How can an infinite universe expand? I understand the expansion of the universe as actually an increase in the ratio of space to matter. Is this a correct understanding?  Otherwise, I don't understand how an infinite structure can expand.

Expansion means that distances are increasing as a function of time. Say if the distance between two galactic clusters is $D$, then in an expanding Universe the distance is governed by some strictly increasing function of time $a(t)$ called the scale factor where

$$D=a(t)D_0$$

where $D_0$ is the distance at the present time and by definition $a(t_{0})=1$.

Cosmology assumes that the Universe is on large scales the same everywhere (homogeneous) and the same in all directions (isotropic) so the above applies to all distances above a certain scale. The scale factor $a(t)$ can be found from the Friedmann equations and initial conditions.

Expansion is possible in Universes of both finite and infinite spatial extent.

As the volume of a (large enough) region of space increases in proportion to $[a(t)]^3$, but the amount of matter remains constant, the matter density changes in proportion to $[a(t)]]^{-3}$. Expansion however also decreases the kinetic energy of of its contents, so the energy density decreases by a greater factor if the contents has kinetic energy.

I understand the expansion of the universe as actually an increase in the ratio of space to matter. Is this a correct understanding?

It isn't wrong. The ratio is increasing. But it isn't a "correct understanding". It's merely an observation of one of the results of the expansion of space.

  If the universe is infinite how can it expand?

I don't know. Nor do I know how big bang cosmology can be reconciled to an infinite universe. If you look around on the internet, you can find articles like this which say this:

"The linear dimensions of the early universe increases during this period of a tiny fraction of a second by a factor of at least 10$^{26}$ to around 10 centimetres (about the size of a grapefruit)".

However in 2013 results from the WMAP mission appeared to confirm that space is flat. Then a non-sequitur crept in. See this article and pay careful attention to this:

"We now know (as of 2013) that the universe is flat with only a 0.4% margin of error. This suggests that the Universe is infinite in extent; however, since the Universe has a finite age, we can only observe a finite volume of the Universe. All we can truly conclude is that the Universe is much larger than the volume we can directly observe."

That's a massive error. It absolutely doesn't suggest that the universe is infinite in extent. Or that the Universe is much larger than the volume we can directly observe. But this myth has legs, and people repeat it ad-infinitum, even though they can't explain how it fits in with Big Bang cosmology. What you tend to hear is that the observable universe was the size of a grapefruit, but it absolutely doesn't satisfy. Moreover there's a dreadful flaw lurking in the shadows. Take a look at the stress-energy-momentum tensor, and note the energy-pressure diagonal. A gravitational field is something like a spatial pressure gradient, and you can think of space as having an innate "pressure". So you can reason that the universe must expand. As to why Einstein didn't, I just don't know. But anyway, for an analogy, squeeze a stress-ball down in your fist, and let go. It expands because of the pressure. However if that material was infinite in extent, the pressure is counter-balanced at all locations. So it can't expand. In similar vein, in my opinion, an infinite universe can't expand.

People claim the universe must be infinite because of the cosmological principle. But this is merely an assumption. There's an assumption that the universe is homogeneous and isotropic, but this isn't fact. You cannot use it to make sweeping claims about an infinite universe that was always infinite. For all we know some observer 50 billion light years away might be looking up at the night sky wondering why half of it is black. Or a mirror-image of the other. Or some kind of edge.

It is said that in days gone by, people could not conceive of a world that was curved. They could only conceive of a world with an edge. Nowadays I rather fancy that there are some people who cannot conceive of a world that is not curved. They cannot conceive of a world with an edge.

Edit:

See The Foundation of the General Theory of Relativity: "the energy of the gravitational field shall act gravitatively in the same way as any other kind of energy". Energy is the source of the stress-energy tensor. Matter is only a source because of the energy-content. Also see Inhomogeneous and interacting vacuum energy which refers to spatial energy. An interesting read is the article Universe 156 billion light-years wide
 featuring Neil Cornish. This isn't entirely accurate, but the compound interest and the hall of mirrors concepts are of interest. As for the non-sequitur, see this interview with Joseph Silk:

"We do not know whether the Universe is finite or not."

I hope nobody will argue with that. Reading on:

"To give you an example, imagine the geometry of the Universe in two dimensions as a plane. It is flat, and a plane is normally infinite. But you can take a sheet of paper [an 'infinite' sheet of paper] and you can roll it up and make a cylinder, and you can roll the cylinder again and make a torus [like the shape of a doughnut]. The surface of the torus is also spatially flat, but it is finite".

This is akin to the old Asteroids game. But the Planck mission found no evidence of any torus. Reading on further:

"So you have two possibilities for a flat Universe: one infinite, like a plane, and one finite, like a torus, which is also flat."

I dispute that. There is a third possibility. A flat finite universe with no intrinsic curvature. If anybody can cite some reliable sources that support the assertion that a flat universe must be infinite, I'd like to see them.


Need help with telescope mount; stars have tails If I take a shot longer than 20 seconds, the stars get a tail. I have a Celestron Advanced GT mount with starshoot autoguide scope. I use a cannon can on a 9" Celestron scope. I can't get a picture without having tails on my stars. It only looks good if I keep the pictures under 20 seconds.

EDIT :

I did the polar alignment. The mount has a polar alignment scope threw the center of it. It's also balanced, the mount has a brids eye level. I use the new PhD software with the starshoot autoguide.. There is no slop in the mount, the gears have almost no backlash.

If you can't get any longer than 20 seconds, chances are it's your polar alignment. The mount you have is a motorised GoTo EQ mount (I believe), but if these aren't aligned properly, then they can still have trouble.

Also, depending on your software, try a 2 or 3 star align to get a better accuracy.

Also check the scope is nicely balanced with the camera on the end, and that your mount can take the weight.

EDIT :

If you try all this and you are still getting trailing, it may be worth going to a professional to get the mount looked at. It may be faulty.

you need a motorized equatorial mount, which tracks the stars in such a manner as to cancel the rotation of the earth and eliminate the tails on your photos. These are commercially available for many common types of telescope mounts.


How to approximate the distance at which the thin disk peaks along a given LOS from Earth for a given galactic coordinate per 1 sq deg? The density of stars drops off exponentially from the center of the galaxy and can be approximated as follows using a cylindrical coordinate system:
$$ \rho(R, z) = \rho_0 \exp \frac{-R}{h_R} \ \exp  \frac{-|z|}{h_z} $$
screenshot
The radial scale length and height of the thin disk are ~ 3000 pc and ~ 200 pc respectively. At the moment rho_0 will just be taken as 1. We observe at a given l and b of 90 and 15 respectively using the galactic coordinate system (just look at l and b for now, the rest are annotations I added that I will describe shortly). The galactic coordinate system uses spherical coordinates:

We want to know where the density of stars peak along at given LOS (from Earth) per 1 sq degree at a given l and b. There are 41253 sq degrees in the entire sky, so 1/41253 will be the fraction we want to multiply the density equation by. However, we must also consider how the volume element changes by distance along the given LOS, so we need to also multiply rho and 1/41253 by this change in volume. This volume element can be represented by $4 \pi r^2 dr$ (the surface area of the sphere and the small change in the distance along the LOS). We are almost done but need a way to transform between coordinate systems by relating galactic coordinates to distance r and R (as seen in the diagram above). Mathematically, they are related as follows (Eq. 3 in Where Does the Disk Stop and the Halo Begin? Kinematics in a Rotation Field Morrison, Flynn and Freeman 1990 Ast. J. 100 (4) 1191-1222):
$$ r^2 = R_☉^2 + R^2 - 2 R R_☉ \cos b \cos l $$
screenshot
So, for a given r (distance from star or LOS to sun), l and b, we can find R (the distance from the G.C. to the star or LOS). This is in cylindrical coordinates, so we need to use cos(b)R and sin(b)R to find R and z in the density equation above (note these are different r's). The volume element needs to be the LOS from Earth, which corresponds is the R in the diagram and equation for the paper (for the given l,b, and r). Putting this all together, I have the following code. Please let me know if this attempt seems reasonable, or if not, where I went wrong.
import numpy as np
import matplotlib.pyplot as plt

C=8500 #Distance from center of galaxy to sun (R_sun_

#Galactic Coordinates
b=np.deg2rad(15) #above plane
l=np.deg2rad(90)

#Scale heights for the thin disk
h_R=3000
h_z=200

nsqdeg_in_sky=41253 #square deg in whole sky

#R: Distance from star to sun
R=np.arange(0,15000,1)

#distance from star to G.C.
#r: distance from star to G.C. kpc
r=np.sqrt(C**2+R**2-2*R*C*np.cos(b)*np.cos(l))
print(r)

#Spherical Galactic Coordinates to cylindrical coordinates
#z component of R
R_z=np.sin(b)*r
#r component of R
R_r=np.cos(b)*r

#density equation describing thin disk
rho_thin=np.e**(-R_r/h_R)*np.e**(-np.abs(R_z)/h_z)

#desnity equation combining volume element and portion of sky
N_thin = (1./nsqdeg_in_sky)*(4*np.pi)*R*R*rho_thin

plt.title("Number of stars per 1 square degree")
plt.plot(R,N_thin,color='blue',label='thin')
plt.legend(loc='best')
plt.xlabel(r'$distance$ [pc]')
plt.ylabel(r'$N_\ast$ [pc$^{-1}$]')

edit to add plot for given l and b above:

If you have a ray from an origin point $\mathbf{x}_0$  with unit direction $|\mathbf{v}|=1$,  $\mathbf{x}(s)=[x(s),y(s),z(s)]=\mathbf{x}_0+\mathbf{v}s$ and look at the density times distance $\rho(s)=\rho_0 e^{-R(\mathbf{x})/h_R}e^{-z(\mathbf{x})/h_z}s^2$ the derivative is
$\rho'(s) =\rho_0\left\{ [-e^{-R(\mathbf{x})/h_R}R'(\mathbf{x})/h_R]e^{-z(\mathbf{x})/h_z}s^2 + [-e^{-z(\mathbf{x})/h_z}z'(\mathbf{x})/h_z]e^{-R(\mathbf{x})/h_R}s^2 +[2s]e^{-R(\mathbf{x})/h_R}e^{-z(\mathbf{x})/h_z} \right\}$
$=\rho(s)\left\{[-R'(\mathbf{x})/h_R]+[-z'(\mathbf{x})/h_z]+[2s]/s^2\right\}$.
Now, $R'(s) = (x(s)v_x+y(s)v_y)/R(s)$ and $z'(s)=v_z$ so the criterion for a maximum or minium is $0 = -(x(s)v_x+y(s)v_y)/h_RR(s)-v_z/h_z + 2/s$ or $$\frac{2}{s} = \frac{[x_0v_x+y_0v_y] +[v_x^2 + v_y^2]s}{h_R\sqrt{(x_0+v_xs)^2+(y_0+v_ys)^2}}+\frac{v_z}{h_z}$$
I can't see how to simplify it further.
I think the right approach here is just to do a numeric solution using golden section search rather than do a lot of algebra that anyway ends with a root finding algorithm. Doing that also allows you to use any stellar density model.


Can planets inside the Earth's orbit around the Sun appear to undergo retrograde motion? Retrograde motion is:

  The apparent motion of the planets when they appear to move backward (from east to west) from the direction that they move ordinarily with respect to the stars.

Can planets inside the Earth's orbit around the Sun appear to undergo retrograde motion?

Both Mercury and Venus undergo retrograde motion, but are visually close to the Sun when this happens, so we don't notice it.

Below is Mercury's right ascension from May 15 to July 15 of this year. Note that it starts off increasing, then decreases (retrograde motion) and then increases again.

*******************************************************************************
Ephemeris / WWW_USER Thu Feb 19 08:02:41 2015 Pasadena, USA      / Horizons
*******************************************************************************
Target body name: Mercury (199)                   {source: DE-0431LE-0431}
Center body name: Earth (399)                     {source: DE-0431LE-0431}
Center-site name: GEOCENTRIC
*******************************************************************************
Start time      : A.D. 2015-May-15 00:00:00.0000 UT
Stop  time      : A.D. 2015-Jul-15 00:00:00.0000 UT
Step-size       : 1440 minutes
*******************************************************************************
Target pole/equ : IAU_MERCURY                     {East-longitude -}
Target radii    : 2439.7 x 2439.7 x 2439.7 km     {Equator, meridian, pole}
Center geodetic : 0.00000000,0.00000000,0.0000000 {E-lon(deg),Lat(deg),Alt(km)}
Center cylindric: 0.00000000,0.00000000,0.0000000 {E-lon(deg),Dxy(km),Dz(km)}
Center pole/equ : High-precision EOP model        {East-longitude +}
Center radii    : 6378.1 x 6378.1 x 6356.8 km     {Equator, meridian, pole}
Target primary  : Sun
Vis. interferer : MOON (R_eq= 1737.400) km        {source: DE-0431LE-0431}
Rel. light bend : Sun, EARTH                      {source: DE-0431LE-0431}
Rel. lght bnd GM: 1.3271E+11, 3.9860E+05 km^3/s^2
Atmos refraction: NO (AIRLESS)
RA format       : HMS
Time format     : CAL
EOP file        : eop.150218.p150512
EOP coverage    : DATA-BASED 1962-JAN-20 TO 2015-FEB-18. PREDICTS-> 2015-MAY-11
Units conversion: 1 au= 149597870.700 km, c= 299792.458 km/s, 1 day= 86400.0 s
Table cut-offs 1: Elevation (-90.0deg=NO ),Airmass (>38.000=NO), Daylight (NO )
Table cut-offs 2: Solar Elongation (  0.0,180.0=NO ),Local Hour Angle( 0.0=NO )
**************************************************************************************************************
 Date__(UT)__HR:MN     R.A._(ICRF/J2000.0)_DEC  APmag  S-brt            delta      deldot    S-O-T /r    S-T-O
**************************************************************************************************************
$$SOE
     2015-May-15 00:00     04 42 01.78 +24 08 32.2   1.53   4.36 0.69142867651373 -28.9445515  18.6988 /T 129.3762
     2015-May-16 00:00     04 43 23.87 +24 00 54.9   1.74   4.49 0.67512627148213 -27.4926615  18.0144 /T 132.4768
     2015-May-17 00:00     04 44 24.75 +23 51 35.3   1.96   4.63 0.65967892310678 -25.9844651  17.2466 /T 135.5875
     2015-May-18 00:00     04 45 04.61 +23 40 37.1   2.21   4.77 0.64511881456032 -24.4211059  16.3963 /T 138.7127
     2015-May-19 00:00     04 45 23.80 +23 28 04.3   2.47   4.91 0.63147747327791 -22.8037380  15.4651 /T 141.8554
     2015-May-20 00:00     04 45 22.86 +23 14 01.5   2.75   5.06 0.61878570842679 -21.1337209  14.4555 /T 145.0181
     2015-May-21 00:00     04 45 02.49 +22 58 33.7   3.05   5.20 0.60707343838108 -19.4128031  13.3707 /T 148.2014
     2015-May-22 00:00     04 44 23.61 +22 41 46.8   3.37   5.33 0.59636941752453 -17.6432855  12.2149 /T 151.4048
     2015-May-23 00:00     04 43 27.36 +22 23 47.7   3.71   5.45 0.58670087720487 -15.8281545  10.9939 /T 154.6255
     2015-May-24 00:00     04 42 15.08 +22 04 44.1   4.07   5.56 0.57809309959244 -13.9711803   9.7148 /T 157.8583
     2015-May-25 00:00     04 40 48.34 +21 44 44.8   4.46   5.63 0.57056894619759 -12.0769753   8.3871 /T 161.0937
     2015-May-26 00:00     04 39 08.89 +21 23 59.8   4.87   5.66 0.56414836550867 -10.1510071   7.0239 /T 164.3154
     2015-May-27 00:00     04 37 18.70 +21 02 40.1   5.29   5.62 0.55884790663755  -8.1995645   5.6456 /T 167.4924
     2015-May-28 00:00     04 35 19.87 +20 40 57.9   n.a.   n.a. 0.55468026744915  -6.2296724   4.2893 /T 170.5572
     2015-May-29 00:00     04 33 14.63 +20 19 06.0   n.a.   n.a. 0.55165390565664  -4.2489596   3.0439 /T 173.3313
     2015-May-30 00:00     04 31 05.30 +19 57 18.0   n.a.   n.a. 0.54977273920552  -2.2654833   2.1722 /T 175.2574
     2015-May-31 00:00     04 28 54.24 +19 35 48.1   n.a.   n.a. 0.54903595776444  -0.2875219   2.2105 /L 175.1837
     2015-Jun-01 00:00     04 26 43.78 +19 14 50.1   n.a.   n.a. 0.54943796056976   1.6766529   3.1266 /L 173.1900
     2015-Jun-02 00:00     04 24 36.24 +18 54 38.3   n.a.   n.a. 0.55096842790710   3.6190086   4.3897 /L 170.4268
     2015-Jun-03 00:00     04 22 33.81 +18 35 25.8   5.29   5.65 0.55361252502737   5.5319758   5.7564 /L 167.4122
     2015-Jun-04 00:00     04 20 38.57 +18 17 25.2   4.88   5.70 0.55735122910170   7.4086481   7.1443 /L 164.3125
     2015-Jun-05 00:00     04 18 52.42 +18 00 48.1   4.49   5.68 0.56216176249744   9.2429487   8.5180 /L 161.1913
     2015-Jun-06 00:00     04 17 17.09 +17 45 44.4   4.12   5.62 0.56801810956371  11.0297524   9.8584 /L 158.0785
     2015-Jun-07 00:00     04 15 54.09 +17 32 22.7   3.77   5.53 0.57489158958468  12.7649573  11.1530 /L 154.9902
     2015-Jun-08 00:00     04 14 44.76 +17 20 50.1   3.45   5.42 0.58275145605987  14.4455044  12.3929 /L 151.9356
     2015-Jun-09 00:00     04 13 50.19 +17 11 11.8   3.15   5.31 0.59156549247635  16.0693469  13.5713 /L 148.9204
     2015-Jun-10 00:00     04 13 11.32 +17 03 31.6   2.87   5.18 0.60130057734526  17.6353760  14.6833 /L 145.9474
     2015-Jun-11 00:00     04 12 48.85 +16 57 51.6   2.61   5.06 0.61192319592983  19.1433125  15.7250 /L 143.0178
     2015-Jun-12 00:00     04 12 43.37 +16 54 12.5   2.37   4.94 0.62339988172426  20.5935736  16.6938 /L 140.1313
     2015-Jun-13 00:00     04 12 55.28 +16 52 33.7   2.15   4.81 0.63569757625371  21.9871244  17.5876 /L 137.2863
     2015-Jun-14 00:00     04 13 24.86 +16 52 53.2   1.94   4.69 0.64878390050511  23.3253217  18.4055 /L 134.4805
     2015-Jun-15 00:00     04 14 12.31 +16 55 08.3   1.75   4.58 0.66262733523404  24.6097559  19.1469 /L 131.7103
     2015-Jun-16 00:00     04 15 17.69 +16 59 15.2   1.57   4.46 0.67719731085661  25.8420965  19.8117 /L 128.9719
     2015-Jun-17 00:00     04 16 41.05 +17 05 09.2   1.40   4.35 0.69246421084326  27.0239479  20.4002 /L 126.2604
     2015-Jun-18 00:00     04 18 22.35 +17 12 45.4   1.25   4.24 0.70839929524826  28.1567176  20.9133 /L 123.5707
     2015-Jun-19 00:00     04 20 21.53 +17 21 57.8   1.10   4.14 0.72497455264057  29.2415000  21.3516 /L 120.8972
     2015-Jun-20 00:00     04 22 38.49 +17 32 40.5   0.96   4.04 0.74216248879516  30.2789728  21.7164 /L 118.2337
     2015-Jun-21 00:00     04 25 13.15 +17 44 46.7   0.83   3.94 0.75993585914434  31.2693054  22.0086 /L 115.5738
     2015-Jun-22 00:00     04 28 05.39 +17 58 09.6   0.71   3.84 0.77826734979511  32.2120717  22.2296 /L 112.9109
     2015-Jun-23 00:00     04 31 15.13 +18 12 42.2   0.59   3.75 0.79712920966069  33.1061670  22.3805 /L 110.2380
     2015-Jun-24 00:00     04 34 42.27 +18 28 17.0   0.48   3.65 0.81649283449812  33.9497246  22.4625 /L 107.5478
     2015-Jun-25 00:00     04 38 26.75 +18 44 46.2   0.37   3.56 0.83632830264123  34.7400316  22.4769 /L 104.8329
     2015-Jun-26 00:00     04 42 28.49 +19 02 02.2   0.27   3.47 0.85660386201288  35.4734443  22.4247 /L 102.0858
     2015-Jun-27 00:00     04 46 47.47 +19 19 56.7   0.17   3.38 0.87728536859778  36.1453054  22.3070 /L  99.2988
     2015-Jun-28 00:00     04 51 23.65 +19 38 21.2   0.07   3.30 0.89833567803594  36.7498651  22.1247 /L  96.4642
     2015-Jun-29 00:00     04 56 17.02 +19 57 07.2  -0.02   3.21 0.91971399456822  37.2802130  21.8789 /L  93.5742
     2015-Jun-30 00:00     05 01 27.57 +20 16 05.6  -0.12   3.13 0.94137518555319  37.7282290  21.5704 /L  90.6210
     2015-Jul-01 00:00     05 06 55.30 +20 35 07.1  -0.20   3.05 0.96326907552925  38.0845640  21.2001 /L  87.5974
     2015-Jul-02 00:00     05 12 40.17 +20 54 01.8  -0.29   2.96 0.98533974152847  38.3386660  20.7688 /L  84.4961
     2015-Jul-03 00:00     05 18 42.16 +21 12 39.8  -0.38   2.88 1.00752484088954  38.4788698  20.2775 /L  81.3104
     2015-Jul-04 00:00     05 25 01.19 +21 30 50.5  -0.46   2.80 1.02975501346900  38.4925685  19.7269 /L  78.0344
     2015-Jul-05 00:00     05 31 37.13 +21 48 23.0  -0.54   2.73 1.05195341081967  38.3664848  19.1182 /L  74.6631
     2015-Jul-06 00:00     05 38 29.80 +22 05 05.9  -0.63   2.65 1.07403541451894  38.0870574  18.4526 /L  71.1926
     2015-Jul-07 00:00     05 45 38.91 +22 20 47.8  -0.71   2.57 1.09590861358459  37.6409536  17.7315 /L  67.6206
     2015-Jul-08 00:00     05 53 04.08 +22 35 16.7  -0.79   2.49 1.11747311584451  37.0157141  16.9564 /L  63.9467
     2015-Jul-09 00:00     06 00 44.78 +22 48 20.6  -0.87   2.41 1.13862226817234  36.2005225  16.1294 /L  60.1725
     2015-Jul-10 00:00     06 08 40.36 +22 59 47.7  -0.95   2.34 1.15924385212199  35.1870766  15.2529 /L  56.3023
     2015-Jul-11 00:00     06 16 49.99 +23 09 26.1  -1.03   2.26 1.17922180051076  33.9705111  14.3295 /L  52.3431
     2015-Jul-12 00:00     06 25 12.69 +23 17 04.6  -1.11   2.17 1.19843844390623  32.5502953  13.3626 /L  48.3050
     2015-Jul-13 00:00     06 33 47.27 +23 22 32.8  -1.19   2.09 1.21677724378686  30.9310023  12.3560 /L  44.2011
     2015-Jul-14 00:00     06 42 32.40 +23 25 41.0  -1.28   2.00 1.23412590569533  29.1228358  11.3140 /L  40.0476
     2015-Jul-15 00:00     06 51 26.58 +23 26 21.1  -1.37   1.91 1.25037969990154  27.1418048  10.2415 /L  35.8640
    $$EOE
**************************************************************************************************************
Column meaning:

TIME

  Prior to 1962, times are UT1. Dates thereafter are UTC. Any 'b' symbol in
the 1st-column denotes a B.C. date. First-column blank (" ") denotes an A.D.
date. Calendar dates prior to 1582-Oct-15 are in the Julian calendar system.
Later calendar dates are in the Gregorian system.

  Time tags refer to the same instant throughout the universe, regardless of
where the observer is located.

  The dynamical Coordinate Time scale is used internally. It is equivalent to
the current IAU definition of "TDB". Conversion between CT and the selected
non-uniform UT output scale has not been determined for UTC times after the
next July or January 1st.  The last known leap-second is used over any future
interval.

  NOTE: "n.a." in output means quantity "not available" at the print-time.

 R.A._(ICRF/J2000.0)_DEC =
   J2000.0 astrometric right ascension and declination of target center.
Adjusted for light-time. Units: HMS (HH MM SS.ff) and DMS (DD MM SS.f)

 APmag S-brt =
   Target's approximate apparent visual magnitude & surface brightness. For
planets and satellites, values are available only for solar phase angles in the
range generally visible from Earth. This is to avoid extrapolation of models
beyond their valid (data-based) limits.
   Units: MAGNITUDE & VISUAL MAGNITUDES PER SQUARE ARCSECOND

 delta  deldot =
   Range ("delta") and range-rate ("delta-dot") of target center with respect
to the observer at the instant light seen by the observer at print-time would
have left the target center (print-time minus down-leg light-time); the
distance traveled by a light ray emanating from the center of the target and
recorded by the observer at print-time. "deldot" is a projection of the
velocity vector along this ray, the light-time-corrected line-of-sight from the
coordinate center, and indicates relative motion. A positive "deldot" means the
target center is moving away from the observer (coordinate center). A negative
"deldot" means the target center is moving toward the observer.
Units: AU and KM/S

 S-O-T /r =
    Sun-Observer-Target angle; target's apparent solar elongation seen from
observer location at print-time. If negative, the target center is behind
the Sun. Angular units: DEGREES.

    The '/r' column is a Sun-relative code, output for observing sites
with defined rotation models only.

         /T indicates target trails Sun (evening sky)
         /L indicates target leads Sun  (morning sky)

    NOTE: The S-O-T solar elongation angle is the total separation in any
direction. It does not indicate the angle of Sun leading or trailing.

 S-T-O =
   "S-T-O" is the Sun->Target->Observer angle; the interior vertex angle at
target center formed by a vector to the apparent center of the Sun at
reflection time on the target and the apparent vector to the observer at
print-time. Slightly different from true PHASE ANGLE (requestable separately)
at the few arcsecond level in that it includes stellar aberration on the
down-leg from target to observer.  Units: DEGREES

 Computations by ...
     Solar System Dynamics Group, Horizons On-Line Ephemeris System
     4800 Oak Grove Drive, Jet Propulsion Laboratory
     Pasadena, CA  91109   USA
     Information: http://ssd.jpl.nasa.gov/
     Connect    : telnet://ssd.jpl.nasa.gov:6775  (via browser)
                  telnet ssd.jpl.nasa.gov 6775    (via command-line)
     Author     : Jon.Giorgini@jpl.nasa.gov

**************************************************************************************************************

Unsure where the original poster obtained an idea that objects inside the Earth’s orbit, as seen from Earth, “move ordinarily with respect to the stars” from west to east.

First of all, of which “west” and “east” do we speak? To be specific, let’s understand these for the region of sky not far from the meridian. That is, when the object X culminates, then nearby objects having the right ascension greater than RA(X) are on the east from the meridian (not yet culminated), whereas those having RA < RA(X) are on the west from the meridian (already culminated).

For the Sun itself, every educated person can remember that its diurnal motion is slightly slower than diurnal motion of the inertial celestial sphere (fixed stars) – that is, with respect to those stars Sun goes from west to east. Its right ascension increases monotonically. For other objects inside the Earth’s orbit, their right ascension is within 90° of the Sun’s, hence in average they move “from west to east” as well, and quicker do it when are behind the Sun.

But definitely the terrestrial picture is opposite for e.g. Venus when it overtakes Earth in their orbital motions near the inferior conjunction.

During this part of the orbit the right ascension of Venus decreases – seen on this animation near the end of cycle. “Experts” asserting that it’s impossible to see mistake – Venus is not very close to Sun during it, and anyway perfectly observable (albeit rare) during Venus’s transits before the Sun.


Are the planets Mercury, Saturn, Uranus and Neptune losing mass? My understanding is that Venus, Earth, Mars, and Jupiter are all losing mass due to their gravity, temperature, and the solar wind.  But what about Mercury, Saturn, Uranus, and Neptune?  Mercury has a magnetic field and so typically would repeal the worst of the solar wind and no atmosphere to lose, only an exosphere.  I do not know the calculations to determine if Saturn, Uranus, and Neptune are losing mass though.

Anything that's working, spending energy is losing mass,  Einstein Mass-energy equivalence of SR along with radiant energy(electromagnetic and gravitational radiation) would be effecting all the planets, but at a different rate for each one.  https://en.m.wikipedia.org/wiki/Radiant_energy
  And.   https://en.m.wikipedia.org/wiki/Mass_in_special_relativity
  These should help I think


Astronomy as a profession for moderately good software engineer I'm a software engineer (a moderately good one). I like programming, mathematics and I love everything about astronomy.

What would be best way to go ahead?

A profession in astronomy can entail various things.  Unlike what other people are saying, you don't have to have a PhD in astronomy to do work in this field, to be an astronomer.  There are other options, but you will have to work hard.

Study as much physics and mathematics as possible.  Astronomy is essentially a branch of physics.  If you're still in school, then take classes in these areas.  If you're not in school anymore, then there are plenty of online resources to continue your education.  This will allow you to understand what people are talking about when they discuss technical things.
Decide what exactly you're interested in.  There are lots of different areas of study within astronomy.  Decide what appeals to you the most.  Do you like planets?  Galaxies?  Cosmology?  All of these different focuses require more specialized knowledge.
Your skills are already marketable.  Being an astronomer can mean lots of different things.  There are some astronomers who hardly ever look through telescopes, and spend most of their time reducing data.  A large part of astronomy is computer programming.  As a software engineer, you may already have the skills you need to work with astronomy research.
Keep an open mind.  There are lots of jobs in astronomy research where you might not be called an "astronomer".  For example, as a software engineer, you could find work with NASA supporting space research.  Your job title might not be "Astronomer", but you'll still be in the field.
Be prepared for hard work.  Being an astronomer can be difficult, not just because of the subject matter, but also because of the effort you have to put in to get a job.  It can be relatively difficult to find an employer looking for people with your particular skillset.  My recommendation would be to keep an eye on the NASA Jobs website to see when they're hiring.  You can also search for internships on NASA's OSSI site.
Try some amateur astronomy. There are lots of ways you can get involved in astronomy without having it as a job.  A quick Google search for "Amateur astronomy projects" yields many different results.
Join a local astronomy group. If you live in a reasonably sized city, there should be an astronomy group near you.  Find them and join them.
Don't give up! Having astronomy as a profession can be difficult, but it is also rewarding.  If you truly want to do this, you'll need to work at it.  Just don't give up!

Considering your conditions and what you've already achieved. You could most likely start by doing a masters degree and then doing a PhD in astronomy if you want to do research.
The reason why I say to start out on a masters is because I assume you have a firm understanding of physics considering what you've done as I said. If you don't have enough background in physics, you could probably do a couple bridging courses and then do your masters.
That's a simple way to put it.
Regards, Ethan


Assume that a star has a uniform mass density. Then how does the temperature scale as a function of the distance to the center? Assume that a star has a uniform mass density. Then how does the temperature
scale as a function of the distance to the center?
Note: I am not looking for an answer that involves derivations or integrations.
Here is what I did:
The star is in equilibrium so the pressure at distance $r$ must equal the gravity force at distance $r$. Pressure is given by $P=nKT$, and so $P\times Area=F_g$.
Then $nKT \times 4\pi r^2=F_G$. The mass of the part of the star contained within radius $r$ from the center is $\frac{4}{3}\pi r^3\times m_sn$ where $m_s$ is the mass of each particle and $n$ is the number of particles per volume.
Then we can rewrite $nKT \times 4\pi r^2=m(\frac{G\frac{4}{3}\pi r^3\times m_snF}{r^2})$ and simplify to $KTr=\frac{m}{3}Gm_s$. Isolating $T$ we get that $T=\frac{mGm_s}{3Kr}$.
This seems to imply that the temperature decreases the further we are from the core, however I am not sure what $m$ should stand for in $F_g=mg$, should $m=m_s$, i.e be the mass of a single particle? Or does $m$ itself depend on the radius in which case the answer I got is invalid.
Can someone verify that the answer I got is right?

I am not sure what you mean by looking for an answer without derivations? Of course you have to solve the maths to get a mathematical answer.

The handwaving answer is that the star has to be hotter in the middle, because at constant density the pressure depends on temperature and you need a negative pressure gradient to support the weight of the gas above.

The answer is:

The equation of hydrostatic equilibrium is
$$\frac{dP}{dr} = -\rho g,$$
where $\rho$ is the mass density and $g$ is the gravity at radius $r$, which by Newton's shell theorem is due to the total mass interior to $r$.

This equation is easily derived by considering the force equilibrium on a slab of gas in the star (your error above was in not recognising that it is the difference in pressure across the slab that provides the upward force).

If the density (and composition and ionisation state) are constant, then this equation can be written as
$$ \frac{\rho k_B }{m} \frac{dT}{dr} = -\rho \frac{4\pi G r^3 \rho}{3r^2},$$
where $m$ is the mass per particle and a perfect gas has been assumed.

Integrating with respect to radius and assuming $T(r=0)=T_0$, we obtain
$$T = T_0 - \frac{2\pi G\rho m}{3 k_B}r^2$$

Finally, if the star has radius $R$, then this will be radius at $T=0$, since nothing could be supported beyond that. In which case
$$T = \frac{2\pi G\rho m}{3 k_B}(R^2-r^2)$$

Or in terms of mass $M$
$$T = \frac{GM m}{2 k_B R^3}(R^2-r^2)$$

Constant density is of course a poor approximation. The above treatment gives a central temperature of 7 million Kelvin for the Sun, assuming $m\simeq 10^{-27}$ kg.

Based on info from CalTech,  a constant density is unrealistic.  I wouldn't place much faith in any equations derived from such an assumption.


Why aren't there galaxy-sized balls of iron out there? Since we know that stars burn out, collapse, or blow up when too many of their atoms fuse into iron, then how come, with all the ancient galaxies we've been observing, there isn't one galaxy whose stars have become giant balls of iron, which would eventually collide into one galaxy-sized iron globule in space?

In short, you can't make a galaxy-mass ball of iron because there's no way to support the ball against the inward crush of its own gravity. First, it would collapse into a neutron star, but even neutron stars can't hold themselves up beyond about 2-3 solar masses.  Your galaxy-sized object would collapse into a black hole.  But we do see these!  All massive galaxies are believed to have supermassive black holes at their centres. These are up to 10 billion solar masses in size, which is comparable to smaller galaxies.

In any star, be it a main-sequence star like the Sun, a giant, or even a white dwarf or neutron star, the inward force of gravity must be balanced by some outward force or pressure.  (In fact, we basically assume this when we construct stellar models, although it's a very good assumption.) In main-sequence stars, this pressure is provided by the nuclear reactions happening inside them but as a star evolves, it gets a bit more complicated. In some situations, the nuclei get so close together that they their electrons begin to share quantum states. And because no two electrons can be in the same state, they have to occupy higher energy states. This in effect exerts a kind of pressure (called degeneracy pressure), because the electrons are forced to move around faster than they otherwise would. This is what supports white dwarfs and the cores of some evolved stars. e.g. low-mass red giants.

Degeneracy pressure is fine up to a limit known as the Chandrasekhar limit, which is about 1.44 solar masses (depending on the composition of the object in question). Anything above this would collapse. First, this collapse would yield a neutron star, where things are a bit different. But a similar principle holds, and there's a degeneracy pressure because the neutrons themselves also occupy one quantum state each.  The details are much more difficult here, but the overall consensus is that neutron degeneracy pressure can support objects up to about 2-3 solar masses. After that, there's nothing left to balance gravity, and the star collapses into a black hole.

Our sun will eventually become a white dwarf. A star 10 times it mass will become a neutron star. A star 100 times the mass of the sun will become a black hole. So if 100 stars came together into one mass, it would collapse into a black hole. It is currently believed that a super massive black hole (at least 1000 times the mass of our sun) is at the center of our galaxy and also in the Andromeda galaxy. This is where all the iron might be found from this region of space.


How many planets are there in this solar system? So, in school (that's a long time ago) they have been teaching us there are 9 planets in our solar system.

Mercury
Venus
Earth
Mars
Jupiter
Saturn
Uranus
Neptune
Pluto

But every now and then I keep reading stories about another "dwarf planet" (Eris, discovered in 2005) that - depending on what source tells the story - is another planet according to the astronomical definition, while other sources say that it isn't a planet. Some even say Pluto isn't a planet anymore either.
The result: I'm confused due to the contradicting stories. Even Wikipedia isn't clear about Eris and only writes (emphasis mine):

NASA initially described it as the Solar System's tenth planet.

Initially? So, is it a 10th planet or not? Fact is, there is another "something" out there and it surely seems to look like a planet. Yet, some people keep stating there are 9 planets in our solar system, while others say there are more than 9 planets, and then again there are people stating that the latest definition of "planet" has kicked out Pluto too so there are actually fewer than 9 planets in our solar system.
Trying to get a definite, official, and astronomically correct answer I can actually rely on, I'm therefore asking: How many planets are there in this solar system?
EDIT
The "Definition of planet" at Wikipedia doesn't really help either, as it states:

Many astronomers, claiming that the definition of planet was of little scientific importance, preferred to recognize Pluto's historical identity as a planet by "grandfathering" it into the planet list.*
* Dr. Bonnie Buratti (2005), "Topic — First Mission to Pluto and the Kuiper Belt; "From Darkness to Light: The Exploration of the Planet Pluto"", Jet Propulsion Laboratory. Retrieved 2007-02-22.

So, if you link somewhere to provide proof, it would be great if you could point me to a more trusted source than Wikipedia. Ideally, an astronomical trusted source and/or paper.

In addition to Undo's fine answer, I would like to explain a bit about the motivation behind the definition.

When Eris was discovered, it turned out to be really, really similar to Pluto. This posed a bit of a quandary: should Eris be accepted as a new planet? Should it not? If not, then why keep Pluto? Most importantly, this pushed to the foreground the question

  what, exactly, is a planet, anyway?

This had been ignored until then because everyone "knew" which bodies were planets and which ones were not. However, with the discovery of Eris, and the newly-realized potential of more such bodies turning up, this was no longer really an option, and some sort of hard definition had to be agreed upon.

The problem with coming up with a hard definition that decides what does make it to planethood and what doesn't is that nature very rarely presents us with clear, definite lines. Size, for example, is not a good discriminant, because solar system bodies come in a continuum of sizes from Jupiter down to meter-long asteroids. Where does one draw the line there? Any such size would be completely arbitrary.

There is, however, one characteristic that has a sharp distinction between some "planets" and some "non-planets", and it is the amount of other stuff in roughly the same orbit. This is still slightly arbitrary, because it's hard to put in numbers exactly what "roughly" means in this context, but it's more or less unambiguous.

Consider, then a quantity called the "planetary discriminant" µ, equal to the ratio of the planet's mass to the total mass of other bodies that cross its orbital radius and have non-resonant periods (so e.g. Neptune doesn't count as sharing Pluto's orbit) up to a factor of 10 longer or shorter (to rule out comets, which has little effect in practice). This is still a bit arbitrary (why 10?) but it's otherwise quite an objective quantity.

Now take this quantity and calculate it for the different bodies you might call planets, comparing it to both the objects' mass,

and their diameter,

or with an arbitrary horizontal axis, in order of decreasing discriminant,

Suddenly, a natural hard line emerges. If you look only at the mass and the diameter of the objects (shown in the insets above the plots), then there is a pretty continuous spread of values, with bigger gaps between the gas giants and the terrestrial planets than between Mercury and Eris/Pluto. However, if you look at the planetary discriminant, on the vertical axis, you get a very clear grouping into two distinct populations, separated by over four orders of magnitude. There's a finite set of bodies that have "cleared their orbits", and some other bodies which are well, well behind in that respect.

This is the main reason that "clearing its orbital zone" was chosen as a criterion for planethood. It relies on a distinction that is actually there in the solar system, and very little on arbitrary human decisions. It's important to note that this criterion need not have worked: this parameter might also have come out as a continuum, with some bodies having emptier orbits and some others having slightly fuller ones, and no natural place to draw the line, in which case the definition would have been different. As it happens, this is indeed a good discriminant.

For further reading, I recommend the Wikipedia article on 'Clearing the neighbourhood', as well as the original paper where this criterion was proposed,

  What is a planet? S Soter,  The Astronomical Journal 132 no.6 (2006), p. 2513. arXiv:astro-ph/0608359.

which is in general very readable (though there are some technical bits in the middle which are easy to spot and harmless to skip), and from which I took the discriminant data for the plots above.

Edit: I must apologize for having included, in previous versions of this post, an incorrect plot, caused by taking data from Wikipedia without verifying it. In particular, the planetary discriminant for Mars was wrong (1.8×105 instead of 5.1×103), which now puts it below Neptune's instead of just below Saturn's, but the overall conclusions are not affected. The Mathematica code for the graphics is available at Import["http://goo.gl/NaH6rM"]["http://i.stack.imgur.com/CQA4T.png"].

... and, as a final aside: Pluto is awesome. It was visited in July 2015 by the New Horizons probe, which found a world that was much more rich, dynamic, and active than anyone expected, including what appear to be churning lakes of solid nitrogen ringed by mountains of water ice, among other marvels.

(Note the image has been colour-enhanced to bring out the variety of surface materials; the true-colour version of this image is here.) I, personally, don't feel it's at all necessary to 'grandfather' Pluto into the list of planets to really feel the awe at the amazing place it is - it's perfectly OK for it to be a cool place with cool science, that is also not a planet.

Definitions don't change facts. The fact is that there are 10 known planets in our system. From closest to farthest from the Sun they are:

1.) Mercury

2.) Venus

3.) The Earth

4.) Mars

5.) Jupiter

6.) Saturn

7.) Uranus

8.) Neptune

9.) Pluto

10.) Eris

Note: Pluto's orbit overlaps with Neptune and Eris' orbit overlaps with Pluto, so Neptune is sometimes farther than Pluto and Pluto sometimes farther than Eris.

In 2016 an eleventh planet was postulated from the orbits of smaller Kuiper belt objects. However, since then alternative theories explaining the orbits have been proposed so that planet Eleven doesn't have to exist. In 2019 it was proposed that planet Eleven, if it exists, may actually be a primordial black hole, having a diameter of about 2 inches (5 cm). As such it could have own planets, but these belong to the black hole, not directly to the Sun.

If you want to know why the 2006 definition of planets and so-called dwarf planets, which was selected by ~4% of IAU astronomers, has no scientific hold, google something like "Alan Stern Pluto planet". Dr Stern explains very well how unscientific the definition is and more.


How much mass will the Sun have when it becomes a white dwarf? In 4 billion years, when our Sun sheds all of its outer gas layers and turns into a white dwarf, how much mass will the white dwarf have compared to what the sun has today?

Will the planets still orbit in the same way, or will the reduced mass cause the planets' trajectories to change, so that they eventually leave the solar system?

Short answer:

The Sun will lose about half of its mass on the way to becoming a white dwarf. Most of this mass loss will occur in the last few million years of its life, during the Asymptotic Giant Branch (AGB) phase. At the same time the orbital radius of the Earth around the Sun will grow by a factor of two (as will the outer planets). Unfortunately for the Earth, the radius of the Sun will also reach to about 2 au, so it will be toasted.

There is the possibility that the decreased binding energy and increased eccentricity of the Earth and the outer planets will lead to dynamical instabilities that could lead to planetary ejection. This is highly dependent on the exact time dependence of the late, heavy mass loss and the alignment or otherwise of the planets at the time.

Long answer:

Stars with mass less than about 8 solar masses will end their lives as white dwarfs on a timescale which increases as their main sequence initial mass decreases. The white dwarfs that are formed are of lower mass than their progenitor main sequence stars, because much of the initial mass of a star is lost through stellar winds (particularly during the thermally pulsating asymptotic giant branch phase) and final ejection of a planetary nebula.  Thus, the current distribution of white dwarf masses, that peaks between $0.6$ and $0.7 M_{\odot}$ and with a dispersion of $\sim 0.2 M_{\odot}$, reflects the final states of all main sequence stars with $0.9 <M/M_{\odot}<8 M_{\odot}$, that have had time to evolve and die during our Galaxy's lifetime.

The most reliable information we have about the relationship between the initial main sequence mass and final white dwarf mass (the initial-final mass relation or IFMR) comes from measuring the properties of white dwarfs in star clusters of known age. Spectroscopy leads to a mass estimate for the white dwarf. The initial mass is estimated by calculating a main sequence plus giant branch lifetime from the difference between the age of the star cluster and the cooling age of the white dwarf. Stellar models then tell us the relationship between main sequence plus giant lifetime and the initial main sequence mass, hence leading to an IFMR.

A recent compilation from Kalirai (2013) is shown below. This shows that a star like the Sun, born with an initial mass of $1M_{\odot}$ (or maybe a per cent or two more, since the Sun has already lost some mass), ends its life as a white dwarf with $M = 0.53 \pm 0.03\
M_{\odot}$. i.e. The Sun should lose approximately 50% of its initial mass in stellar winds and (possibly) planetary nebula ejection.

A comprehensive treatment of what happens to solar systems when the central star loses mass in a time-dependent way is given in Adams et al. (2013). The simplest cases are initially circular orbits where the mass loss takes place on much longer timescales than the orbital period. As mass loss proceeds, the gravitational potential energy increases (becomes less negative) and thus the total orbital energy increases and the orbit gets wider. Roughly speaking, $aM$ is a constant, where $a$ is the orbital radius, which is a simple consequence of conservation of angular momentum: so the Earth would end up in a 2 au orbit.

However, in the presence of a non-zero eccentricity in the initial orbit, or in the case of rapid mass loss, such as that which occurs towards the end of the AGB phase, then things become altogether more unpredictable, with the eccentricity also growing as mass loss proceeds. This has a knock-on effect when considering the dynamical stability of the whole (evolved) solar system and may result in planetary ejection. The faster the mass loss, the more unpredictable things get.

The radius of an AGB star can be calculated using $ L = 4\pi R^2 \sigma T_{eff}^{4}$. Stars at the tip of the AGB branch have luminosities of $\sim 10^{4} L_{\odot}$ and $T_{eff} \simeq 2500\ K$, leading to likely radii of $\sim 2$ au. So it is quite likely that unless the Earth is ejected or has its orbit significantly modified by some dynamical instability that, like the inner planets, it will end up engulfed in the outer enevelope of the AGB star and spiral inwards...

Even should it narrowly escape this immediate fate, it is then quite likely that tidal dissipation will rapidly extract energy out of the orbit and the Earth will spiral in towards the envelope of the giant Sun... with the same result.

Well, put simply, the Sun will certainly lose at least a fourth of its mass. This is because most of the mass of the Sun is centered in its core. And since a white dwarf is just the remnant core of a star . . .
Oh, and before the Sun becomes a white dwarf, it goes through the "red giant" phase, where it grows to about the size of the orbit of mars. All the planets will burn up, or stop orbiting, and they will cease to exist when the Sun's nova occurs. Happy ending . . .


Does the Milky Way's movement through space have spacetime implications? The way I usually see spacetime explained, it says that if you just sit still, you are moving at 0kph and so you are moving at maximum speed through time (ie. time for you is flowing at maximum speed).  However, sitting still we're moving at the speed of our galaxy through space, which is about 600km/s.  The speed of light is about 300,000km/s.  So for simplicity's sake, let's say the galaxy is actually moving through space at 1,000km/s.  Does this in fact mean that if I sit still on Earth, time for me is actually flowing 1/300th slower than if I were sitting still with respect to the universe?  And does it also mean that if the Milky Way were travelling at 99% the speed of light, time for me would be flowing at 1% the rate of the flow of time for someone that was sitting still with respect to the universe?

Your question has a misunderstanding at its root:  You say, "So for simplicity's sake, let's say the galaxy is actually moving through space at 1,000km/s". The problem here is that the whole point of Relativity is that the idea of movement with respect to space is meaningless -- the only thing that has meaning is movement with respect to other bodies.

Time dilation is also relative. If you're moving at 1000kps relative to Fred, Fred sees your clocks as running slower, and in his "reference frame" they are.  But you, in your reference fame, see Fred's clocks as running slower! (This is the essence of the famous Twin Paradox.)

You're being led astray by your intuitions about space and time. (And don't be ashamed about that. Newton, who was probably the greatest physicist ever, even greater than Einstein, and Kant (one of the great philosophers) and many, many others had the same intuitions. Yet experiment has taught us that they were wrong.)  Neither space nor time is absolute, and an understanding of Special Relativity (which is much easier to understand than it was to discover in the first place!) is required to ask as well as to answer your question.

As far as you are considering the proper motion of the galaxy towards the Great Attractor the speed is less than a percent of the speed of light c.
Although astonishing this speed doesn't require a relativistic correction.
Of course if we were to sit on a galaxy moving at, say, 0.9 × c, then we would be like the travelling twin of the famous "paradox" as compared to someone still nearby and watching us.
However for calculating the time dilation  you can't go by % proportion as in your question but apply the relativistic formula

$$\Delta t^\prime = \frac {\Delta t}{\sqrt{1-\frac{v^2}{c^2}}}$$
which expresses the fact that the moving observer's period of the clock $\Delta t^\prime$ is longer that the period $\Delta t$ in the frame of the clock itself.

that I copied from https://en.m.wikipedia.org/wiki/Time_dilation
It might be worth pointing out that the formula is valid for both observers and thus they will see each other's clock ticking slower.
Related but more general questions and discussion by following How does space time differ between galaxies?
in which the effect (lack of) of galaxies on spacetime is attacked from a general relativity point of view (how does galaxy's mass influence the spacetime curvature).
edit after receiving comments I am not mirroring anymore the wording "moving through space" as it leads to think, or that the authors of the Q and the A think, of absolute motion in absolute space. I also add that the conclusions for the observer on the galaxy and the one still coincides and both are right (unless the galaxy for some reason turns back ;)
Nevertheless
I would like to point out that moving through space is a perfectly sensible sentence in both common speaking and in physics. Through what something should move when its spatial coordinates are changing if not space? :)


Can dark matter decrease the Jeans length? I am wondering if there are any models of interstellar cloud collapse out there which take dark matter into account.

If dark matter has local perturbations or density fluctuations, then that may be a large enough effect to cause collapse of an interstellar cloud. This might increase the mass and density of a cloud enough to overcome the kinetic energy of hydrogen atoms leading to collapse.

Are there dark matter density variations at the scale of interstellar clouds?

This is an area of active research. The current Cold Dark Matter (CDM) paradigm predicts bottom-up formation of halos, whereby smaller halos coalesce in to the larger halos we observe indirectly today through X-ray measurements of clusters. This suggests that small dark matter halos wouldn't survive to the present day.

There has been research in to the possibility of mini haloes surviving, but interactions with dense baryonic structures, such as stars, as well as streaming by an irregular galactic potential tend to wipe out any small overdensities in the distribution.

However, if such mini haloes did exist in the early universe they could have contributed to the collapse of primordial gas to form Population III stars. Simulations have been carried out to investigate such a model, with Population III stellar masses comparable to those expected $(\sim100 M\odot)$.

Dark Matter is a kind of "fill in the blank" at the moment.  No one really has any definitive theory that has wide acceptance on what "dark matter" may be, never mind how it interacts with other matter.  There's still a little resistance to even saying it exists at all, although it's pretty much the mainstream now that it does.  But what it is, that's still up for grabs.

But I did find this from Google :

http://iopscience.iop.org/article/10.1086/306334/fulltext

It presumes a certain type of model for dark matter, but may be what you're looking for.


Why does gas form a star instead of a black hole? When a space gas gets pulled together a star is formed. On the other hand, when a massive star dies, it collapses to a black hole.

You would think that the initial mass of the gas would be bigger than the star that had existed for billion years and lost mass in the process.

So what stopped the space gas from forming a black hole in the first place?

Essentially gas does, it just happens to form a star first.

Mass is not the only factor in creating a black hole. You also need for this mass to reach a high density. In the process of doing this, a star usually forms. The energy producing processes in the star interior produce a pressure that balances the gravitation attraction. This prevents a star from reaching a critical density needed for black hole formation. As these energy producing processes run out of useable fuel the star will eventually collapse onto itself creating a black hole.

So you can't just take a large volume of gas and create a black hole. Other physical processes occur.

When that space gas gets pulled together to form a star, the original gas and now a star their mass is the same (unchanged), but the size reduced, due to gravity. When the star collapses to be a black hole, again their mass the same but size shrunk, due to gravity.
It is not a valid way to ask in this manner:

  So what stopped the space gas from forming a black hole in the first place?

The space gas does eventually shrink to become a black hole [if the initial mass is good enough for it to exist as a star then collapsed as a black hole], only the observer taking section by section of the whole process to observe [depends on which section is observed]. It's simple, when replace the object for analogy to this: heating the ice. Ice melts to become water, then gas. Now the OP is like asking: why ice doesn't turn to gas when heated but water? It sounds logical but in fact it is the construct of the words mystified a simple analogy.


Why does Saturn have both moons and rings? From my understanding, a ring can form around a planet when a moon gets too close to its Roche limit, and gets ripped appart by the planet's gravity pull. That makes sense to me, but I don't understand why Saturn has both moons and rings at the same place.

I know the F ring is supposedly created by Enceladus' rejections, but I don't get why Pandora and Prometheus are not ripped appart as well as the ancient objects that formed the ring they are in.

I couldn't find specific explanations for this. Does it have any link to the moon's density?

You're right that density is the important thing here. The Roche limit is the distance from the main body $d$ such that
$$d=1.26R_M\left(\frac{\rho_M}{\rho_m}\right)^{\frac{1}{3}}$$
where $_M$ denotes the main body and $_m$ denotes the satellite.

As you can see from the chart on the Wikipedia page, Pandora and Prometheus are both at least one and a half times the Roche limit from Saturn. Therefore, they're in no danger of being ripped apart any time soon.

This answer is just misleading.  The rings are not formed by the moons.  They are clearly formed by a force.  And I don't think its gravity.  Gravity forms spheres or spherical shells if as you suggest its the roche limit.

FYI saturns rings are about 30 feet thick and extend for millions of miles.  Gravity does not confine things to thin discs.  There is an magnetic force at work here.  And it has been demonstrated by plasma physicists.