Does a galaxy have a minimum size to produce stars? Or could you have a few dozen stars clustered together by themselves?

Yes, stars can form outside galaxies if the conditions are right. An impressive example is D100, a galaxy that is moving through a cluster so fast that the ram pressure from the ambient gas forces galactic gas out of it leaving a long tail. That tail is dense and cold enough to allow star formation, and there are newly formed clusters in it.

In principle star formation can happen anywhere where there is dense, cold gas that can collapse gravitationally. Stars can sometimes help by forcing gas together by radiation pressure. There are intergalactic gas clouds, but they are usually too hot.  Dark matter halos are not great at capturing hot gas, but intra-cluster gas is losing energy from Bremsstrahlung so over time some of it would accumulate there.

Space is large, so no doubt there is some star formation happening in corners where gas clouds have cooled off remotely from galaxies. After all, that was how stars and galaxies once started.


In astronomy Is there a name for the exact time between solar noon and sunset? Like there are names for various events such as sunset, sunrise, solar noon, solar midnight, astro, nautical and civil twilight, what exactly is name for time between sunset and solar noon. Afternoon it is but what exactly between these two points? Is golden hour the only subcategory for the afternoon?

I did come across the word midafternoon. similarly, midnight, midmorning and midevening. but then again midafternoon would mean 3 o clock or does it mean point between solar noon and sunset? because midnight in general is more acceptable as 12 o clock. is it?

No, there is no such word better than "afternoon". I can't think of a situation that would require technical language for this period of time, the exact point halfway. When astronomers want to be technical about time they use UTC or UT1 or AT1 or any of the other "times". When they want to talk about more loosely defined periods they use "morning", "afternoon", "evening" just like everybody else.

"Mid-afternoon" is an ill-defined period, from about 2pm to about 4pm


How far into the future can we go by traveling close to a black hole? If we sent someone on a path that passed as near as possible to a black hole without getting pulled in, how far into the future would they go due to time dilation?  Let's assume the black hole is 5 solar masses (I am assuming the mass will affect the calculation).  I would like to know from two considerations:

An observer that can survive anything (e.g. ideal case)
The forces that a human can tolerate (e.g. a human would not be able to get as close to a black hole as would be ideal)

The way that you have specified the question, the answer is as far as you like. You simply put your spaceship into any orbit around the black hole and wait.

A more sensible question is what is the largest time dilation factor that can be accomplished - i.e. that maximises your travel time into the future for a given amount of proper time experienced on the spaceship.

This in turn is governed by how close to the black hole you can come and still tolerate the tidal forces. If you don't put a limit on this (your first case), then the answer is again infinite; you can hover as close to the event horizon as you like, using an enormous amount of rocket fuel, and the time dilation (see below) can be arbitrarily large.

Your second case is more realistic. Roughly we can say that the tidal acceleration across a body of length $l$ is given by $GMl/4r^3$, where $M$ is the black hole mass and $r$ is the distance from the black hole. If we make this acceleration equal to say $1 g$, and your body length $l \sim 1$m, then for a $5M_{\odot}$ black hole $r \simeq 2500$ km (well outside the Schwarzschild radius of 15 km).

If you could "hover" at this radius, then the time dilation factor would be
$$\frac{\tau}{\tau_0} = \left( 1 - \frac{2GM}{rc^2}\right)^{1/2},$$
where $\tau$ is the time interval on a clock on the spaceship and $\tau_0$ is the time interval well away from the black hole.

For $M=5M_{\odot}$ and $r = 2500$ km, this factor is 0.9970.

If the spaceship is in a circular orbit, the factor is $(1 - 3GM/rc^2)^{1/2} = 0.9955$.

These factors are perhaps not as big as you might have imagined!


What is the difference between these various Angular Resolution formulas? I have recently learned about angular resolution and its dependence on the ratio λ/D. There seem to be three different equations for angular resolution that I have come across. One of these has the number 1.22 (radians), while the other contains the number 206,000 (arcseconds). A third equation states that angular resolution can be found just with the ratio λ/D. I have attached a picture of these three different equations - please pardon the bad merging!
Naturally, this has caused some confusion on what situations each of these equations apply to. Would one of these be more relevant to an interferometer — while another is for a single telescope? I'd like to share that these equations come from the textbooks "Foundations of Astrophysics" (Ryden & Peterson) and "Understanding Our Universe" (Palen, Kay & Blumenthal). If anyone has come across these equations, I'd be grateful for some advice on how to distinguish them (i.e. what situations or conditions they apply to).

$$\theta_\text{min}[\text{rad}] = 1.22 \frac{\lambda}{D}$$
$$\theta = 2.06 \times 10^5 \frac{\lambda}{D}~\text{arcseconds}$$
A telescope's resolution is determined by the ratio $\lambda/D$.

$2.06 \times 10^5$ is the radian arcsecond conversion, equal to $180 \times 60 \times 60 / \pi$ .
The $1.22$ is a geometrical factor applicable to circular apertures, as outlined in Wikipedia.


What are the smallest star clusters affected by Galaxy Rotation Curve? We all know that Milky Way exhibits Galaxy Rotation Curve which makes it rotate very differently that Newtonian mechanics would expect. But Milky Way is a pretty large galaxy, and the one we only see inside, so maybe it's not that good of an research object.

What are the smallest galaxies or star clusters whose rotation is clearly non-Newtonian? Are dwarf galaxies' rotation curves affected? Are globular clusters inside galaxies such as ours affected, or is their movement decisively Newtonian?

Yes, dwarf galaxy rotation curves are affected -- in fact, they tend to require relatively more dark matter than is required to explain the rotation curves of giant galaxies like the Milky Way. The smallest known systems are so-called ultrafaint dwarf galaxies (UFDs), which can have as little as $\sim 10^{4}$ solar masses worth of baryons (in the form of stars), but apparently have as much as 100 times this in the form of dark matter -- versus the roughly 10:1 (DM:baryons) case for galaxies like the Milky Way. This means that some of these systems have fewer stars than typical globular clusters.

But globular clusters, on the other hand, do not show signs of dark matter/non-Newtonian motion: you can explain the motions of their stars as being due to the normal, Newtonian gravity of all the baryonic mass (i.e., the stars, plus a plausible amount of stellar remnants like neutron stars and stellar-mass black holes).


What is the distribution of organic compounds in the Milky Way Galaxy? I imagine the answer is unknown for the entire galaxy, but I would appreciate a notion of the scales and locations of regions where astronomers have ventured at least a guess to the distribution of organic matter in interstellar dust.

Updated

I am specifically looking for plots showing the distribution of some selected organic compounds (doesn't really matter which ones right now) within a given section of the sky.  Or, if possible, within a given arc and range with respect to the galactic center (again, if possible).

The answer depends on which you are talking about (you mentioned both organic compounds and organic matter, which are two completely different things).

Organic compounds we do not have a clear definition of, but is most agreed upon to contain carbon atoms, either on their own (C), bonded with at least one hydrogen atom (C-H), or bonded with at least one other carbon atom (C-C). Compounds such as Methane (C-H4), Ethanol (C2-H6-O), and  Glucose (C6-H12-O6) are textbook examples of organic compounds. Methane has been discovered on several objects within our solar system, including Mars, Venus, and Saturn. Methane has also been discovered on the exoplanet HD 189733b (though it is the only example we have of methane on an exoplanet), and often exists in interstellar clouds, showing that it is quite common throughout our galaxy.

Organic matter refers to matter composed of organic compounds, as part of an organism. While we are searching for organic matter as proof of extraterrestrial life, we have yet to discover any organic matter or clear evidence for such matter outside of our own planet.