natural satellites - Why does Jupiter have so many moons?

Bigger is better.

Most moons, especially those of gas giants, are not "formed", they are just "captured" (unlike our Moon, which could have been captured, but probably was formed in a much more exciting way).

Jupiter is the most massive planet in the solar system. It stands to reason that it has a larger region of gravitational influence (where its influence outweighs the force due to the other planets and the sun). So, it's easy for it to capture rocky masses.

If you have a look at the contours on the following image (Ignore the Lagrange points marked on it, I only want the contours)

the circular area around the Earth is more or less the area (there's a velocity dependence here which I'm not getting into) in which a moon-like body can form a reasonably stable orbit. The size of the small "well" will increase as the planet moves farther from the sun, and also when the planet is more massive.

Jupiter is both pretty far away from the Sun, and is very massive. This leads to a huge sphere of influence.

The asteroid belt may have something to do with this too, but I doubt it (it's pretty far away). However, if we assume the "half-baked planet formation" theory for the formation of the belt, Jupiter may have leeched off much of the mass that would have otherwise become part of that planet during the formative period.

orbit - Are there ever any simultaneous transits of both Mercury and Venus as seen from the Earth?

EDIT: As it turns out, I'm not the first or even the second person to run calculations like this:

Meeus' work (second link) mentions the 13425 CE event in "Table 1. Simultaneous and near-simultaneous transits of Mercury and Venus, years 1 to 300,000"

Within the limits of DE431 (7 May 13201 BCE to 7 May 17091 CE), there is no
time at which both Mercury and Venus transit the Sun.

The closest we get to this:

• On 16 Sep 13425 CE at 11:57pm UTC, Venus starts transiting the Sun. This
  transit ends the next morning (17 Sep 13425 CE) at 7:30am.




• Less than 9 hours later, at 4:27pm, Mercury starts transiting the
  Sun. This transit ends at 10:26pm.

The program I used to compute this:

https://github.com/barrycarter/bcapps/blob/master/ASTRO/bc-solve-astro-13227.c

The list of transits I computed while solving this:

https://github.com/barrycarter/bcapps/blob/master/ASTRO/mercury-transits.txt.bz2
https://github.com/barrycarter/bcapps/blob/master/ASTRO/venus-transits.txt.bz2

Although I believe this answer is correct, Stellarium does not agree with
me, and HORIZONS doesn't compute positions past 9999 CE, so don't put too
much faith in this answer, since there's no good way to confirm it. I
believe that I'm correct and Stellarium is wrong this far in the future, but
it could be the other way around.

Even if my calculations are correct, the uncertainty in calculating the
relevant positions (Sun, Merucry, Venus, Earth) this far in the future is
high. On their own transit pages, NASA only computes Venus transits from
2000 BCE to 4000 CE, and Mercury transits from 1601 CE to 2300 CE, even
though they could've made the same calculations I made from 13201 BCE to
17091 CE:

This suggests NASA isn't confident enough of Mercury/Venus (and Earth/Sun)
positions to predict that far in the past or future, so my results may be fairly inaccurate.

observational astronomy - Why is there a gap in this image of supernova discoveries?

The coordinate system in this image is RA and Dec. It is a coordinate system which uses the Earth's equator (projected onto the sky) as its midline.

The inverted U is the Milky Way. The Milky Way is full of dust and gas, and blocks our view of galaxies (and supernovae) behind it. There is enough dust in the plane of the galaxy to block our view in that direction. For example the galaxy IC 342 is one of the nearest galaxies, and would be brilliant if it were not close to the galactic plane. There may be other galaxies that are completely hidden.

Our galaxy's bulk not only hides supernovæ that are in other galaxies, it also hides most of the supernovæ that occur in the Milky Way

Missing Terms in Weinberg's treatment of perturbations on Newtonian Cosmology

I was reading Appendix F of Steven Weingberg's book "Cosmology". In this Appendix he works out the perturbations to a cosmological fluid described by non-relativistic hydrodynamics and Newtonian gravity.

It turns out that the first order perturbations satisfy,

$$
frac{partial delta rho }{partial t } + 3 H delta rho + H vec{X} cdot nabla delta rho + bar{rho} nabla cdot vec{v} = 0, qquad tag{1}
$$

$$
frac{partial delta vec{v}}{partial t } + H vec{X} cdot nabla delta vec{v} + H delta vec{v} = - nabla delta phi, qquad tag{2}
$$

$$
nabla^2 delta phi = 4pi G delta rho. qquad tag{3}
$$

Weinberg applies the following Fourier transform to these equations,

$$ f(vec{X},t) = int exp left( frac{i vec{q} cdot vec{X}}{a} right) f_{vec{q}}(t) mathrm{d}^3vec{q} $$,

where $f(vec{X},t)$ is a place holder for $delta vec{v}, delta rho, $ and $delta phi$.

The resulting equations he gets are,

$$
frac{mathrm d delta rho_{vec{q}}}{mathrm d t } + 3 H delta rho_{vec{q}} + frac{ibar{rho}}{a} vec{q} cdot delta vec{v}_{vec{q}} = 0 qquad tag{1'}$$

$$
frac{mathrm d delta vec{v}_{vec{q}}}{mathrm d t } + H delta vec{v}_{vec{q}} = -frac{i}{a} vec{q} delta phi_{vec{q}} qquad tag{2'}$$

$$
vec{q}^2 delta phi_{vec{q}} = -4pi G a^2 delta rho_{vec{q}} qquad tag{3'}$$.

For the most part these new equations can be obtained by making the substitution $nabla rightarrow i vec{q}/a$.

---

My question : There doesn't seem to be any terms in the transformed equations which correspond to the terms $ H vec{X} cdot nabla delta rho$ and $H vec{X} cdot nabla delta vec{v}$. Weinberg makes no comment about their absence. Is anyone aware of a legitimate mathematical reason for these terms to disappear in the transformed equations?

Can life survive on the equator of cooled and fast rotating white dwarf or neutron star?

I am going to attempt a weak answer, mods feel free to delete it, but I'm fairly certain I'm right.

Shortly - no. It's not possible. Even if you balance gravity and centrifugal force perfectly at ground level at the equator, they will very, very quickly become imbalanced as soon as you move north, south, or up from there. So quickly in fact that the gradients may be too big even for a human being not moving at all. Maybe if you're laying down, with your body oriented along the equator, but even then I think the gradients would be too big.

Maybe bacteria would survive, briefly.

I'm sure the math could be done quite easily to estimate the gradients. This is based entirely on intuition.

black hole - Interstellar movie: What is the "portal" to the other galaxy?

Yes, it is a wormhole indeed. This has been indicated quite clearly in the movie as well, when Dr. Romily explains with a pen and paper to Dr. Cooper. The explanation goes like this :

Imagine a sheet of paper to be 2-D space, then a line joining two points on the sheet of paper is the shortest distance possible to reach that point, but if due to some disturbance, the space is bent ( achieved by folding the sheet of paper), you can pierce a hole in the sheet after aligning the two points together. That is a 2-D wormhole (which is a circle indeed). So what happens when you consider a 3-D wormhole? It becomes a sphere.

Now, you can see the other end of a 2-D wormhole (which is effectively the other hole in the sheet visible from the first hole, and one can see beyond the hole in the other direction as well). Same happens with the 3-D spherical wormhole where you can see to the other side of the wormhole as well.

Now, the means to bend spacetime : Space time can be bent by having a very big mass placed inside the space time (read Einstien's General Relativity ). So, it is safe to assume that the wormhole is a space time disturbance created due to a massive object, but it is not a blackhole.

gravitational waves - What will eLISA be trying to observe?

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

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

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

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