text:
How dark/bright are polar nights?
During the polar night directly at the pole the Sun sets once and—after a period of twilight—stays down for months at a time.
The only sources of non-human-made light are the stars, the auroras, and the sunlight reflected by the moon.
Yet when I see various photos of, for example, Amundsen-Scott base during this period, the sky (and the ground illuminated by it) is often quite bright. Here is a timelapse of August, which should be way outside even the civil twilight for this period: https://youtu.be/cf14SL7m8Ys
Is there an issue with photo-sensitivity of the cameras?
Are these sources of illumination actually bright enough to turn the middle of the night into something more twilight-like (if not day-like) when perceived by a human eye? If so, how much of the polar night is actually dark, percentage-wise?

That's a timelaps video, most likely created from a sequence of photos. It's easy to choose sensitivity and ISO settings on normal cameras such that the night sky at the darkest night is over-exposed. The moon light looks like sunlight - as it simply is reflected sunlight.
E.g. see https://cloud.planetmaker.de/index.php/s/yJ7AXWG4dEC6csW - it's a an exposure of several seconds taken several hours after sunset in November in Kiruna, Sweden... the over-exposed part in the center are headlights of a car, the rest is illuminated by the nearly-full moon at that time.
As children of modern civilisation living in urban areas, we are simply not accustomed anymore to how bright even nights can be when you had sufficient time to adopt - and cameras can do that instantly. In arctic (or generally snowy) regions, the snow adds a lot to the overall brightness as it reflects the little available light at night extremely well.

When I go to bed at night my bedroom is just about totally black, with just a few linty glowing lights from electronic devices.  When I wake up in the middle of the night I find that my eyes have adapted and the room looks like various shades of gray instead of black to my now dark adapted eyes and I can see where I am going quite well.  And apparently a camera can be instantly adjusted to make the same transformation in how brightly lite the scene looks.

tl;dr: You are seeing Astronomical twilight growing brighter through the month of August 2017 at the south pole!
For more on the Moon's motion see the excellent answer to How does the Moon move in the "night" sky as seen from the poles?

Here are two screenshots and a calculation of the altitude and azimuth of the Moon and the Sun throughout the month of August 2017 from the Amundsen-Scott South Pole Station.
The solid blueish line is for the Moon, the dashed orangish line is for the Sun. You can see that while the Moon dips below the horizon for the middle third of the month, the Sun is always below the horizon. However  twilight is approaching and the Sun is only about 10 degrees below the horizon near the end of the month.
Astronomical dawn begins when the Sun at 18 degrees below the horizon and the Sun is above -18 degrees for the entire month!
Also note that while the Sun and Moon spend much of the month at very different azimuth positions, opposite each other around the 8th and nearly the same azimuth around the 22nd.
I've also taken two screen shots and calculated the positions of the Moon and Sun at their timestamps. It looks like the camera is pointed at an azimuth of about 250 degrees, the Moon is clearly above the horizon at the beginning of the month and a nice twilight glow from 14.4 degrees below the horizon reveals the Sun's position.

                 left                              right
            ------------------------        ------------------------
object           The Moon                           The Sun
UTC ISO     2017-08-02T02:34:56Z            2017-08-13T18:50:03Z
alt(elev)   <Angle 15deg 34' 47.4">         <Angle -14deg 25' 57.5">
azimuth     <Angle 256deg 39' 49.6">        <Angle 258deg 40' 32.9">
distance    <Distance 0.00269439 au>        <Distance 1.01304 au>

Here is the Python script used for this information, which is based on the excellent Skyfield package.
from skyfield.api import Loader, Topos
import numpy as np
import matplotlib.pyplot as plt
from skyfield.api import load

halfpi, pi, twopi = [f*np.pi for f in (0.5, 1, 2)]
to_degs, to_rads = 180/pi, pi/180

load = Loader('~/Documents/fishing/SkyData')  # avoids multiple copies of large files
ts = load.timescale() # include builtin=True if you want to use older files (you may miss some leap-seconds)
eph = load('de421.bsp')

earth, sun, moon = [eph[x] for x in ('earth', 'sun', 'moon')]
AS = earth + Topos('90.0 S', '0.0 E', elevation_m = 2835)

hours = np.arange(24*31)
days = hours/24 + 1
times = ts.utc(2017, 8, 1, hours)

malt, maz = [thing.radians for thing in AS.at(times).observe(moon).apparent().altaz()[:2]]
salt, saz = [thing.radians for thing in AS.at(times).observe(sun).apparent().altaz()[:2]]

plt.figure()
plt.subplot(2, 1, 1)
plt.plot(days, to_degs * malt, '-')
plt.plot(days, to_degs * salt, '--')
plt.plot(days, np.full_like(days, fill_value=-18.),
         '-k', linewidth=0.5) # twilight
plt.xlim(1, 31.98)
plt.ylabel('altitude (deg)')
plt.subplot(2, 1, 2)
plt.plot(days, to_degs * maz, '-')
plt.plot(days, to_degs * saz, '--')
plt.xlim(1, 31.98)
plt.ylabel('azimuth (deg)')
plt.xlabel('days')
plt.suptitle('Moon from Amundsen-Scott August 2017')
plt.show()

t1, t2 = ts.utc(2017, 8, 2, 2, 34, 56), ts.utc(2017, 8, 13, 18, 50, 3)

print(t1.utc_iso(), AS.at(t1).observe(moon).apparent().altaz())
print(t2.utc_iso(), AS.at(t2).observe(sun).apparent().altaz())


text:
What was the size of the universe around 400 million years after the Big Bang?
I was trying to find out how large the universe was around 400 million years after the Big Bang as this is when some of the first galaxies were formed but I am having trouble finding out the answer or working it out.
I found some information on the size of the universe after 380,000 years when the CMB was created and by calculating the red shift from the time of creation a size of roughly 80 million light years is given by reducing the observable universes size by a factor of 1100 but I am not sure how to use this to calculate the universes size after 400 million years and I am not sure if as the above calculation is for the observable universe how that related to the actual size.
Is the size of the universe in terms of its length or diameter known or can be calculated after 400 million years from the Big Bang event?

The issue:  By "universe" do you mean "Universe", or "the observable universe".  As far as we can tell, the universe may be infinitely large and always has been. Or it might not be.  Observations limited by the speed of light can't tell.

@JamesK When you say the Universe may have always been infinite is that based on the Big Bang occurring in an existing space? I do mean the whole universe not just the observable but i may be misinterpreting theories and getting confused but i mean the distance that all the matter and energy that came out of what we call the Big Bang had spread to.

No, ...... The whole universe was hot and dense and infinite.  Now it is cold, rare and infinite.  The big bang didn't occur in the universe.  The big bang is the universe

@JamesK Where does the singularity and inflation come in at the early stages if it was already infinite?

There are two issues: what counts as "the size of the universe", and how to rescale this. The second issue is easier.
If you have something at distance $d_1$ at time $t_1$ it will be at distance $d_2 = d_1 a(t_2)/a(t_1)$ at time $t_2$, where $a(t)$ is the scale factor at the relevant time. To further make things nice, $a(t)=1/(1+z)$ where $z$ is the redshift at the time.
For $\approx 400$ myr and a flat universe, $z=11.35$ and $a=0.081$ (from this calculator).
Note the word "flat" above: flat and open universes are by default unbounded. So in a literal sense they are always infinitely large. However, it is entirely possible that they have a nontrivial topology (imagine a universe that repeats like a hall of mirrors; various cases are possible), except that there is at present no evidence for this possibility - if there is something like going on it is likely larger than our current cosmological horizon.
The observable universe on the other hand is finite, and grows both because light from more and more distant objects arrive and because of the expansion makes their distance larger. This is why the size is not just $ct$, it depends on the expansion. The calculator gives 367.92 Mpc as the distance to the horizon at $z=11.35$, or 1.2 giga-lightyear.

Thanks, So if I am understanding this correctly all the mass of the observed universe would crammed into a space of 1.2 billion light years?

Not quite: the 1.2 billion light years is the observable universe back then. However, you can divide the radius of our observable universe by 11.35 to get the radius back then of the region that would expand to become the current observable universe; since matter is not moving fast most matter in these two spheres is the same.

Ok thanks so that puts 90 billion light years diameter of superclusters and filaments into around 8 billion light years of space.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

sort: votes


text:
Are the Venusian "continents" likely to have existed before the global resurfacing event?
A topographic map of the surface of Venus shows the large highland areas Ishtar Terra, Aphrodite Terra and Lada Terra that have occasionally been described as "continents", plus various other smaller regions such as Beta Regio. Are these "continents" likely to have existed before the global resurfacing event, or is the topography of the planet likely to have been completely altered?

Ishtar Terra belongs to the tessera type of terrain, "one of the most tectonically deformed types of terrain on Venus", representing ~7.3 % of Venus' surface (Ivanov & Head 2011, Global geological map of Venus). Tessera are "dated" (relatively, from stratigraphic relationships) from the Fortunian, the oldest period:

The most important features of the Fortunian are large tessera-bearing
crustal plateaus (e.g., Fortuna, Ovda, etc.) that form a specific
class of first-order (a few thousands of kilometers across) highs on
Venus [...] The lowest stratigraphic position of tessera and its
relative abundance show that this unit is of extreme importance as a
probable ‘window’ into the geological past of Venus.

Stratigraphy of defined and mapped geologic units on Venus. The
visible geologic history of Venus consists of the Fortunian,
Guineverian, and Atlian Periods. The major portion of the surface of
Venus (about 70%) was resurfaced during the Guineverian Period. See
text for discussion. T is the mean model absolute age of the surface

Further works by Ivanov & Head detail the history of volcanism (2013) and tectonism (2015) on Venus. They consider tessera themselves as a resurfacing process:

The globally observed stratigraphic relationships among the tectonic and volcanic units that make up the absolute majority (~96%) of the surface of Venus divide the observable portion of its geologic history into three different episodes, each with a specific style of resurfacing. These are as follows: (1) Global tectonic regime, when tectonic resurfacing dominated. Exposed occurrences of these units comprise about 20% of the surface of Venus. (2) Global volcanic regime, when volcanism was the most important process of resurfacing and resurfaced about 60% of Venus. (3) Network rifting-volcanism regime, when both tectonic and volcanic activities were about equally important. During this regime, about 14% of the surface of Venus was modified.

So, to summarize, "continent-like" (tessera) terrains like Ishtar Terra have existed for a long (yet unknown) time, and have survived the resurfacing from volcanism that covers most of the surface today. But they are themselves considered to represent a (tectonic) resurfacing process that has affected older (and unknown) pre-Fortunian terrains.

A global correlation chart that shows the three major regimes of
resurfacing on Venus: global tectonic regime, global volcanic regime,
and Network-rifting-volcanism regime.

I had no idea that Venusian stratigraphy was so detailed, thanks for all the references!

Wiki describes the global resurfacing event as follows :-

NASA's Magellan spacecraft mission discovered that Venus has a geologically young surface with a relatively uniform age of 500±200 Ma (million years). The age of Venus was revealed by the observation of over 900 impact craters on the surface of the planet. These impact craters are nearly uniformly distributed over the surface of Venus and less than 10% have been modified by plains of volcanism or deformation. These observations indicate that a catastrophic resurfacing event took place on Venus around 500 Ma, and was followed by a dramatic decline in resurfacing rate.

In the same paragraph :-

Based on its surface features, it appears that Venus is tectonically and convectively alive but has a lithosphere that is static.

(Some emphasis on 'lithosphere is static')
The above paragraph makes it less likely that the Venusian 'Continents' could have existed before the resurfacing event, but we cannot be sure that the same rules applied before the event.
I also found this website that says that doesn't really say anything about 'continent'-like things before the resurfacing event, but it says that the event might have :

"turned off" any plate tectonics on the planet, completely solidifying the crust into a single surface.

This means that any further resurfacing events would be unlikely.
Couldn't find any papers on the same though.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
I saw a weird thing around Venus
I pointed my telescope
(50mm aperture, 500mm focal length, 4mm eyepiece, image taken through a window)
at Venus and saw this:

What is it? I don't think this is Uranus or Neptune as my telescope is very bad. I haven't increased the image size too much and it was 6 A.M. (UTC+3, EEST)

Venus is a very bright "morning star" at the moment, but what your image shows is only a blur. The telescope is very far from focus and you can see no details of the the planet. You may also be getting some reflections from the glass of the window
If Venus was in focus you would see a very small "half moon" (or perhaps "waxing gibbous" shape) at the moment. Try to practice focusing on distant landmarks during the day. That makes it easier to get the focus right at night.

Oh, ... and here's me hoping for an imaginative answer given the latest discovery of Phosphine of Venus & the opening of a discussion about the possibility of life on Venus. Something like a Venetian equivalent of either: The War of the Worlds, or Mars Attacks, or even The Day of the Triffids. ...:-(

Along this line: why is it that photos of Bigfoot or UFOs are always really poor photos?

how does the shape of a "waxing gibbous" differ from that of a "waning gibbous"?

in the first derivative(!)

:-) Oh, venus has dates! https://astronomy.stackexchange.com/q/23271/7982


text:
What would you call death by magnetar?
The magnetic field of a magnetar is so strong that it would stretch people's atoms apart. This, would obviously, result in death. But how spaghettification refers to death by extreme tides, and for the most part, refers to death by black hole, what field-specific term would best refer to death by extreme magnetic fields, eg. from a magnetar?

Isn't this better suited for meta stackexchange? I believe there is some astronomy jokes section there and people can try and come up with creative answers such as teslascind there...

"Magnetaaarrgh!"

ESE can be a tough room sometimes. I've had some degree of success but you have to be fairly careful how the question is asked, demonstrate prior research, and stick to real or at least nearly-real words. 1, 2, 3, 4, 5, 6.

I think this question would be better if you added one more sentence asking for a potentially scientific word for what happens to inanimate matter, like a dust particle. You could also ask that as a separate question and that might be better... or not.

Mag-neato.  X-men joke.  Which I wouldn't have explained if there wasn't a minimum character requirement.

Nobody has ever needed such a word, and it doesn't exist.
"Spaghetification" is, of course, a joke word, but each time you make a new joke based off an old one, the less funny it becomes.
If you just want a (poor) joke:  Teslascind

In order to capture @pela's ephemeral comment in a more durable post:

Magnetaaarrgh!

I like it

I didn't know that Spaghetification was a joke though, I though it was the original term

It is a science term: https://en.m.wikipedia.org/wiki/Spaghettification

@WDUK Maybe it was a joke that caught on so well it ended up a scientific term


text:
Cannot identify mistake in calculating orbital eccentricity vector; magnitude equals one instead of zero (with python code)
I have a gravitational nbody simulation, for which I would like to determine various orbital parameters. For each body, I have 3-D vectors (x,y,z -space) for position, velocity, and acceleration. I am trying to follow the steps laid out in this post to obtain the eccentricity of each orbit. Before throwing n bodies into the simulation, I am testing the algorithm on simpler systems, such as a 2-body system in which the orbital path of the Earth around the Sun is nearly a perfect circle. Because the orbit is circular, I expect the eccentricity to be zero; this is not the output I get, so I am hoping someone can help me identify my errors (either in understanding or in code). Specifically, I would like to know what I am doing wrong in trying to calculate the eccentricity.
Sorry ahead of time for the length of this post; most of the code below is to show that the methodology works to obtain vectors of position and velocity; the last part of code (skip down to PROBLEM) is to "show my work" in using these parameters to calculate the eccentricity vectors. Aside from visual inspection, methods from this post were used to ensure that the orbit is circular.
Create circular orbit via Sun-Earth system
First, we'll initialize the initial conditions of our coupled ODEs and relevant simulation parameters.
import numpy as np
import matplotlib.pyplot as plt

## simulation parameters
ndim = 3 ## x,y,z
gravitational_constant = 6.67e-11 ## SI units
nbodies = 2 ## sun, earth
duration = 365*24*60*60 ## duration; 1 years --> seconds; day/yr * hr/day * min/hr * sec/min
dt = 2 * 24 * 60 * 60 ## time-step; 2 days --> seconds
t = np.arange(duration/dt)

meters_to_au = 1.496e11 ## 1.496e11 meters = 1 AU

## BODY 1 (sun)
m_sun = 1.989e30 ## kilograms
x_sun = np.zeros(ndim) ## position (x,y,z); meters
v_sun = np.zeros(ndim) ## velocity (x,y,z); m/s

## BODY 2 (earth)
m_earth = 5.972e24 ## kilograms
x_earth = np.array([meters_to_au, 0, 0]) ##
_v = np.sqrt(gravitational_constant * m_sun / meters_to_au)
v_earth = np.array([0, _v, 0])

## standard gravitational parameters and reduced mass
mu = np.array([m_sun, m_earth]) * gravitational_constant
mred = (m_sun * m_earth) / (m_sun + m_earth)

Then, we solve the coupled ODEs using a simple Euler method.
## initialize SOLUTION SPACE
X = np.zeros((nbodies, ndim, t.size))
V = np.zeros((nbodies, ndim, t.size))
xi = np.array([x_sun, x_earth])
X[:, :, 0] = xi ## position of bodies at time t=0
vi = np.array([v_sun, v_earth])
V[:, :, 0] = vi ## velocity of bodies at time t=0

## ITERATE (i --> k=i+1)
for ti in range(1, t.size): ## t=1, ..., t=end
    ak = []
    for j in range(nbodies):
        dacc = 0
        for k in range(nbodies):
            if j != k:
                dpos = xi[j, :] - xi[k, :]
                r = np.sum(np.square(dpos))
                dacc -= mu[k] * dpos / np.sqrt(r**3)
        ak.append(dacc)
    ak = np.array(ak)
    vk = vi + ak * dt
    xk = xi + vk * dt
    X[:, :, ti] = xk
    V[:, :, ti] = vk
    xi, vi = xk, vk

## GET POSITION VECTORS PER BODY
Xs = X[0, :, :]
Xe = X[1, :, :]

## GET VELOCITY VECTORS PER BODY
Vs = V[0, :, :]
Ve = V[1, :, :]

To verify that the simulation ran as expected, we plot.
## VERIFY -- SHOW POSITION VECTORS
fig, ax = plt.subplots(figsize=(7,7))
ax.scatter(Xe[0, :] / meters_to_au, Xe[1, :] / meters_to_au, marker='.', color='steelblue', s=2, label='Earth')
ax.scatter(Xs[0, :] / meters_to_au, Xs[1, :] / meters_to_au, marker='*', color='darkorange', s=5, label='Sun')
ax.set_aspect('equal')
ax.set_xlabel('X (AU)', fontsize=8)
ax.set_ylabel('Y (AU)', fontsize=9)
fig.legend(mode='expand', loc='lower center', ncol=2, fontsize=8)
plt.show()
plt.close(fig)

PROBLEM
I am more familiar with seeing angular momentum expressed as $L = \vec{r} x \vec{p}$, where $\vec{p} = m\vec{v}$, though I suppose one can interpret the angular momentum below expressed in units of angular momentum per unit mass. In Cartesian coordinates, $\vec{r} = \vec{x} + \vec{y} + \vec{z} = x\hat{x} + y\hat{y} + z\hat{z}$.
## GET ANGULAR MOMENTUM VECTORS PER BODY
Le = np.cross(Xe, Ve, axis=0)
Ls = np.cross(Xs, Vs, axis=0)

## GET ORBITAL ECCENTRICITY PER BODY
Ee = np.cross(Ve, Le, axis=0) / mred - Xe / np.sqrt(np.sum(np.square(Xe), axis=0))
Es = np.cross(Vs, Ls, axis=0) / mred - Xs / np.sqrt(np.sum(np.square(Xs), axis=0))
mag_Ee = np.sqrt(np.sum(np.square(Ee), axis=0))
mag_Es = np.sqrt(np.sum(np.square(Es), axis=0))

## VERIFY -- SHOW ORBITAL ECCENTRICITY VECTORS PER BODY
fig, ax = plt.subplots(figsize=(7,7))
ax.scatter(Ee[0, :], Ee[1, :], marker='.', color='steelblue', s=2, label='Earth')
ax.scatter(Es[0, :], Es[1, :], marker='*', color='darkorange', s=5, label='Sun')
ax.set_aspect('equal') ## x- and y- scales are equal; nearly perfect circle
ax.set_xlabel(r'eccentricity $\hat{x}$', fontsize=8)
ax.set_ylabel(r'eccentricity $\hat{y}$', fontsize=8)
fig.legend(mode='expand', loc='lower center', ncol=2, fontsize=8)
plt.show()
plt.close(fig)

## VERIFY -- SHOW ORBITAL ECCENTRICITY MAGNITUDES PER BODY
rescaled_t = t * dt
fig, ax = plt.subplots(figsize=(7,7))
ax.scatter(rescaled_t, mag_Ee, marker='.', color='steelblue', s=2, label='Earth', alpha=0.5)
ax.scatter(rescaled_t, mag_Es, marker='*', color='darkorange', s=5, label='Sun', alpha=0.5)
ax.set_xlabel('Time', fontsize=8)
ax.set_ylabel('Eccentricity', fontsize=8)
ax.set_ylim(bottom=-0.1, top=1.2)
fig.legend(mode='expand', loc='lower center', ncol=2, fontsize=8)
plt.show()
plt.close(fig)

It is my understanding that eccentricity varies as $0 ≤ e < 1$ for elliptical orbits (circular orbits being $e=0$), $e=1$ for parabolic orbits, and $e>1$ for hyperbolic orbits. So something must be off. Do I need to consider the coordinates from a specific reference frame? Or maybe I missed an assumption for the equations used to hold? Can someone point to the cause of this error? Less importantly, is the equation used to calculate eccentricity generalizable to all orbits or just elliptical ones?

You are doing many things wrong.

You are computing the eccentricity of one body with respect to the center of mass. You need to compute the eccentricity of one body with respect to the other.

You are using reduced mass in np.cross(Ve, Le, axis=0) / mred - Xe / np.sqrt(np.sum(np.square(Xe), axis=0)) This is wrong for multiple reasons. First off, look at the units! The first term, np.cross(Ve, Le, axis=0) / mred, has units of length^3/time^2/mass. The second term, np.sqrt(np.sum(np.square(Xe), axis=0)), is unitless. And you should not be using reduced mass at all. You should be using the combined gravitational parameter (not the reduced gravitational parameter). A gravitational parameter has units of length^3/time^2.

To calculate the eccentricity correct, compute the position of the Earth with respect to the Sun (Xrel = Xe - Xs and the velocity of the Earth with respect to the Sun (Vrel = Ve - Vs). Then compute the cross product of these two (Lrel = np.cross(Xrel, Vrel) to yield the specific angular momentum of the Sun-Earth system. Finally, compute the eccentricity vector via np.cross(Vrel, Lrel) / mu_combined - Xrel / np.sqrt(np.sum(np.square(XRel))), where mu_combined is the sum of the gravitational parameters of the Sun and the Earth.

Finally, as a comment rather than a critique, it is best not use mass and the universal gravitational constant. It is much better to use gravitational parameters. You can find a fairly accurate list of solar system gravitational parameters in the wikipedia standard gravitational parameter article. Conceptually, a body's gravitational parameter is equal to the product of its mass and the gravitational constant. Another way of looking at it is that a body's mass is the body's gravitational parameter divided by the gravitational constant. The problem is that the gravitational constant is only known to four or five decimal places, while a body's gravitational parameter is observable and is known to six or more decimal places.

Thanks for the detailed response; very helpful! I do have one related follow-up question, regarding how to treat (1) and (3) in a system with more than two bodies. If I consider a system with a few planets (say, 3) in circular orbits about a star, then is it best to calculate the eccentricity of each planetary orbit wrt only dominant central mass (ie, not account for perturbations from other bodies)? Also, if we consider an exotic orbit (perhaps a stable horseshoe or an unstable figure-eight), then does one consider orbits piece-wise (wrt to dominant central mass) or in some other way?

@allthemikeysaretaken - You essentially are asking about osculating orbital elements, or perhaps some time-averaged version of them. When one does this, one finds that the five Keplerian elements that are supposed to be constant are not constant. Mercury's orbit precesses by about 575 arc seconds per Julian century. Most of this precession, about 532 arc seconds per century, is attributable to Newtonian influences by other planets. The discrepancy, 43 arc seconds per century, is due to general relativity.

If I understood the linked post correctly, one chooses $t$ as the time elapsed since periapsis (for a 2-body system); one calculates the 3 anomalies for any given time $t$. If so, then the time-dependence of these parameters makes sense (ie, perturbations). I realize this simulation won't be accurate to relativistic order. (I read somewhere on stackexchange about a relativistic correction to the gravitational potential that I wanted to play with; will search for link tomorrow). That said, I'd like to understand this better and every example I've seen is a classic ~circular orbit.

@allthemikeysaretaken - Osculating elements are relatively easy to calculate. (1) Compute the specific angular momentum vector, $\vec h = \vec r \times \vec v$. The $z$ component yields the inclination while the $x$ and $y$ components yield the right ascension of ascending node. (2) Compute the eccentricity vector, $\vec v \times \vec h / \mu - \hat r$. The magnitude yields the eccentricity and the direction points from the central mass to the periapsis point, hence yielding the argument of periapsis. (continued)

(3) Compute the semi-major axis length via the vis viva equation, $v^2 = \mu\left(\frac2r-\frac1a\right)$. (4) The true anomaly is the angle between the eccentricity vector and current location.


text:
How does Saturn's ring rain fit with the earlier "big empty" results?
In 2017, the results from Cassini's flight through the region between Saturn and the inner edge of the rings were described as the "Big Empty". Nevertheless in 2018, the region was described as having a huge amount of "ring rain" consisting of dust grains coated in ices and organics falling into the atmosphere.
How can these be reconciled? Was the earlier determination of the conditions in error, is this a consequence of variability or sampling different types of particles, or something else?


text:
Angular magnification of three lenses telescope?
What is the magnification of a three-lens telescope?
As a child I built a "telescope": objective lens $$f_1 = 20 cm$$ middle lens $$f_2=5cm$$ and eye lens also $$f_3=5cm$$
My drawing of my telescope:

I measured the magnification "with my eyes". It was about 10x-15x.
But what's the formula?
I had found a formula for a terrestrial telescope, but it doesn't really help. It is said that the middle lens is "only used to correct" the image. Then the magnification should be done like a "normal" telescope (see link):
$$
f_ {objectiveLens} / f_ {eyesLens}
$$
In my telescope, the magnification was dependent on the distance between the middle and eye lenses: larger distance -> bigger magnification.
UPDATE 1:
My idea was calculate magnification of objective like
$$
25/f_{objective}
$$
Then calculate magnification for system of middle and eye lenses like for microscope (here: link). And then multiply both magnifications.
Here is my calculations (see drawing):
Magnification of objective:
$$
m_{objective} = 25cm/f_{1}=25cm/20cm=1.25
$$
In my case distance beetwen middle and eye lenses was 20 cm, then L:
$$
L = d_{2}-f_{1}-f_{2}=20-5-5=10
$$
$$
m_{middle+eye}=(-L/f_{1})*(25/f_{2})=-10/5*25/5=-2*5=-10
$$
Then total magnification is:
$$
M_{total}=m_{objective}*m_{middle+eye}=1.25*10=12.5
$$
But i feel i am very wrong.
UPDATE 2:
I have learn a little bit about lenses and i have use lens equation for calculating angle magnification. But i am not sure that's right. Can someone say something about?

First telescopes are always fond memories :-) Mine was a magnifying mirror from the bathroom, an eyepiece from my grandmother's bird-watching binoculars and a big old cardboard tube that was used for newsprint. Here's one way to do this; multi-element optical systems can be quickly analyzed to first order using simple 2x2 matrices for thin lenses and the spaces between them using Ray transfer matrix analysis. This would have six 2x2 matrices and you'd just multiply them out. It's a little tedious by hand, but it can be done.

this matrix analysis can support 2nd and higher order, the matrices are simply enlarged beyond 2x2 for the higher order terms e.g. $x^2, x \theta, \theta^2$ etc.


text:
What is a good focal length for DSO
I found two telescopes that i'd like to buy, but one has 900mm focal length ((Skywatcher Newton 114/900 EQ1)) and the other has 500mm focal length ((National Geographic Dobson N 114/500)).
My questions are:

What is the "recommended" focal length for observing deep sky?
Are there better telescopes in the same price range?

I don't think there is a "recommended" FL for DSO observation. If you only observe visually, the eyepieces you use are just as important und the magnification depends on them. If you are concerned about the field-of-view you will get from these telescopes, I recommend checking out FOV calculator tools such as astronomy.tools. There you can enter your telescope and eyepiece information and show the FOV compared to a deep sky object of your choice.
I personally have been very happy with Skywatcher telescopes (I own an Esprit 100ED Triplet and a 200/1000 Newtonian), however I never really tested them visually as I only use them for imaging.

There isn't really a "best" focal length because it depends on the object you might want to observe.
Angular Size
The Ring Nebula (Messier 57) is only about 1.4' across.  Meanwhile there are larger objects such as the Andromeda Galaxy (Messier 31) -- that's about 3° across (and a pair of binoculars may work better than a telescope because too much focal length means the whole object wont fit into the field of view.
Consider that a telescope that an instrument that can take in the entire Andromeda Galaxy wold render the Ring Nebula as a very tiny dot ... while an instrument that can nicely resolve the Ring Nebula would only be able to view a tiny piece of the Andromeda Galaxy.
Aperture
In the "What telescope should I get?" category, consider that the larger the aperture, the better the ability to resolve detail.  An 8" Dobsonian mounted telescope would resolve more detail than a 4" refractor.  A 16" Schmidt Cassegrain Telescope (aka SCT or "Cat" because an SCT is a "Catadioptric" telescope so they get the nickname "Cat") would resolve significantly more detail... but be so heavy that it really needs a permanent pier in an observatory (the largest telescope I use as a "portable" telescope is a 14" Cat ... but absolutely requires two people to set it up because it is too heavy and unbalanced for just one person to lift it onto the tripod.)  The point here is that there is such a thing as "too big" in that you have to consider if you can transport it or set it up without requiring assistance.
Dobsonian reflector telescopes are very easy to setup and use and provide surprisingly good views.  They often have short-ish focal ratios (e.g. f/5-f/6 or something near that range).  An 8" (200mm) aperture model would have a focal length of around 1000-1200mm.  That tends to be pretty good as an all-around instrument... it's not particularly short but also not particularly long for a telescope.
You can find them made by brands such as (and in no particular order) Orion, Zhumell, Apertura, Sky-Watcher, Explore-Scientific, Celestron, or Meade.  There are other brands that I left off the list such as Obsession -- they make great telescopes but their smallest instrument is a 12.5" Dob with sizes up to 25".
These telescopes typically do not include electronics or motors ... but there are variations that do have those options.  They are typically not well-suited for imaging (but you did not ask about imaging).
Mount
If the telescope is perched atop a tripod, place strong consideration on the stability of the mount.  My first telescope came with a tripod that had aluminum extruded legs.  Merely touching the focus knob resulted in the telescope vibrating enough that viewing through the telescope and trying to focus was a real test of patience because the object would shake so violently that it was difficult to determine if the focus was improving or getting worse.
This is one of the reasons why, when asked for a suggestion on a first telescope, I typically suggest a Dobsonian mounted reflector ... because the mount doesn't have legs and tends to be very stable.  Focusing is usually not a problem.
This is not to suggest that tripod-mounted telescopes are inferior... it's just to suggest that if a tripod-mount will be used, you want a very sturdy tripod.

When this type of question is asked it seems to be almost universally accepted that the poster is looking at equipment that is at the maximum price range they can afford.
The bottom line is you will get what you pay for and that applies to viewing Deep Sky Objects in a very big way.
No traditional eyepiece will give you a view with the color and detail of an electronic eyepiece.
If you think you will eventually have the resources to get into the electronic side of viewing it would be worth your while to look into it now. Better to know what direction you want to go before you start. :)
A Google with "electronic eyepiece for telescope" will yield pages of adds for the products. I use Mallincams which are not readily available on line. You can see what users are doing with those cameras at the Mallincam iO website
https://groups.io/g/MallinCam
They have a photo section with thousands of Deep Sky Object images made with the various cameras. Some are snapshots of live video, some have been processed.
You can also view live broadcasts of Electronic viewing sessions at
https://www.liveskies.org/
There are no schedules at Live Skies. People from all over the world come on line with a broadcast whenever they want to. Generally most broadcasts originate in North America. If it is dark and clear somewhere, someone might be broadcasting on line.


text:
Cross-correlations between Lagrange and Fourier synthesis
In the context of forecast for large surveys, I have to make cross-correlations between 2D (with angular coordinates of Lagrange transformation for GC photometric and Weak Lensing) and 3D (Fourier transform with radial coordinates for GC spectroscopic).
For the moment, only cross-correlation for 2D is achieved (GCph+WL+XC), or what we call 3x2 points (with XC representing the cross-correlations).
For example, a Fisher matrix element for 3x2pt is expressed as :

where $\epsilon$ ranges over G(Galaxy), L(Weak lensing) and GL(Galaxy,Weak lensing)
Could anyone tell me or give links plese about the full cross-correlations between 2D and 3D, i.e GCsp+GCph+WL+XC (4x2 points) in the litterature ?
Beyond a complicated theorical formula, I prefer to know more about the state of the art.
I mean, I would like to know what has been already done in this attempt of cross-correlation 2D+3D.
One told me there are been Bessel-Fourier works or some similar stuff but I would like to have more informations about this.
Any help is welcome

It seems to be a question that is not suitable for this site -- Any luck on https://physics.stackexchange.com/?


text:
Clarifications about distances in cosmology
I would like to get clarifications about some usual notions of distances in cosmology.

First, is the comoving distance the current distance of objects whose light has been reached by us now, i.e. with regards to cosmic time? Or, is it the same with redshift dependent? If yes, this comoving distance is simply the angular diameter distance, isn't it?

Is the angular diameter distance, $D_A$, the distance from an observer to an object which is starting to emit, such that the light is reaching us now?

Is the comoving distance transverse of an object is the size of object in 2D plane?
From the following link (distances in cosmology), it is written:

5) Comoving distance (transverse)
The comoving distance between two events at the same redshift or distance but separated on the sky by some angle $\delta \theta$ is $D_{\mathrm{M}} \delta \theta$ and the transverse comoving distance $D_{\mathrm{M}}$ (so-denoted for a reason explained below) is simply related to the line-of-sight comoving distance $D_{\mathrm{C}}$:
$$
D_{\mathrm{M}}=\left\{\begin{array}{ll}
D_{\mathrm{H}} \frac{1}{\sqrt{\Omega_{k}}} \sinh \left[\sqrt{\Omega_{k}} D_{\mathrm{C}} / D_{\mathrm{H}}\right] & \text { for } \Omega_{k}>0 \\
D_{\mathrm{C}} & \text { for } \Omega_{k}=0 \\
D_{\mathrm{H}} \frac{1}{\sqrt{\left|\Omega_{k}\right|}} \sin \left[\sqrt{\left|\Omega_{k}\right|} D_{\mathrm{C}} / D_{\mathrm{H}}\right] & \text { for } \Omega_{k}<0
\end{array}\right.
$$
with $D_H=\dfrac{c}{H_0}$

6) Angular diameter distance
The angular diameter distance $D_{\mathrm{A}}$ is defined as the ratio of an object's physical transverse size to its angular size (in radians). It is used to convert angular separations in telescope images into proper separations at the source. It is famous for not increasing indefinitely as $z \rightarrow \infty ;$ it turns over at $z \sim 1$ and thereafter more distant objects actually appear larger in angular size. Angular diameter distance is related to the transverse comoving distance by
$$
D_{\mathrm{A}}=\frac{D_{\mathrm{M}}}{1+z}
$$

Question 1) In the part 1) they define the comoving distance by:
$$D_{\mathrm{C}}=D_{\mathrm{H}} \int_{0}^{z} \frac{d z^{\prime}}{E\left(z^{\prime}\right)}$$
But this is not the physical distance. Indeed, we need to multiply $D_{\mathrm{C}}$ by $R(t=0)=R_0$ to get the physical distance between us and the object, don't we?
Question 2) I don't understand in the text above why they say that "but separated on the sky by some angle $\delta\theta$ is $D_M\delta\theta$ and the transverse comoving distance $D_M$ is simply related to the line-of-sight comoving distance $D_{\mathrm{C}}$.
Should it be rather the current size of object is equal to $D_C\,\delta\theta$, i.e. the comoving distance multiplied by the $\delta\theta$ ?, like we do in trigonometry for computing a portion $\delta\theta$ of circle perimeter.
Question 3) Concerning the angular diameter distance, why could we rather take the ratio $\frac{\text{cosmological horizon}}{1+z}$ instead of the ratio $\dfrac{D_{\mathrm{M}}}{1+z}$? Indeed, angular diameter distance is the distance at z given between the observer and the object emitting.
I confuse comoving distance transverse and angular diameter distance as well as comoving distance (coordinates or physical distance with the factor $R(t)$?)

Try to look James Rich-Fundementals of Cosmology 2nd Edition  chapter 3. It will answer most of your questions. I want to give an answer but it will be too long to write.If you still feel confused after looking the book or you want a full answer let me know. Or maybe someone else from here can help.

Here $D_M \equiv S_k(\chi) \equiv r$ where $$S_k(\chi) =
\begin{cases}
sinh(\chi) & k= -1  \
\chi & k = 0 \
sin(\chi) & k = +1
\end{cases}$$

@Layla Thanks. Do you agree that it could bring confusions by calling $D_M$ a "transverse" comoving transverse, don't you ?. The transverse size is rather  $D_M,\delta\theta$ as I said in the trigonometry case,portion $l$ of perimeter is equal to $l=R,\delta,\theta$ with $R$ the radius of the circle : could you validate this point ? Regards

Its confusing yes..I really dont like the Hogg's article about this topic. Especially if you are new to these concepts.

Isn't really anyone who can clarify my questions ? Best regards


text:
Transverse Comoving distance
I don't understand the link between the comoving distance and transverse comoving distance : how can they be equal ?
Here an example of definition that I have found :
Angular Diameter Distance :
The angular diameter distance $D_{A}$ is defined as the ratio of an object's physical transverse size to its angular size (in radians). It is used to convert angular separations in telescope images into proper separations at the source. It is famous for not increasing indefinitely as $z$ -> infinity; it turns over at $z \sim 1$ and thereafter more distant objects actually appear larger in angular size. Angular diameter distance is related to the transverse comoving distance by
$D_{A}=\dfrac{D_{M}}{1+z}(17)$
Some clarifications would be fine.
EDIT 1: @benrg If I have well understood, the transverse comoving distance is simply equal to the comoving distance between object that has emitted at redshift "z" and us that we reveive now this light, isn't it ? If yes, why complicate the things by using the qualified word "transverse" into "transverse comoving distance" ?

Angular diameter distance is the reduced circumference of the circle, centered at our location, on which the object was located when it emitted the light (or the reduced area of the sphere if you prefer). If the universe is spatially flat then this is the same as the metric radius of the circle. In general it's related to the radius by the function $S_k$ defined here, where $k$ is the Gaussian curvature at that time.
Transverse comoving distance is the same thing scaled up by $1{+}z$, i.e., the reduced circumference of the circle on which the same object is located at the present cosmological time, if it had no net peculiar motion in the mean time. It's therefore equal to (radial) comoving distance if the universe is spatially flat.

Thanks, could you take a look please at my EDIT 1 ? Regards


text:
Would the electron cyclotron-maser emission mechanism affect Proxima b's ability to retain an atmosphere?
In a recent arXiv preprint, Pérez-Torres et al. "Monitoring the radio emission of Proxima Centauri" claim the detection of radio emissions synchronised with the orbit of the planet Proxima b. They explain this as the result of electron cyclotron-maser (ECM) emissions, making the Proxima–Proxima b system a scaled-up analogue of the Jupiter–Io or Jupiter–Ganymede systems.
Would this mechanism make it more difficult for Proxima b to retain an atmosphere?

This mechanism, being actively emitting in the radio-wavelengths, is certainly negligible for the overall atmospheric energetics at Proxima b. One can conclude this by taking the band luminosities from the cited paper ($\rm 2.51\times10^{20} erg/s$, p.3, first paragraph) and compare them to the solar constant at the planets orbit, which should amount to $\rm9\times10^{23}erg/s$.
The latter is however only the thermal energy contribution given by the stellar effective temperature, there might be other significant chunks of energy contributed in the UV and X-bands.
The luminosities thus considered would be important for driving both bulk escape and jeans escape. Being negligible in the radio-range, shows that this process is unimportant for the retention of a bulk atmosphere via thermal mechanisms.
In terms of non-thermal mechanisms it's possible to imagine that there would be some resonant absorption of the ECM emission at the planetary radius, driving scattering of heavier species. Considering however the scalings of the cyclotron frequency $f=qB/{2\pi m}$, and the fact that the magnetic field is weaker at the planet compared to the emission region at the star, the scattered particles would have to be lighter than electrons. Those particles do not exist in atmospheres in significant numbers, only in particle accelerators.
Hence we can conclude that non-thermal processes driven by ECM heating will also not play a role for atmospheric escape.


text:
Identifying cluster of exoplanets between Deneb and Vega
I noticed an interesting patern of exoplanets.
Is this a known phenomenon?

What do you mean by "pattern?"

That's the field the Kepler telescope was observing during its primary mission (before the second reaction wheel failure). The shape is due to the detector array.
I.e. this is not a real pattern in the actual distribution of exoplanets, it's a consequence of how the observations were done.


text:
How does one go about naming a lunar crater?
Suppose I discover an uncatalogued crater on the lunar surface, and decide to name it. How would I submit my proposed name to the IAU?


text:
Vmag versus VpgMag in GSC 2.3 catalogue
I tried to find out the magnitude in V for V* V881 Per.
Here are my results:
SIMBAD: $11.09$    (Fluxes->V)
GSC 2.3: $10.9815$  (VpgMag) and  $10.887$   (Vmag)
Which of VpgMag or Vmag in GSC 2.3 correpsonds to SIMBAD V-flux?
For each star GSC 2.3 cites two values for magnitude in V so which of the two should I use?

V881 Per is a variable star, so its brightness changes over time. According to AAVSO.ORG it varies between 10.8 and 11.2 (approximately). Which brightness would you like to use?

Your "GSC 2.3" entry is actually from GSC 2.4.1, as shown in the header lines containing the string "GSC241".
According to the record structure for GSC 2.4.1, Vmag is "V-band magnitude" while VpgMag is "Photographic V-band magnitude". The magnitudes are associated with a code that describes what bandpass/filter they are using. In your case, the VpgMagCode field is 1, which corresponds to the emulsion+filter IIaD+W12 (au used by the Palomar Quick-V survey). Confusingly the VMagCode field is set to 3, which corresponds to a Johnson B-magnitude. Perhaps this is an error in the catalogue and should be 4 (Johnson V-magnitude)? The source code of 19 indicates the magnitude is from APASS (the AAVSO Photometric All-Sky Survey). The APASS website provides some further information on the magnitudes used:

Johnson B and V were chosen to extend the Tycho calibration to fainter magnitudes and to match the many archival published datasets. These passbands can be easily used to create sequences for any VSX / GCVS star. Sloan passbands provide a direct link to the SDSS, SkyMapper, PanSTARRs etc. surveys, and provide a homogeneous catalog of brighter stars.  The Johnson B and V magnitudes are on the Vega system; the Sloan u',g',r',i',z_s and Z magnitudes are on the AB system.

A few more details on the Johnson magnitude system can be found on this web page: the system is set up so that the star Vega has a V-magnitude of 0.03, and the colors (e.g. B-V) are 0. Details of the Johnson-Cousins UBVRI filters can be found on the website of the Leibniz-Institut für Astrophysik Potsdam.
The details of the magnitude in SIMBAD can be obtained by going to the reference listed there: Norton et al. (2007). This paper indicates that the magnitude is from the SuperWASP photometric survey. The details of the filter the SuperWASP survey uses can be found on the SVO Filter Profile Service. Norton et al. (2007) describe the value in section 3 of the paper:

the mean SuperWASP magnitude,
defined as $-2.5 \log_{10} (F/10^6)$ where $F$ is the mean SuperWASP
flux in microVegas; it is a pseudo-V magnitude which is comparable to the Tycho V magnitude

A microVega is a millionth of the flux from Vega. Note that the above formula implies a V-magnitude for Vega of 0, in contrast to the Johnson system.
As for which one you should use, it depends on what you are doing. If you're comparing with other SuperWASP results (or perhaps objects in the Tycho survey), use the SuperWASP V-magnitude. If you're comparing with other magnitudes in the Johnson system, go with Vmag from GSC 2.4.1.

Vmag is the visual magnitude in the V wavelength, viz. 551 nm, whereas Vpgmag is the photographic magnitude in the same wavelength. Photographic emulsions—and now CCD and CMOS detectors—have a different sensitivity curve than the human eye, hence the difference.

Thanks a lot! Would you suggest any particular (introductory) bibliography on this?

https://en.wikipedia.org/wiki/Photographic_magnitude is very introductory…


text:
Starlight: are the rays that we see here on Earth parallel?
Are the rays of light emitted from stars that are perceived by an eye here on Earth parallel?
Do they travel in a straight line from the star to the eye, never crossing, and equidistant?
Do they wiggle around and cross as they move through space?
Do the rays meet at a point on the retina of the eye?
Assuming that a star is pretty far away, it seems reasonable to think that they've traveled zillions of miles without being parallel.

I'll mention quickly that while ray optics is an extremely useful approximation because it is simple and handy, when pushed to its limit it breaks down. It might be better to think of light from a star as a series of photons and each of those as a wave that completely fills the aperture of a telescope or the pupil of an eye. There are indeed things that can slightly distort the wavefront in space but once it reaches our atmosphere it gets distorted much more

That distortion is perceived as astronomical seeing or "twinkling" because it is constantly changing.

Whether or not rays from a distant point meet at the same point on your retina depends on how well your eyes are able to focus. I can tell you for a fact, that they do not meet at a single point on my retinas unless I view the star through some corrective optics.

Mathematically parallel? No. Do the trigonometry. Assume that spacetime is flat (but, see answers below for more on that subject) and assume an isocoles triangle whose base is the distance between your eyes, (or the distance between your eye and my eye—depends on what, exactly, you are asking) and whose height is the distance to the star. The angle between the two sides will be really, freakingly, incredibly close to zero; but it will not actually be zero.

Light rays travel along things called geodesics in spacetime (specifically they travel along a certain class of geodesics called 'null geodesics').
If spacetime is flat, which will be the case in the absence of gravity, then those geodesics are just straight lines as we normally think of them, and so yes, light rays travel along straight lines in that case.
But there is gravity and spacetime is therefore not flat.  This means there there is really no definition of 'straight line' any more: geodesics are the best thing we have (that's why they're called geodesics rather than straight lines).
But in very many cases the deviation caused by spacetime not being flat is pretty small: for instance it's possible to see the difference in apparent position of stars if the light from them passes close to the Sun, but it's a small deviation.
In some cases – if the light passes close to very massive objects, the deviation can be very large (and in fact if the object is massive enough, it can be as large as you like).
The general effect of light travelling along geodesics which are not well-approximated by straight lines is called 'gravitational lensing' and it's now a well-confirmed thing which is used in practice to detect things like exoplanets (amazingly, in this case it's the lensing of the light from stars by the planet which is observed, which is just astonishing.)
So a brief answer is yes, almost always, but occasionally not at all!

Light will travel in a straight line through the 4D Space-time continuum. It does not wiggle around and is only deflected by gravity. Light from distant stars reaches us more or less parallel due to the great distance involved.
But depending on how accurately you wish to make the measurement, there is scope for some light from that star to reach us very slightly off parallel. However the difference would be extremely small indeed and might not be measurable  so this is more of a theoretical prediction. The angle would be depend on the diameter of the star and its distance from us.
So to be clear photons of light from a distant star will all arrive arrive
more or less parallel. But some of them may be off parallel by a miniscule amount.

So you don't think that the rays are distorted as they move through space and time? I think they get bent all over the place. A one-year ray might be many miles off-target for Earth, until it gets bent by some force. Kinda like a sniper bullet gets moved around by the wind and heat rising, until it hits its target.

Light travels in a straight line through space-time, but space-time is itself distorted by gravity so light appears to bend around massive objects. So it depends what you mean by a straight line.

"It depends what you mean by a straight line" XD Love that quote!!!

Note even if there were no gravitational effects light from a distant star might still not be parallel due to the fact that light may originate from either side of the star on converge on the observer.

What do you mean by a straight line? Are you talking about a 3D or 4D straight line?

The light rays will certainly be very close to being parallel because the star is so far away and because gravitational distortions are so small in our neighborhood.
First, rays from opposite edges of the star will arrive at angles that differ by the diameter of the star divided by the distance to the star (~10^-8 radians).
Second, the effect of gravitational lensing can cause the wave function corresponding even to a single photon to be split. For example, the Einstein Cross has an apparent size of 10^-5 radians. Gravitational lensing (exactly as for conventional lenses) can even cause light-rays to cross. This is true even for the wave-function for a single photon originating on distant star or galaxy billions of light-years away.

I've been pondering this question since it was first posted in Space SE, and my feeling has been that rays initially parallel would never cross, at least parallel rays that begin 6 mm apart would never cross before reaching a somebody's pupil. But now I'm not so sure about even that.


text:
Exact ecliptic position of Earth and Moon from Mars as photographed by Curiosity Rover?
I've been studying the idea of travelling through space and navigation - in particular: Apollo sextant (used by astronauts to plot the position of their spacecraft relative to stars). And, in order to appreciate the fundamental characteristic of the solar system, in terms of planes and orbits, I've also been reading about the route of the recently discovered interstellar Comet 2I/Borisov, which passed through the ecliptic plane from above and exited below the solar system.
The goal is to visualise how planes (planets) might appear from different parts of the solar system (in terms of Line Of Sight) - parallel to the ecliptic plane and from different compass points... I find the eyes.nasa.gov/apps/ website for Mars really useful for moving through and around the solar system.
Finally, considering that the Moon has a 5° tilt to the Earth, and Mars is on the same ecliptic plane as Earth - where exactly was Mars and Earth and the Moon when this photo was taken by Curiosity Rover as detailed on this space.com article?
I guess Mars was positioned near Earth September Equinox, and Earth positioned just after the Perihelion, and full Moon.
I can't quite understand how the moon seems to be much lower than Earth.
EDIT: answering the general position of the Earth and Moon, which is the first step towards determining the exact.
Interestingly, the light seen from Mars is that reflecting from the surface of the Pacific ocean.
On the question of "exact", there is a difference, obviously, between knowing the exact and general position. And the first knowing is most suited for the purposes of science, navigation, research - whereas the second form of knowing (generally) for the sake of natural curiosity. But, I think there is a great deal to be gained from appreciating both the exact and general position (us and the Earth) in space - I think this can broaden our perspective.
This is (in general) the position of the Earth and Moon as seen from Mars...
Firstly, the mountain range in the background is the same as the range visible in the 2nd image below. We can compare and notice the same peak and shape of the summits.

The Dingo Gap is situated at Gale Crater.
Mars, Earth, Moon, in space...
In general... January 31, 2014 - 10:30pm. This broad view looking downwards at the plane.

And again...

And from beneath the plane... showing us just how far ahead Mars was relative to Earth.

Finally, considering the exact position of the Moon in order to answer the question for why the Moon appears to be positioned beneath Earth.
In fact, the Moon, at the time, was midway through its 5° tilt orbit around Earth. And, from certain angles and from such distance from Mars, could have appeared beneath Earth, as shown in the photo taken by Curiosity.

Because our galaxy is at an angle, it is difficult to determine where planets are relative to one another - consequently, from one perspective, as shown in the above image, Earth appears to be ahead of Mars. The only absolute way to determine the general position (visually) of planets is from beneath or above the ecliptic plane(?).

You may find info here at NASA/JPL Horizons ephemeris; https://ssd.jpl.nasa.gov/horizons.cgi

BTW, although the inclination of Mars's orbit to the ecliptic plane is quite small (1.851°), given its orbital radius, it can get quite a distance above or below the ecliptic. (Maybe around 7 million km at perihelion).

Space.com says it was taken on January 31, 2014, “80 minutes after local sunset,” but let’s use 0 h UT—it doesn’t matter much: Mars moves by about ~0.5°/d and the Earth ~1°. The Moon travels about 13.2° per day, but we’ll use an approximation since we don’t know the exact time, although you could find it from Curiosity’s position on Mars. Using JPL Horizon as suggested by Adrian Howard, we have Earth long = 100.8825, lat = -0.0002; Moon long = 101.0055, lat = +0.0136; and Mars long = 359.7556, lat = -1.4133. This should get you going. ;-)

There are a lot of resources in the answer to Where can I find the positions of the planets, stars, moons, artificial satellites, etc. and visualize them? you may find that some of them are easy enough to do your self, and in that case you are always welcome to post an answer to your own question. What form do you want your positions in? There's an Ecliptic coordinate system where you can have X, Y, Z in kilometers (or meters or AU or whatever)

Horizons will do that for you, it will even do it for Curiosity's exact latitude and longitude on Mars. Or if you want someone to run it for you the first time so you can see how it works, then that will be great as well. Here's a tutorial for using Horizons, and here's an example of how to specify latitude and longitude on another solar system body.

@PM2Ring ... I see. The moon at its nearest and furthest, at 5°, would be quite a distance. It's a question of distance and position of each part individually, and  collectively.

@Dylan Indeed! And on 2014-Jan-31 the Moon's ecliptic latitude was around 5°N. According to Horizons, its distance above the ecliptic plane at 00:00 on that date was a little over 30,698 km. As the others have indicated, Horizons is great for this kind of thing. It gives you direct access to the ephemeris data produced by JPL. You can generate tables via its Web interface, and it's not hard to write scripts that access it in various ways.

I've only been learning how to use Horizons for a month or two, so I'm no expert, but here's the Python script I used to get that Moon position data:

link

NASA Photojournal item PIA17936 says:

Researchers used the left eye camera of Curiosity's Mast Camera (Mastcam) to capture this scene about 80 minutes after sunset on the 529th Martian day, or sol, of the rover's work on Mars (Jan. 31, 2014).

Between sols 528 and 532, Where is Curiosity? shows it in a place called Dingo Gap at Martian longitude 137.41°, latitude -4.63°.
At that location on Mars in Stellarium, sunset on 2014-01-31 occurs around 21:00 UT, so the picture was taken around 22:20 UT.
JPL HORIZONS gives these J2000 heliocentric positions at that time:

lon. (l)
lat. (b)
dist. (r)

Mars
168.784°
1.613°
1.662 au

Earth
131.767°
-0.001°
0.985 au

Moon
131.732°
0.010°
0.983 au

and these Mars-centric positions:

lon. (λ)
lat. (β)
dist. (Δ)

Earth
22.935°
-2.535°
1.058 au

Moon
22.826°
-2.522°
1.059 au

In other words, the Moon should appear about 0.11° west of Earth in the Martian sky, or to the right if Mars's north celestial pole is up:

But looking toward the western horizon (zenith up) from a tropical latitude, celestial north is to the right, so the Moon appears below Earth in the photo.

Beautiful answer! There is certainly something to be said for using real programs that run with easy-to-use GUIs and generate spectacular images instead of downloading tables of state vectors in barycentric coordinates from Horizons and trying to make plots with Python.

@uhoh Under the hood it's all the same. If you have such a WIP I look forward to seeing it.

Many+ thanks Mike... I was just checking the Stellarium website. Personally, I think there's great benefit to using the data+programming approach, alongside GUI software. The 'Where is Curiosity': wow! I'll look through all of this info, and aim to draw it out using the elliptical layout to illustrate.

@Dylan I hope you'll post your illustration as an answer.

@MikeG - I must admit, I was under the impression that your heliocentric positions table states where each were, what I'd try to do next is (understand them) plot it out on paper. If I can add this to the question later, definitely.


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What if you 'placed' the Chicxulub asteroid onto the Earth?
We all know what sort of devastating impact an asteroid could have if it collided with the Earth.
But I'm curious to know what the effects would be if you removed the speed element from it, and were somehow able to casually 'place' a giant asteroid onto our planet?
I think I read somewhere that the Chicxulub asteroid was most likely around 9 miles long, and that the end of it would have extended beyond our atmosphere if it hit the Earth lengthways.
But would this have any effect if it was travelling at, say, 2mph? There would be nothing burning in the atmosphere or shockwaves etc.
Would it have any affect at all apart from squashing whatever was underneath?
If the answer is no... how about placing a moon-sized object onto the Earth?

Note:  To remove the speed would require magic.  The absolute minimum speed that an asteroid could have on hitting the atmosphere is almost 11km/s

I’m voting to close this question because this requires magic.

@Pel Please consider editing your question: I guess you are asking what the effect of placing a huge object on the Earth's surface would make, do you? If so, I would write it that way, e.g. "What happens if we could bring a body of $80 {\rm km}$ diameter and a mass of $10^{16} {\rm kg}$ on Earth's surface?"

You could ask on Worldbuilding Meta if they would be willing to take a question about this. You'd probably have to substantially reword the question to define the parameters for which you'd like to see an answer and give context about why you want to know.

Shame it's been closed... I've seen (and asked) other ridiculous questions like this in the past and there's been no issues

@Pel ha! better not advertise that you've "asked other ridiculous questions", or someone may go back and vote to close them as well! I always feel bad if my questions are closed. It's not a perfect system but since this site is completely open to the internet there needs to be some limits. For this one, you could consider asking a question in Earth Science SE about placing a very large body on Earth and how it would compress the Earth below it, or about the atmospheric forces on it. At 2 MPH it would displace a lot of air and make wind, but not make a shock wave.

@Pel in Space SE there may be some questions about megastructures extending from the surface to space, I'm not sure. But you might have fun asking this type of question in Worldbuilding SE

@uhoh thanks! I didn't realise there was  an Earth Science SE actually... although I have to disagree with some of the others about my question requiring 'magic' – I'm sure in hundreds, or perhaps thousands of years we'd have the technology to do such things with asteroids (if we actually wanted to ha)

@uhoh ...here is one of my other equally strange / curious questions I was referring to if you'd like to take a look  https://astronomy.stackexchange.com/questions/35101/how-long-could-an-ocean-last-in-space

@uhoh oh wow I've actually just noticed you commented on my other question back when I first posted it haha!

@Pel ya I wanted to let you and other readers know that the question wasn't so terribly unreasonable :-) Btw your comment about magic reminds me of the third of Clarke's three laws

An asteroid resting on Earth would be a mountain. Or, for smaller asteroids, a pile of gravel.
Mountains are limited in altitude by the strength of stone to resist compression: a too tall mountain would sink down as the base crumbled and spread out. The limit on Earth is about 10 km. Besides the strength issue mountains are also floating ("isostasy") on the fluid but denser interior of Earth, so even if it doesn't fall apart it may sink down partially.
So the impactor would not hold together that well. A big ball of rock 10 km in diameter would first definitely sink down several kilometres into the crust, and almost definitely break. That means that something like $10^{15}$ kg will descend several kilometres, giving an energy $mgh \approx 10^{19}$ J (and there are estimates 100 times larger for the mass). That is "just" like a big hurricane, but will mostly be released as localized seismic energy and heating. The end result would be a massive pile that likely destabilizes the crust in the vicinity a lot: far less than an actual high velocity impact, but still a major disaster.
If you do the same with the moon you will break up the crust, boil off the oceans and wipe out all life on the planet.

Lol wow - sounds interesting! Thanks for satisfying my curiosity


text:
Demo to get Matter Power Spectrum in cosmology
I would like please to demonstrate the expression of Power spectrum in Cosmology :
First, I have the relative contrast:
$$\delta_{i}(\vec{x}, z) \equiv \rho_{i}(\vec{x}, z) / \bar{\rho}_{i}(z)-1\quad(1)$$
After, we decompose this relative contrast on Fourrier basis :
$$\delta_{i}(\vec{x}, z)=\int \frac{\mathrm{d}^{3} k}{(2 \pi)^{3}} \tilde{\delta}_{i}(\vec{k}, z) \exp (\mathrm{i} \vec{k} \cdot \vec{x})\quad(2)$$
and finally, how to find the following expression (3) from (1) and (2) :
$$\left\langle\tilde{\delta}_{i}(\vec{k}, z) \tilde{\delta}_{i}\left(\vec{k}^{\prime},z\right)\right\rangle=(2 \pi)^{3} \delta_{\mathrm{D}}\left(\vec{k}+\vec{k}^{\prime}\right) P_{i}(\vec{k}, z)\quad(3)$$
?
Any help is welcome.

This is just a Fourier transform: (let $\boldsymbol{x}=\boldsymbol{r}_2-\boldsymbol{r}_1$)
$$
\begin{aligned}
\langle \delta(\boldsymbol{k}_1)\delta(\boldsymbol{k}_2) \rangle&=\int\int d^3r_1d^3r_2\langle  \delta(\boldsymbol{r}_1)\delta(\boldsymbol{r}_2) \rangle e^{-i\boldsymbol{k}_1\cdot\boldsymbol{r}_1}e^{-i\boldsymbol{k}_2\cdot\boldsymbol{r}_2}\\
&=\int d^3r_1 e^{-i\boldsymbol{k}_1\cdot\boldsymbol{r}_1}\int d^3r_2\langle\delta(\boldsymbol{r}_1)\delta(\boldsymbol{r}_2)\rangle e^{-i\boldsymbol{k}_2\cdot\boldsymbol{r}_2}\\
&=\int d^3r_1 e^{-i\boldsymbol{k}_1\cdot\boldsymbol{r}_1}\int d^3x\langle\delta(\boldsymbol{r}_1)\delta(\boldsymbol{r}_1+\boldsymbol{x})\rangle e^{-i\boldsymbol{k}_2\cdot(\boldsymbol{r}_1+\boldsymbol{x})}\\
&=\int e^{-i(\boldsymbol{k}_1+\boldsymbol{k}_2)\cdot\boldsymbol{r}_1}d^3r_1\int\xi(\boldsymbol{x})e^{-i\boldsymbol{k}_2\cdot\boldsymbol{x}}d^3x\\
&=(2\pi)^3\delta_D(\boldsymbol{k}_1+\boldsymbol{k}_2)P(\boldsymbol{k}_2)
\end{aligned}
$$
Here, $\langle\delta(\boldsymbol{r}_1)\delta(\boldsymbol{r}_2\rangle)$ is two-point correlation function (2pcf) in real space. If we assume our universe is statistically homogeneous, $\langle\delta(\boldsymbol{r}_1)\delta(\boldsymbol{r}_2\rangle)$ should have the form $\xi(\boldsymbol{r}_1-\boldsymbol{r}_2)$. So power spectrum is the Fourier transform of 2pcf.
In addition, if we assume our universe is statistically isotropic (not true in redshift space), 2pcf can be $\xi(|\boldsymbol{r}_1-\boldsymbol{r}_2|)$ and power spectrum can be $P(k)$.

Zhao : Thanks. which trick do you use in integrals to pass from first line to second line in your demonstration  :

$\begin{aligned}
\langle \delta(\boldsymbol{k}_1)\delta(\boldsymbol{k}_2) \rangle&=\int\int d^3r_1d^3r_2\langle  \delta(\boldsymbol{r}_1)\delta(\boldsymbol{r}_2) \rangle e^{i\boldsymbol{k}_1\cdot\boldsymbol{r}_1}e^{i\boldsymbol{k}_2\cdot\boldsymbol{r}_2}\
&=\int e^{i(\boldsymbol{k}_1+\boldsymbol{k}_2)\cdot\boldsymbol{r}_1}d^3r_1\int\xi(\boldsymbol{r}_1-\boldsymbol{r}_2)e^{i\boldsymbol{k}_2\cdot(\boldsymbol{r}_1-\boldsymbol{r}_2)}d^3(r_1-r_2)\
&=(2\pi)^3\delta_D(\boldsymbol{k}_1+\boldsymbol{k}_2)P(\boldsymbol{k}_2)
\end{aligned}$ ? Best regards.

@youpilat13: Thanks for your comments, I fixed my Fourier notation and improved the derivation. Hope this helps!

Zhao. Thank you, it is clearer. Best regards


text:
Exoplanet dip in transit light curve when the planet passes behind the star
In the animation below, I don't understand why the brightness slightly decreases when the planet is behind the star.

Where does this effect come from?

Just before the planet goes behind the star, we see the light directly from the star as well as the light reflected from the planet's surface.  When the planet is behind the star, we no longer see the reflected light.
Note that the top of the line also curves to reflect how much of the planet's surface is illuminated from the star.  The most illumination is when it is almost "full" just before it passes behind the star.  The least illumination is when it is "new" just before it passes in front of the star.
Edit (to address Fraxinus's comment below): Here is an example from Zhang et. al: Phase Curves of WASP-33b and HD 149026b and a New Correlation between Phase Curve Offset and Irradiation Temperature.

The above graphic shows processed data from the Spitzer space telescope.  From the paper:

Observations were timed to begin before a secondary eclipse and end
after the following secondary eclipse,

These light curves help aid in estimates of planetary albedo, day/night temperature, and atmospheric composition.

Is the second "dip" really visible for any real star?

@fraxinus Yes, I added an example above.

I'm constantly amazed at the precision that we achieve for measurements of distant stars. Intuitively I'd expect this difference to be lost in the noise.

@Barmar I definitely cherry-picked an example to show this effect.  Wasp-33b is a super-hot Jovian giant with an orbital period of only 1.22 days, so it absorbs and emits tremendous amounts of energy.  Most exoplanets' secondary eclipse light signatures would be lost in the stellar noise variations.  I don't think aliens from another star system would be able to detect Jupiter's pass behind our Sun!


text:
Could a star become a planet?
Could a star become a planet? I am asking this because the gas giants are ¨Failed Stars¨ and they are classified as planets in our solar system.

Your question in the title is "Could a star become a planet" and the question in the body is "Could a planet become a star".    Which question to you mean to ask?

@Joseph Casey   Brown dwarfs are not classified as planets in our solar system becuae they aren't classified as planets and aren't in our solar system..  There are no known brown dwarfs in our solar system.  Brown dwarfs ae not classified as planets or as stars.  Brown dwarfs are over about 13 times the mass of jupiter and are massive enough to fuse deuterium for at least part of their life times.  Objects more than about 75 to 80 times the mass of Jupiter are stars.

@JamesK I meant in the title, sorry my mind has been crazy today

I am editing to revert @fasterthanlight's edit (there are no brown dwarfs in our solar system) and to make the question in the body be in line with the question raised in the title.

Are black holes, neutron stars, and white dwarfs that orbit a currently fusing star planets? Once a brown dwarf has consumed all of its deuterium, does it become a planet? Per the IAU definition of a planet, there are only eight planets in the entire universe. One reason the IAU restricted it's definition of what qualifies as a planet to objects that orbit the Sun was to avoid the fuzzy boundaries between super-Jupiters and brown dwarfs, and between brown dwarfs and red dwarfs. Classifying black holes, neutron stars, and white dwarfs as planets seems ludicrous, but they aren't "stars".

A star could become a planet, but this almost never happens. It would probably have to lose almost all of its mass early in its life, perhaps by a black hole. Then, the exposed core would have to be stopped from becoming degenerate and further mass loss would have to continue. It is possible, but it is not realistic and  will not happen in a human lifetime, even if you observed every star in the universe (including white dwarfs).
A brown dwarf definitely can become a planet. A brown dwarf early in its life acts like a star, fusing deuterium and reaching temperatures of almost 2000 K. But as it ages it cools down and stops fusing deuterium. Then it acts like a planet.
In theory, a brown dwarf directly could become a planetary-mass object through mass loss. This could happen from a giant impact or another body accreting from it.
Contrarily, a planet (brown dwarf) can become a star if it accretes enough mass. For example, the most massive brown dwarfs have masses of about ~80 Jupiters, while the least massive stars have a mass of about ~60 Jupiters (there is an overlap). Therefore, a "planet" could become a star under special circumstances.
As @User123 said in a comment, all stars start as a "planet," without burning hydrogen. While it accretes more material, fusion ignites in its core and it becomes a star.

We could say, that even Sun started out as a planet: without burning hydrogen and helium. So the answer is pretty obvious.

@User123 Thanks for the feedback :)   However, isn't there a term called a "fusor" which is more general than a star? A fusor is any object capable of fusion but does not fuse. See https://en.wikipedia.org/wiki/Fusor_(astronomy)

Actually at first there was only a small "rock", with another small colliding. So we can say that it started as a "planet" (in parentheses). By the way, thank you for the new word I have never heard.

@fasterthanlight  Your statement that the lowest mass star has a mass of about 90 Jupiters might be incorrect.  This list http://dictionary.sensagent.com/List%20of%20least%20massive%20stars/en-en/ of the lowest mass stars includes many stars with masses lower than 90 Jupiters, someless massive than the most massive brown dwarfs on the list.  The lowest mass star on the list is listed with a mass of only 52 Jupiters.  If that is correct, there are stars with much less than 90 Jupiter's mass.

@M.A.Golding Oops! Edited that.

Thank you all for your help, you guys are awesome!

This is a great answer, but since it has no supporting links whatsoever it's hard to judge the veracity of the information. It's best if we included sources for facts rather than hoping our memory is 100% correct, that way future readers will be able to better judge the accuracy and to read further. (at)MAGoldring has provided one such link already, consider citing it as one such source? Thanks!


text:
Significantly blueshifted Lyman-$\alpha$
Lyman-$\alpha$ line in galaxies is notably known to show a double peaked profile, mainly due to its scattering in a moving medium, see e.g., the very recent Matthee et al. 2021, The X-SHOOTER Lyman-α survey at z=2 (XLS-z2) I: the panchromatic spectrum of typical Lyman-α emitters.
Reading through some literature I learned that most of the time the blue peak is reduced and the red one is enhanced because of the presence of outflowing medium, commonly present in star-forming galaxies. The peak separation is of the order of a few hundreds of $\rm km ~s^{-1}$. In case of inflowing medium (rarer) the situation is inverted.
I have in hands a spectrum of a galaxies with a Lyman-$\alpha$ absorbed at the systemic redshift and with a narrow component blueshifted by $>1000 ~\rm km ~s^{-1}$, whit no red peak. Is there any known mechanism that could explain this?

Does the absorption go all the way to zero flux? Could you show the spectrum? What redshift is it? Which instrument did you use? Are you Tony P.?


text:
Could an orphan/rogue planet have a moon?
Is there some reason why an orphan or rogue planet wouldn't have a moon? Let's say it started out in a normal system and then just got flung out by an unstable orbit.

If it didn’t have a moon in the beginning, of course!

It depends on the (bigger) problem that you are trying to solve. If you want an inhabited moon, then having it rotate and orbit a planet may be a useful way to generate thermal energy through tidal forces creating internal friction. What's the underlying world-building issue that you're trying to solve here? Else fasterthanlight's suggestion might be worth following.

The planet most definitely can have a moon.  After all, if a passing star messes with its orbit and it ends up flying out into space, it could retain the moon the same way as it does orbiting a single star.  There is a limit (Hill sphere, 1.5 million km versus 0.38 million km for the Moon) so getting too close to a star could take away the moon.  It may be harder to plot out a solution for ejecting the planet without such a close encounter.

Is this a duplicate of https://astronomy.stackexchange.com/questions/1236/is-it-possible-for-a-moon-to-orbit-a-planet-floating-free-in-the-galaxy-rather-t ?

Does this answer your question? Is it possible for a moon to orbit a planet floating free in the galaxy rather than orbiting a star

Could an orphan planet have a moon?

Yes.

s there some reason why an orphan or rogue planet wouldn't have a moon?

Yes, this answer lists some possibilities.
What you didn't ask was:

Is there some reason why no orphan nor rogue planet could ever have a moon?

and based on the previous two, No.
Any non-contact gravitational acceleration that could pull a planet away from its star has some chance of accelerating its moons sufficiently to stay with the planet.
If the perturber were a few times the Moon's orbit farther from the planet, staying with the planet might even be likely.
Looking at this obliquely from a delta-v perspective, Jupiter orbits the Sun at 13.1 km/s, and its inner Galilean moon Io orbits Jupiter at 17.3 km/s.
Without doing the math, I think it's possible that you could give Jupiter a 13.1 km/s kick impulsively in an optimal direction and Io would stay bound to it, but now in a much more elliptical orbit.

Note that the Jupiter example uses a very theoretical scenario: A sudden acceleration of Jupiter which doesn't also affect any of its moons. A more realistic scenario would be a very massive body passing by and affecting Jupiter with its gravity. That would accelerate Jupiter's moons in the same way as it would affect Jupiter itself.

@Philipp I propose that the perturber pass within "few times the Moon's orbit farther from the planet" so while it may affect both there will be some difference between the two. I'm making a hand-waving argument that Io will remain bound by looking at the problem "obliquely from a delta-v perspective" but I didn't go so far as to say the perturbation had to be an impulse. I think I'll just go model it in Python and compare to any other models that might arise.

Certainly an orphan planet could retain its moon.
One doesn't need to apply impulse to a planet in order to make it go rogue.  One could just as easily apply impulse to the star it orbits instead (though the impulse would have to be orders of magnitude greater). As soon as the difference in velocities between a planet and star exceed escape velocity, the planet has gone rogue.
If a star had a close pass to our Sun, the acceleration introduced to the Sun could cause the differential velocity of the Sun and Jupiter to exceed escape velocity, while leaving the local orbital system of Jupiter relatively intact.
I used Universe Sandbox to create a simulation demonstrating this possibility and showing a moon can be retained on a rogue planet.  I uploaded the result to youtube:

+n! n-factorial upvotes for the answer as simulation as video! Is it possible to mention the name of the software used?

Also, I'm still curious if my prediction that the pair can be ejected by something (probably much smaller than Sirius, like a 2x Jupiter perhaps) passed close to them rather than the Sun, leaving Io perturbed but still bound to Juipiter. I guess it would have to be fairly fast and moving in roughly the same direction as Io was (relative to Jupiter) at closest approach perhaps as well as the same direction that Jupiter was moving relative to the Sun. I guess that would have to be another rogue Jupiter, a less likely event.

@uhoh  The software used is a game/simulator called Universe Sandbox.  I purchased my copy from the Steam Store:https://store.steampowered.com/app/230290/Universe_Sandbox/

I think I can take out Sirius and add another Jovian to my simulation pretty easily.  I will see if I can get an ejection with a close Jupiter pass that retains Io.

Great! :-) Oh, it's right there in your answer, I'm sorry I missed that. you may be interested in adding answers to Is a General Relativity approximation available in Universe Sandbox (1 or 2)? and Universe Sandbox 2 accurate representation of gravity and Is Universe Sandbox 2 realistic? in fact I'm adding a universe-sandbox tag to many questions now so they are more searchable.

@uhoh Your intuition is spot on.  A very close pass from a double Jupiter mass rogue at 20km/s can budge Jupiter out of a solar orbit, while Io's orbit around Jupiter is retained, albeit with a much higher eccentricity.  Tidal heating makes Jupiter quite toasty as it passes near the Roche boundary of the larger Jovian.  Let me know if it is worthwhile to post the simulation video capture.  By the way, I believe these simulations can be shared to others that own the same game.

Why don't I ask a new question based on the assumptions in my answer here, something like "I made handwaving delta-v arguments here, but how do these hold up under scrutiny of F=ma?" That provides space for a new answer. It can be answered with a simple description of initial and final states and perhaps a screenshot or two, but of course a video would be quite interesting to readers here I'm sure!

I don't know of a way to share setup files within Stack Exchange unless it can be posted as text like a script for example. For myself I don't really run programs; if I need a trajectory I set it up in python. Of course I can't make supernovas that way. :-)

There are quite a few possible reasons for a 'rogue' planet not to have a moon

In the formation of it's original solar system, the planet never had a moon form. This could likely occur if the planet is a rocky planet close to its sun, where such planets are less likely to form moons when the system is young
It never captured a moon during its tenure in its original solar system. Evidence suggests that many moons are 'captured', perhaps being early planets or asteroids that have irregular orbits and get caught in a planets gravity field.
It did have a moon/moons, but when the planet was 'pulled' from its original orbit (presumably by something large and close) it was flung off into space and the moons were destroyed/pulled apart, absorbed in either the large body or planet, or flung off themselves into a different trajectory.

However, keep in mind that a rogue planet would need to have its orbit distorted so much as to 'escape' the gravitational field of not just its sun, but its sun's other planets. Meaning for this to occur the pull of gravity to fling it out must be large indeed - perhaps a Jupiter-type planet pulling a small Mercury-type planet, enough to severely change direction but not enough that tidal forces pull it apart.

You can pretty much ignore the effects of the other planets. Their mass is typically insignificant compared to that of the host star,  and each pair of planets tends to have relative velocities far in excess of either planet's escape velocity at the mutual distance.

A possible case of a rogue planet with a moon has been found. See this paper and the general synopsis from Nature. In this case there are 2 possible solutions when modeling the observations:

Few Jupiter mass planet with a half earth mass moon fairly close to us
A low mass star (0.12 Msun) with a super earth (very uncertain mass) far away.

Unfortunately there is currently no way to distinguish between these 2 possiblities. Detecting the faint star may be possible with future optical telescopes.
This was an answer forStackExchange astronomy


text:
Does meteor direction change if you look from a different country?
So here is the question: Me and a friend of mine from Norway we saw a meteor but que said he saw it in the opposite direction as me and I was looking from Spain, does that make any sense? Or should we see it going the same direction?
Thanks in advance


text:
Could ancient tectonic activity on Mars be due to it having oceans of sulphuric acid?
Could Mars' ancient tectonic activity have ceased due to its oceans having been composed of sulphuric acid, which warmed up and evaporated due to the Theia event as ice chunks landed on its surface? Strongly conductive liquid sulphuric acid on early Mars' surface could have been responsible for generating its core spin causing tectonic action which then all died off as a result of its oceans evaporating and thus removing the conductivity responsible for the core spin, whereas Earth's oceans didn't evaporate (because they are water) so can phase to ice ages to prevent overheating.
Why else would Mars have evidence of ancient alternately polarised extrusion from now ceased tectonic activity?

It doesn't sound very likely to me

The hypothesized Theia impact resulted in a cloud of vaporized and molten rock, not flying chunks of ice, and the early solar system was regularly bombarded with ice in the form of comets. And of course the impact heating would dwarf anything from mixing of sulfuric acid and water. Apart from that, why would you expect ancient Mars to form sulfuric acid oceans in the first place? How would a conductive ocean generate core spin? How would core spin result in tectonic action? Where would that sulfuric acid have gone? This is just a long string of unconnected ideas and unsupported assumptions.

@puffin there is also earth science that has the geophysics tag https://earthscience.stackexchange.com/questions/tagged/geophysics. Would suspect to be suitable there it would need a fair bit more detail on the proposed mechanisms collecting the sulpher, generating core spin and then evaporating.

The problems with the basic assumptions remain: the proposed explanation is a chain of non sequiturs. Apart from the implausibility of such sulfuric acid oceans or the problems with the ice idea, conductivity of oceans is unrelated to core spin and core spin is unrelated to plate tectonics. It seems they really are trying to ask why plate tectonics on Mars stopped and/or its magnetic field faded, and I suggest focusing on that. The answer almost certainly does not involve sulfuric acid oceans or Theia impact ice bombs.

See Wikipedia. Surely it's worth comparing. Plate tectonics don't just happen; otherwise they'd be happening on lots of planets.
There has to be a mechanism behind them.

First, tectonic activity have not ceased on Mars, although the planet is in a "stagnant lid" stage, which represents the last stage before a body becomes tectonically dead. See the answer to Is there convection in Mars' mantle?
Second, tectonic activity is caused by internal heat. As a planet gradually loses its internal heat, tectonic activity slows down, until it eventually stops. There is no need for more convoluted explanations.