Diameter of Astronomical Object Using Magnification I took a picture of Jupiter through my telescope, using an iPhone. I know the magnification of the telescope lens I used, and can Google to get the magnification of the iPhone lens (the magnification without zooming). Using a standard Photo Editor, I also know the pixels across of Jupiter in my image.

Can I, using all of this information above, calculate how many kilometres one pixel represents, and consequently the diameter of Jupiter?

You could use your image to estimate the angular size of Jupiter that night. To find the actual diameter of Jupiter you would also need to know the distance that Jupiter is from Earth, and since Jupiter and Earth are both orbiting the sun, this distance changes.

You can find the distance with software (Eg Stellarium) or using Nasa's Horizons web interface (You will need to use the column labeled "delta" which gives distance in astronomical units). To find the diameter you will need to do some trigonometry. (Diameter = distance $\times$ angular size in radians)

Alternatively you could use the published Diameter of Jupiter (70000km) and use your observations to calculate the distance.

To calculate the distance to Jupiter without knowing it's diameter requires more than a single observation. You will need to have:

A long enough series of observations to establish the orbital period and elements of Jupiter.
Knowledge of Kepler's laws of planetary motion
Knowledge of the distance of Earth from the sun, and to get this you can fund a expedition to the South Pacific during a transit of Venus.
A sharp pencil to do all the maths.

I'm being a little glib... The point is that it took more than 150 years of research between the broad acceptance of heliocentric solar system, and our knowledge of its scale, or the actual sizes of planets.


Information Paradox with Hawking's Radiation A theory about Blackholes storing the information from the universe.

Being said that Blackholes are condensed mass sufficient enough to break every particle into different elementary particles, broken particles will eventually reach the center(or they break down the time they reach there).

Some say that the information is stored on the surface instead, which can be carried by the Hawking's Radiation.

Questions:

How exactly can Hawking's Radiation carry the information?
How is it possible for information to exist at the event horizon?

(If the mass was swallowed by the blackholes then every single particle should be inside the event horizon as they brought towards the black hole at very high velocity and it's becoming difficult to imagine that some particles were left at the horizon itself).

Maybe: If the information is on the event Horizon then it might not have broken to elementary particles!

How exactly can Hawking's Radiation carry the information?

It does not, if you mean information about the interior of the black hole.

The relevant theorem is the No Hair Theorem.  Basically you can only get gross statistical properties of black holes, nothing else and no information about the interior structure (meaning any info inside stored in any way, stays there or is lost as randomized radiation).

If you accept the No Hair Theorem then Hawking radiation would be essentially random in character, giving you no more information about the black hole than the No Hair Theorem allows you to know.  This would mean "information" being lost as a black hole evaporates.

  How is it possible for information to exist at the event horizon?

I don't think it's possible for anything to exist at the event horizon for any time.  It's a surface of zero thickness (as I understand it) and no actual substance.  Nothing can stop there and no real thing could be said to be "on" the horizon, just passing through it.

One of the complication of discussing black holes in relation to things like quantum-scale elementary particles is that while we can't say exactly where a particle is (the uncertainty principle) "classical" general relativity (i.e. without quantum theory) doesn't incorporate this at all and so the idea of a particle having a precise position is perfectly reasonable in classical general relativity.  But that's unlikely to be the case in whatever theory we end up with that links general relativity and quantum field theories.

I suspect you are hearing things related to the Holographic principle.  This is pretty much beyond my real understanding in any depth, but note that it is intimately linked with trying to connect String Theory, gravity (and hence a kind of quantum gravity), Thermodynamics and a notion of "information" within those multiple contexts.  I personally don't intend to try and read too much into this as it's beyond cutting edge theory for me (beyond my "event horizon" I suppose :-) ).  However I'd suggest being wary of reading too much into this idea given the present state of the theories involved.


Do interstellar asteroids decelerate and eventually stop? I do wonder, do interstellar asteroids eventually stop at one point in space after they gradually decelerate (or) even do they decelerate?

Though there is no air like on earth and thus asteroids will not be affected by frictional forces, do they have friction with gravitational forces against their trajectories?

If it is so, they might eventually stop. Are there any known objects like that?

An object moving through the interstellar medium will experience a weak drag force. If the drag force is $f(v)$ the velocity will decline as $dv/dt=-f(v)/M$ where $M$ is the mass. In fluids $f(v)\propto v^2$; this appears to work even for low density gas. Note that this differential equation leads to a velocity that declines forever, slowly approaching zero but never reaching it.

We can make a simple ballpark estimate of the timescale of slowdown by estimating how long it takes for it to run into its own mass of interstellar material. If it has radius $r$ it will encounter $\pi r^2 v \rho$ kg of ISM per second. So for mass $M$ the timescale of slowdown is $\tau \sim M/\pi r^2 v \rho$.

If we take a $r=1$ km spherical asteroid of mass $M= 8.8802\cdot 10^{12}$ kg (assuming density 2.12 g/cm$^3$) moving at 26.33 km/s and use $\rho=10^{-20}$ kg/m$^3$ (this varies a lot) then the timescale is about $10^{22}$ seconds, or 340 trillion years. So after maybe a quadrillion years the asteroid would be nearly at rest relative to the local gas if nothing disturbed its trajectory...

However, it is fairly likely for the asteroid to encounter a star during this time. Stellar density is about $\rho_*=0.14$ per cubic parsec, and the timescale for getting within 100 AU from a star is $\sim 1/\pi (100 AU)^2 v \rho_* \approx 10^{11}$years. This is likely to act as a gravity assist giving it some of the star's relative velocity (on the order of kilometers per second). So the true answer is that the asteroid will never settle down as long as there are stars in the galaxy.


Does the light from a torch reach to the moon? If a torch is switched on and pointed towards the moon, would the light from the torch reach to the moon?

Visible light from a torch has a 50% (blue light) to 90% (red light) chance of making it through the Earth's atmosphere, depending on the amount of dust and aerosols in the atmosphere and how close the Moon is to the horizon.

Light travels in straight lines. The Moon has an angular diameter of only 0.5 degrees as seen from the Earth. Most ordinary torch beams diverge by much more than this, so only a small fraction of the photons from a torch would pass through a circle of angular diameter 0.5 degrees. Hence only a small fraction of the emitted photons would hit the Moon.

Finally, you said that the torch was pointed "towards the Moon". Actually, you pointed it to where the Moon was 1.5 seconds ago, and the torch light reaches the distance of the Moon a further 1.5 seconds later. Thus the Moon has travelled 3 seconds in its orbit. However, since the orbital period (360 degrees) is a month, then it only covers an angle of 1.5 arcseconds in 3 seconds, so this shift is not relevant for your experiment (though it is for those bouncing lasers off reflectors on the Moon, which proves that light does in fact reach the Moon).


What object in the universe is most opaque to neutrinos? I had this thought, and my first guess was "high density = lots of absorption, so I guess it's neutron stars" but this physics.se question about that has a great answer which covers why that's incorrect.

So what object will absorb the highest fraction of neutrinos passing through it, or at least be a good candidate? Feel free to assume a certain energy range of neutrinos. Exclude black holes because they just absorb everything and that's not as interesting.

... So what object will absorb the highest fraction of neutrinos passing through it, or at least be a good candidate? Feel free to assume a certain energy range of neutrinos. Exclude black holes because they just absorb everything and that's not as interesting.

Neutrinos have amongst the smallest mass and travel at nearly the speed of light, this property, along with their weak interaction allows them to travel through all but the densest objects.

You've asked for an answer that excludes trapping by gravity, ridiculously long objects should also be excluded. That leaves reasonable sized (existing) objects of extreme density.

If the progenitor star is in the mass range 8–25 M$_\odot$ the core slowly contracts to a neutron star, and it is therefore called proto-neutron star (PNS). Once it contracts and becomes particularly dense it becomes opaque to neutrinos. It also need to release a huge amount of angular momentum, through neutrino emission, to balance the Equations of State (EoS). In the first tenths of seconds after core-bounce, the PNS is turbulent and wobbly, but during the next tens of seconds, it undergoes a more quiet, “quasi-stationary” evolution (the Kelvin-Helmholtz phase), which can be described as a sequence of equilibrium configurations.

This phase is characterized by an initial increase of the PNS temperature as the neutrino degeneracy energy is transferred to the matter and the PNS envelope rapidly contracts, and then by a general deleptonization and cooling. After tens of seconds, the temperature becomes lower and the neutrinos mean free path is greater than the stellar radius. The PNS becomes transparent to neutrinos, and a “mature” neutron star is born.

The creation of a proto neutron star is explained in "Neutrino Emission from Supernovae" (Feb 28 2017), by H.-Th. Janka. It has this simple illustration on page 4:

  Fig. 2 Evolution of a massive star from the onset of iron-core collapse to a neutron star. The progenitor has developed a typical onion-shell structure with layers of increasingly heavier elements surrounding the iron core at the center (upper left corner). Like a white dwarf star, this iron core (enlarged on the lower left side) is stabilized mostly by the fermion pressure of nearly degenerate electrons. It becomes gravitationally unstable when the rising temperatures begin to allow for partial photo-disintegration of iron-group nuclei to $\alpha$-particles and nucleons. The contraction accelerates to a dynamical collapse by electron captures on bound and free protons, releasing electron neutrinos ($v_e$), which initially escape freely. Only fractions of a second later, the catastrophic in-fall is stopped because nuclear-matter density is reached and a proto-neutron star begins to form. This gives rise to a strong shock wave which travels outward and disrupts the star in a supernova explosion (lower right). The nascent neutron star is initially very extended (enlarged in the upper right corner), and contracts to a more compact object while accreting more matter (visualized by the mass-accretion rate $\dot{M}$) within the first second of its evolution. This phase as well as the subsequent cooling and neutronization of the compact remnant are driven by the emission of neutrinos and antineutrinos of all flavors (indicated by the symbol $v$), which diffuse out from the dense and hot super-nuclear core over tens of seconds. (Figure adapted from Burrows, 1990b)

Notice that the image in the lower right blocks $v$ (neutrinos).

Text on page 2:

  "... [Lots of interesting information] ... [shortest possible quote] ... With neutral-current scatterings of neutrinos off nuclei and free nucleons being possible, it was recognized that the electron neutrinos, $ν_e$, produced by electron captures can escape freely only at the beginning of stellar core collapse (which starts out at a density around 10$^{10}$ g cm$^{−3}$), but get trapped to be carried inward with the in-falling stellar plasma when the density exceeds a few times 10$^{11}$ g cm$^{−3}$. At this time the implosion has accelerated so much that the remaining collapse time scale becomes shorter than the outward diffusion time scale of the neutrinos, which increases when scatterings become more and more frequent with growing density. Shortly afterwards, typically
  around 10$^{12}$ g cm$^{−3}$, the electron neutrinos equilibrate with the stellar plasma and fill up their phase space to form a degenerate Fermi gas. During the remaining collapse until nuclear saturation density (about 2.7×10$^{14}$ g cm$^{−3}$) is reached, and the
  incompressibility of the nucleonic matter due to the repulsive part of the nuclear force enables the formation of a neutron star, the entropy and the lepton number (electrons plus electron neutrinos) of the infalling gas (stellar plasma plus trapped neutrinos) remain essentially constant. Since the change of the entropy by electron
  captures and $ν_e$ escape until trapping is modest, it became clear that the collapse of a stellar core proceeds nearly adiabatically (for a review, see Bethe, 1990). The proto-neutron star, i.e., the hot, mass-accreting, still proton- and lepton-rich predecessor object of the final neutron star, with its super-nuclear densities and extreme temperatures of up to several 10$^{11}$ K (corresponding to several 10 MeV) is highly opaque to all kinds of (active) neutrinos and antineutrinos. Neutrinos, once generated in this extreme environment, are frequently re-absorbed, re-emitted, and scattered before they can reach semi-transparent layers near the “surface” of the proto-neutron star, which is marked by an essentially exponential decline of the density over several orders of magnitude. Before they finally decouple from the stellar medium closely above this region and escape, neutrinos have experienced billions of interactions on average. The period of time over which the nascent neutron star is able to release neutrinos with high luminosities until its gravitational binding energy (Eq. 1) is radiated away therefore lasts many seconds (Burrows and Lattimer, 1986; Burrows, 1990a).".

In the study "Observing Supernova Neutrino Light Curves with Super-Kamiokande: Expected Event Number over 10 s" (Aug 22 2019) by Yudai Suwa, Kohsuke Sumiyoshi, Ken'ichirō Nakazato, Yasufumi Takahira, Yusuke Koshio, Masamitsu Mori, and Roger A. Wendell they investigated neutrino properties observable by Super-Kamiokande up until 20 s after the bounce using the database of Nakazato et al. (2013). It includes this text and accompanying graphic:

Page 4:

  "While the neutrino-radiation hydrodynamics (RHD) simulations account for the neutrino emission before shock revival, the neutrino light curves from the PNS cooling simulations are reasonable for times after the shock revival. On the basis of these considerations, the neutrino light curves of the early and late phases are interpolated by an exponential function assuming shock revival at either $t_{revive}$ = 100, 200, or 300 ms after bounce. In Figure 1, a typical neutrino light curve obtained by this procedure is displayed."

Page 6:

  Figure 1. Neutrino luminosities (top panels) and average energies (bottom panels) as a function of time after bounce for the
  13M⊙, Z = 0.02, trevive = 300 ms model.


How much has the apparent magnitude of the Sun changed? Did the apparent magnitude of the Sun change in the past 1000 years or so? If so, for what reasons?

No. The apparent magnitude of the sun is constant to within 0.2%  How long it has been that way is only speculation since measurements aren't available. 1000 years is a very short time period in the life cycle of a star. Its safe to say yes its been constant for 1,000 years. There is a small variation based on the 11 year sun spot cycle but this variation is reported at less than 0.1%. There is also a larger variation due to the difference in Earth-to-Sun distance caused by the earths elliptical orbit. This effect is about 7%. This is no due to the sun but rather due to the eccentricity of Earths orbit. There are also minor effects with a cycle time or much greater than 1,000 years. These also due to variations in Earths orbit.


What is convection overturn? When I read some papers about MHD dynamo theory, there is always a $\tau$ that means convection overturning time. What does convection overturn mean?

The convective overturn time is the typical timescale for a convective cell to rise in a gas. Imagine a "lava lamp" - it's the time for one of the blobs to rise from its lowest to highest point.

I am most familiar with its use in stars, where convective energy transport is modelled using a mixing length. This posits that the typical height travelled by an adiabatic convection cell is some multiple $\alpha$ of the pressure scale height $H_p$ (the typical length on which the internal pressure changes significantly).

The convective overturn (or turnover) time is then $\alpha H_p/v$, where $v$ is the velocity at which convective cells rise.


Does rotation affect gravitational lines of force I was reading about rotating pulsars and other more-massive objects spinning at absurd speeds (hundreds of revolutions per second), and there was one point of which I was not clear.

Does a rotation of a mass change the gravitational lines of force? In other words, does a body which is rotating change the direction of gravitational pull from this:

to this?

An spherical electric charge has the same electric field lines whether spinning or not. The difference between those two cases is entirely in the magnetic field. Thus, one should expect as similar thing to happen for gravity.

The parametrized post-Newtonian formalism, weak-field GTR has the metric
$$\mathrm{d}s^2 = -(1+2\Phi)\,\mathrm{d}t^2 + 2\mathcal{A}_j\,\mathrm{d}t\,\mathrm{d}x^j + (1-2\Phi)\delta_{ij}\,\mathrm{d}x^i\,\mathrm{d}x^j$$
where $\Phi\equiv -U$ is essentially the Newtonian gravitational potential, while $\mathcal{A}_j\equiv-\tfrac{7}{4}V_j - \tfrac{1}{4}W_j$ in terms of the other PPN potentials. For the four-velocity $U^\alpha\equiv{\mathrm{d}x^\alpha}/{\mathrm{d}\tau} = (U^0,\vec{U})$, the geodesic equation for time-independent $\Phi$ and $\mathcal{A}$ becomes, to linear order in $\Phi$,
$$\frac{\mathrm{d}\vec{U}}{\mathrm{d}\tau} = U^0(\vec{G} + \vec{U}\times\vec{H})\text{,}$$
where the gravitoelectric field is $\vec{G} = -\nabla\Phi$ and the gravitomagnetic field is $\vec{H} = \nabla\times\vec{\mathcal{A}}$, here $\nabla$ being used in the ordinary sense of $\nabla_i = \partial_i$, i.e. with respect to the Euclidean metric $\delta_{ij}$. If $\Phi$ and $\mathcal{A}$ are not time-independent, then $\vec{G} = -\nabla\Phi - \partial_t\vec{\mathcal{A}}$, paralleling electromagnetism, but there will be an extra term in the analogue of Lorentz force that has no electromagnetic counterpart.

Much the same thing can be done in any stationary spacetime, including the rotating Kerr black hole. See also Costa and Natário (2014) for a much more general treatment of several gravito-electromagnetic analogies.

References:

Costa, L. F. O, Natário, J. "Gravito-electromagnetic analogies". Gen. Rel. Grav. 46, 1792 (2014)[arXiv:1207.0465]


Rayleighs units conversion I have a question regarding the units conversion from photon counts on a detector to Rayleighs. I am pretty new to the subject but I would like to know more.
The problem is the following :
I know that $1\, \mathrm{R} = 10^6\ \text{photon}\, {\rm s}^{-1}\, {\rm cm}^{-2}$, but that area should I use in order to obtain the rights conversion. I have an example in which the full area of the HST aperture is used (ish 2.4m in diameter) as a reference but what if I have some optical elements, like a slit for spectroscopic observations?
Should I still consider the same area?
In specific I would be happy to have a systematic way of doing the conversion from counts to Rayleighs using the correct reference area.

In case of light losses in your optical system you obviously have to correct for this. Anything attached in front of your detector could potentially reduce your photon count, so you may have to re-calibrate this to interpret your count rates correctly. You have to examine your setup and make either some theoretical or, if possible, experimental analysis of the effect.


Why do meteorites explode? Some meteorites just blaze away, streaking through the sky. This is clearly explained, as they either slow down enough for the flame to cease, or ablate in their entirety, turning to vapor.

But some, like the Chelyabinsk meteorite, explode violently.

What effect is responsible for this?

Can it be that with their irregular surface acting as a turbine they enter a spin of increasing speed, and once the centrifugal force reaches a certain level, it causes the meteorite to break apart rapidly, hurling parts of it in all directions at speeds far exceeding its speed so far?

The Chelyabinsk meteor didn't explode.

In fact meteors in general usually don't explode, this is a common misconception.
When an object at orbital speeds ($\sim11 km/s$) enters the atmosphere, this speed corresponds to extremely high mach-numbers (M$\sim 30$). This generates a strong shock-front in front of the meteor which causes the meteor to heat up enourmously.
To give a feeling of how important this heat-up is, you can look up on Wikipedia that already at Mach-numbers of 2-3 there are severe engineering problems in not letting aircraft overheat. At Mach-numbers of 30, the air in the shock is so hot that it locally ionizes the air and easily evaporates the solids away from the falling meteor. Upon reaching the ground one hears this shock as loud thunder.
The other thing that happened with the Chelyabinsk meteor is that it broke up in mid-flight. This caused the total solid surface of the meteor that is exposed to the shock to momentarily increase strongly. More surface exposure to the heat causes more surface to glow. This lasts until the smaller fragments that resulted from the breakup evaporate themselves.

Taken together, the mach-shock and the increase in glowing surface area upon break-up give the impression of an "explosion".
Also for an explosion a reservoir of energy would need to be suddenly tapped, which is just not there.


What is meant by sun's path through zodiac I read somewhere that sun's daily course and complete cycle is visible in the ecliptic belt by zodiac constellations.

I didn't understand what that means closest I could find online is: https://www.youtube.com/watch?v=x1DqwC7lRpk

Is the sun's route visible during the night in some areas? Can someone explain?

The stars are not visible during the day, but they are still there (we just can't see them because of the brightness of the sun). We can't see the stars, but we can calculate where they would be if we could see them.

Over the course of the year the sun appears to move relative to the stars. This is actually due to the orbit of the Earth around the sun. The apparent motion of the sun is quite simple: it moves around the sky in a great circle, always following the same path, and taking exactly one year to complete. The route it takes is called the ecliptic. And most of the constellations that lie on the ecliptic are called the zodiac (they are very important to astrologers, but not especially important to astronomers).

The sun's route is not visible to the naked eye from Earth, but it can be calculated with great precision.


What symbol is used for the moon when doing calculations? When doing calculations the mass and radius of the Earth can be represented by $M_{\oplus}$ and $R_{\oplus}$ respectively.

The planets all have their own symbols which can easily be found online.

But what about the Moon ?

Usually something like $M_☽$ is given but is that standard ? Are there other versions of expressing the moon ?

There is no authorative standard - you can use what you want, as long as it is clear what you mean. Besides that: yes, ☽ is reasonably standard. Often also written indices like $M_{moon}$ or $M_m$ are used.

Also see Wikipedia with its many references.


Is this render of a ringed planet's shadow accurate? I've noticed that when it's the equinox on this planet I'm making, the sun seems to cast a double shadow on the planet. How come the planet's ring seems to make two shadows with a space in between, instead of one "line" shadow. Is this accurate to real life, or a quirk of the rendering engine I'm using?

This might happen if you model the ring as opaque but of very little thickness and in the same plane as the vector between the planet and the sun  Seen edge on from the equator, the ring would block none of the sun's light, but from a little north or south of the equator, the ring would block more of one hemisphere of the sun.

So I would check first if your geometry has things lined up too perfectly before considering a fault in the rendering engine.


landing on a comet is more important than landing on an asteroid? Philae landed on comet 67P this year. And it means a big breakthrough for the science and technology of human being.

However Hayabusa mission of Japan not only landed on an asteroid but also returned an asteroid sample to Earth in 2010.

Technologically ,which is more difficult?
And why is Philae more important?

This is a very brief answer but I hope it helps.

Distance

Hayabusa collected samples from a near-earth asteroid while Philae landed on a comet after traveling a distance of over 6.4 billion kilometers.

The Type Of Mission

Hayabusa's did not literally land on the asteroid. It just touched down and collected samples. While, Philae had to land, steady itself, drill and study the composition. Plus, the asteroid on which Hayabusa landed was a near-earth asteroid and so it's nature could have been known easily. On the other hand, the Rosetta team came to know about the exact nature of 67P just weeks before the landing.

The Complexity

The Rosetta mission used about 6 gravity assists and performed 2 flybys. It was not an easy task to plan a journey of 10 years. Plus, when the probe reaches the comet, radio signals would take a lot of time to reach it and thus it would be very difficult to control the probe if something went wrong. Plus, the surface of an asteroid is vastly different from the surface of a comet. The Hayabusa Mission was not the first to touch down on an asteroid. It had been done once by NEAR Shoemaker and so we knew what to expect. See this for a detailed report.

I think this sums up how the Rosetta mission was a major breakthrough.


Long tube for eyepiece I have two short telescopes. The tubes that the eyepieces fit in are different; one is short and one is long, such as the tubes on the Celestron first scope and the Jason Comet chaser. Why is this so and what is the advantages of one over the other? They both take 1.25 eyepieces.

The reason that some telescopes have long eye-piece tubes is that the focal plane of the objective has been placed far enough outside the main tube assembly to allow erecting prisms or binocular adapters to be fitted, if desired.


Can the gravitational potential in a spiral galaxy be positive? Thinking about this question I wanted to start with a rough model of the average gravitational potential of the Milky way. I ran across D. P. Cox and G. C. Gomez 2002 Analytical Expressions for Spiral Arm Gravitational Potential and Density which I think I can understand at least enough to script it. They started with an analytical density distribution and approximated its potential with an analytical form, later they solved for the exact density which generates it, but I am not using that.

I calculated the first example discussed in Section 4 and at least at first glance it seems to agree with their figures. At large distances the potential tends to zero which is good, but the modulation is both positive and negative! This is true both from my script and in their figures.

Without a repulsive force, I don't think the potential can be positive. What am I missing?

Note 1: I've plotted for $z=0$

Note 2: I'm using kg meters and seconds for units, so the plotted potential is in m^2/s^2.

import numpy as np
import matplotlib.pyplot as plt

def PHI(r, phi, z):
    term_1 = -4 * pi * G * H * rho_0
    term_2 = np.exp(-(r-r_0)/Rs)
    gamma  = N * (phi - phi_0 - np.log(r/r_0)/np.tan(alpha))
    K      = n * N / (r * np.sin(alpha))
    KH     = K * H
    beta   = KH * (1 + 0.4*KH)
    D      = (1 + KH + 0.3*KH**2) / (1 + 0.3*KH)
    term_3 = ((C/(K*D)) * np.cos(n*gamma)) * (np.cosh(K*z/beta))**-beta # sech is just 1/cosh

    return term_1 * term_2 * (term_3.sum(axis=0)) # sum over n

G      = 6.67430E-11 # m^3 / kg s^2
parsec = 3.0857E+16 # meters
mH     = 1.007825 * 1.660539E-27 # kg
pi     = np.pi

N     = 2                        # number of arms
alpha = 15 * pi/180.          # pitch angle
Rs    = 7000 * parsec            # radial dropoff
rho_0 = 1E+06 * (14./11) * mH    # midplane arm density
r_0   = 8000 * parsec            # at fiducial radius
H     = 180 * parsec             # scale height of perturbation

C = np.array([8/(3*pi), 0.5, 8/(15*pi)])[:, None, None]
n = np.array([1, 2, 3])[:, None, None]

# plot it
hw = 30000 * parsec
x = np.linspace(-hw, hw, 200)
X, Y = np.meshgrid(x, x)
r = np.sqrt(X**2 + Y**2)
phi = np.arctan2(Y, X)
z = 0.

phi_0 = 0.

potential = PHI(r, phi, z)

if True:
    plt.figure()
    plt.imshow(potential)
    plt.colorbar()
    plt.gca().axes.xaxis.set_ticklabels([])
    plt.gca().axes.yaxis.set_ticklabels([])
    plt.title('+/- 30 kpc')
    plt.show()

I think your recreation is essentially correct. For example, if you look at Figure 1 in that paper, you can see that the potential goes from positive to negative as a function of the azimuthal angle ("phase").

What you're missing is that the potential and density functions they define are perturbations, which are intended to be added to an axisymmetric galaxy model. The idea is that the axisymmetric disk is modulated by their perturbation, so that the total density is less than average (but not less than zero!) where their perturbation is $< 0$ and greater than average where their perturbation is $> 0$. (Similarly, the total potential is $< 0$ everywhere; in the regions where their perturbation is positive, the total potential becomes less negative, but never $> 0$.)

As they point out (pp.4-5), "The impressions invoked by the density distributions of Figures 3 and 5 can be somewhat misleading. These densities must be regarded as perturbations to an azimuthally uniform stellar disk with the same vertical scale height."

And: "In Figure 10, a disk component with the same radial dropoff and scale height as the perturbation, and just sufficient amplitude to make the net density everywhere positive, has been added to the perturbation density." and "In Figures 12 and 13, various amounts of perturbation density are shown added to a representative full stellar disk."


A curious relationship between lunar periods and the solar year Earlier today I was exploring the astronomy features on WolframAlpha, when I stumbled upon an intriguing numerical relationship involving the values for the lunar synodic and sidereal months. Using the values

$$P_{syn}=1\,\text{synodic month}=29.530588\,\text{days},~~P_{sid}=1\,\text{sidereal month}=27.321661\,\text{days},$$

I noticed that this algebraic combination is very close to one year:

$$\frac{1}{P_{sid}^{-1}-P_{syn}^{-1}}=365.256396\,\text{days}\\
\frac{1}{P_{sid}^{-1}-P_{syn}^{-1}}-1\,\text{tropical year}=0.0142053\,\text{days}\approx20\,\text{minutes}.$$

I really doubt this this is a coincidence. Is there a simple way to understand what makes this approximate equality hold?

The synodic month is the "average period of the Moon's revolution with respect to the line joining the Sun and Earth".  However, the Earth also moves in its orbit around the Sun during this month.  From our vantage point, the Sun has appeared to move in the sky with respect to the background stars, in the same direction as the Moon moves in the sky with respect to the background stars.

Your calculation deals with the sidereal month, so your result is the calculation of the sidereal year.

When the Sun returns to the same spot in the sky, that is the sidereal year, whose length is 365.256363004 days, very close to your calculation.  So why is that off from the tropical year by 20 minutes?  Because the tropical year (the year that keeps the seasons in place throughout the calendar year) is slightly shorter than the sidereal year.

  The tropical year is about 20 minutes shorter than the time it takes Earth to complete one full orbit around the Sun as measured with respect to the fixed stars (the sidereal year).

You just found the difference between the sidereal year and the tropical year, and that was no coincidence.

Addition

The number of sidereal months in a sidereal year is one more than the number of synodic months in a sidereal year.  That is because the Earth goes around the Sun once of year (of course), leading to one less synodic month than sidereal month.

Here, $P_{syn}$ is the synodic period and $P_{sid}$ is the sidereal period, and $Y_{sid}$ is the sidereal year, all in days.

$$\frac{Y_{sid}}{P_{sid}} = \frac{Y_{sid}}{P_{syn}} + 1$$

Dividing both sides by $Y_{sid}$ yields:

$$\frac{1}{P_{sid}} = \frac{1}{P_{syn}} + \frac{1}{Y_{sid}}$$

Solving for $Y_{sid}$...

$$\frac{1}{P_{sid}} - \frac{1}{P_{syn}} = \frac{1}{Y_{sid}}$$

or

$$P_{sid}^{-1} - P_{syn}^{-1} = \frac{1}{Y_{sid}}$$

Multiplying both sides by $Y_{sid}$ and dividing both sides by $P_{sid}^{-1} - P_{syn}^{-1}$ yields

$$Y_{sid} = \frac{1}{P_{sid}^{-1} - P_{syn}^{-1}}$$


Ocean floors on ocean planets? I know that models of ocean planets typically have deep liquid water that transitions into exotic forms of ice. What I am wondering is if it makes sense to have an "ocean floor" composed of debris from bombarding asteroids during the early formation stages?

Super-pressurized ice probably wouldn't be static.  It would (almost certainly) move around and shift with the seasons and planetary climate changes and all that good stuff, so I'm extremely doubtful that anything would stay "littered" on the floor in this scenario.

Other factors would be whether the planet had life, which can drop shells or scales and other stuff, which falls to the ocean floor, and volcanic activity which likely happens on most planets and would eject some matter to cover whatever was on the surface over time.

Plate tectonics, obviously another factor though I'm not sure how plate tectonics would affect a thick layer of ice.

This planet is too hot for life (probably) and likely too large for plate tectonics, but I'm sure the hot ice still shifts and moves around and the interior heat has to escape from time to time, either cracks in the surface or volcanoes.   I think you'd need a nearly dead surface like the moon or mars for things to "collect" like that.   I don't think an exotic ice planet would cooperate in that way.

update,

  what I am trying to do is come up with a good scenario where life "as we know it" could arise and flourish on an ocean planet. To me that meant needing some sort of ocean floor that wasn't just some exotic form of ice. I've seen standard models of ocean planet interiors, but hadn't really considered that there will be a wide variety (ie. the interior of Ganymede is probably quite different than Europa). So, in the right conditions I suppose volcanism and plate tectonics may be the answer (super-earths may be more tectonically active than Earth)

Lots to cover here, I'll give it a shot briefly.

Life may have arisen on earth near the ocean floor but not on the ocean floor.  it was the hot vents of gas that Extremophiles lived in.   I don't think they require the ocean surface, just the volcanic jets with methane and other gasses that they could use.  (I think they use Methane, not 100% sure).   I think they require liquid water, but not ground.

Since the deep below the surface of planets tends to be hot, exotic ice would usually be quite hot, I would think, just solid under very high pressure, I suspect that only Extremophiles could live near exotic ice in those high temperatures.   High pressure is less of an obstacle but high temperature might limit the options for life to the very primitive/tiny (I think).    The good news is, nearer the surface, you might get lower temperatures and that might open the door to more advanced life.   I think what's on the surface is generally more interesting than what's on the floor.    Ganymede and Europa both have frozen surfaces with occasional breaches from tidal forces.   A larger planet might have exotic ice deep down but more earth like oceans towards the surface.   I think that would be a lot more interesting to us.

I don't know how many varieties of water worlds there are.   Neptune is a water world of sorts though it's usually categorized as an ice giant and it probably doesn't have what we would consider oceans but it has a lot of water.

On, Plate Tectonics, it might not be essential to life on water worlds.   Volcanism can exist without plate tectonics and Volcanism is probably essential to life on water worlds.   Don't get me wrong.  Nothing beats plate tectonics for circulation and recycling of materials and preventing unwanted buildup of waste, so plate tectonics is really cool and better than just volcanism, but I think active Volcanism would be sufficient for life in oceans.   It's really hard to say anything definite about this subject though, as it's based on theory.

It's probably possible to have volcanic eruptions even though dozens or maybe even hundreds of miles of exotic ice because the heat has to go somewhere, eventually, assing it's likely to build up over time, so either by circulation of eruption, the heat has push through at some point.  This even happens on so called "dead" planets like Mars or even the Moon.   Mars still has the occasional volcanic eruption, just not very often.

But water worlds certainly can have plate tectonics.  There's nothing in the water that would prevent it from happening.  Plate Tectonics is, as I understand it, primarily a factor of the size of the planet.   Gas planets - different story, but planets with a hard surface, Earth sized, a tiny bit smaller to a fair bit but not much bigger are good candidates for plate tectonics (I think).  There's some debate on how large, I think, still going on.   But I remember reading that ocean/water worlds might even be more likely to have plate tectonics.   Plate tectonics is definitely something we'd look for if we ever get a close enough look at other planets in different solar-systems (exoplanets).

Just my thoughts on this.  Not meant to be complete or definitive.


How far away are the events that caused the gravitational waves that have been detected? A certain number of gravitational wave events have been detected. Is it possible to know how far away the mergers that caused those gravitational wave events are?

Yes, it is possible to calculate (within an error range) the distance of observed gravitational wave events.  It is known that a variety of parameters will affect how the amplitude and frequency of the observed gravitational waves will change over time as recorded in the "chirp" event from the interferometers: the parameters include distance of the event, the mass of each of the colliding objects, the angular momentum of each of the colliding objects, the orientation of the objects' angular momentum vectors with respect to each other and with their orbital plane.  With general relativity, you can build a model that calculates the expected "chirp" given a value for all these parameters; when a chirp is observed, it is possible to determine the combination of these parameters that result in a chirp that best matches the observation.

The effect of a larger distance parameter is to decrease the amplitude of the expected waves from colliding objects of a given mass, as well as to "slow down" the entire event due to cosmological red shift.

From GWTC-1: A Gravitational-Wave Transient Catalog of Compact Binary Mergers Observed by LIGO and Virgo during the First and Second Observing Runs

  Gravitational waves from compact binaries carry information about the
  properties of the source such as the masses and spins. These can be
  extracted via Bayesian inference by using theoretical models of the GW
  signal that describe the inspiral, merger, and ringdown of the final
  object for BBH [23–30] and the inspiral (and merger) for BNS [31–33].
  Such models are built by combining post-Newtonian calculations
  [34–38], the effective-one-body formalism [39–44], and numerical
  relativity [45–50].


Is it safe to view moon at daytime through telescope I have a little 76mm Newtonian Reflector. I want to see the moon at daytime. Is it safe to look at sky at daytime when sun is shinning or there is any risk of sun.

As long as you don't directly look at the sun through your telescope, then there is no problem. But don't try to see the moon during solar eclipse without special filters (easily available for your telescope's eyepieces)

In  my telescope, I have seen the moon day time when there was very light cloud cover. It was awesome scene. You can feel that the moon is spherical by observing its terminator and the lighted edge. You can easily make out the dark side edge.


If we accelerate a neutron star to move at speeds comparable to light would it become a blackhole because of the added energy? Could a neutron star turn into a blackhole if it is accelerated to near luminal velocities? And if it slows down after a while, does it become a neutron star again?

In special relativity, changes of reference frame are symmetric between observer and observee.  If moving a neutron star gave it enough kinetic energy to become a black hole, then you could effect this change by jumping into a moving reference frame yourself.

What matters is the neutron star's mass in its rest frame.


time on earth compared to completely stationary in space I recently started trying to understand the theory of relativity and I have the following question.  Because earth is orbiting the sun at approx 67,000 mph (and its my understanding time slows down the faster you are moving) what are the effect of time on earth compared to a completely stationary observer watching earth's orbit, from outer space?  How would the twin paradox work with one twin on earth and the other twin completely stationary in outer space?

Time dilation is a relative phenomenon, not a local phenomenon.

As adrianmcmenamin mentioned, Earth's motion around the sun is cause for no more time dilation than any other point in the universe. Consider that the sun is travelling around the galaxy at 200 km/s, and the Milky Way is approaching Andromeda at 130 km/s. We do not know at what speed we are approaching the great attractor, and I hope that at this point it becomes obvious that there is no difference in "us approaching the great attractor" to "the great attractor approaching us". Nothing is considered "stationary to the background" in an infinite, expanding universe as there is no "background" to measure against.

Consider the gedankenexperiment of comparing a clock on Earth to a clock orbiting the galaxy. Now compare those clocks to clocks outside the galaxy. You might consider the clock that ticks the most time away to be the "most at rest clock". Now I ask you, how will you compare the clocks? You must move at a significant portion of the speed of light for a very long time to get information from one clock to the next. Thus, the messenger would suffer the effects of Special Relativity, rendering the comparison meaningless.

To further the point, consider the development of International Atomic Time. I'll quote directly from wikipedia:

  In the 1970s, it became clear that the clocks participating in TAI were ticking at different rates due to gravitational time dilation

Consider, those were clocks on Earth suffering from measurable time dilation due only to their latitude and altitude differences and the effect of moving information across the distances between them!


How can we be 13.2 billion light years from another galaxy? How did we get 13.2 billion light years from the faraway galaxies being discovered with Hubble?

If we all started near each other at the Big Bang, and we all travelled slower than the speed of light, and the universe is about 13.9 billion years old, how could we possibly be 13.2 billion light years away from recently discovered galaxies?

I know "inflation" was supposedly superluminal, but I bet no one thinks that matter dispersed several billion light years apart during that operation.

The distance you are referring to is the "light travel distance". It is the length of time that light has been travelling from the distant galaxy to us.

When the light was emitted, our Galaxy was much closer to that distant galaxy. What has happened in the meantime is that the space between the galaxies has expanded.

Furthermore we can ask where is that galaxy right now (where now is a time measure since the big bang) and it turns out that the galaxy would be much further away than 13.2 billion light years - more like 40 billion light years if you were to somehow measure that distance with a very long ruler.


Why is sunset on 21st of June not the latest? Why is the sunset not the latest on 21 June, but only on 27 June. Purely geometrically regarding the orbit of the two celestial bodies this should not be like this?

Is there somewhere a good source, that explains this situation ?

It's explained in this Wikipedia article. Basically, the length of a day is defined by both the Earth's rotation itself and Earth orbiting around the sun. Because the Earth's orbit around the Sun isn't a circle but an ellipse (making it moving faster when it's closer to the Sun), and the Earth rotates in a plane different than it orbits the Sun, an apparent solar day isn't always 24 hours; in June, it's slightly longer. That means every day, "noon" (the time of day when the Sun is at its highest point) happens 'later' (compared to the time on your watch), so sunset (and sunrise) are also later. The first few days after June 21st, this effect is more significant than the effect of the Sun moving south (causing the day* to become shorter and the sunrise falling earlier). The same effect causes the earliest sunrise (in your location) to be on June 15th, so before the solstice.

*: day as in day vs. night, the period between sunrise and sunset, so not 24 hours


what happens when we keep going radially away from earth? Is the generally accepted belief in the science community, that if one sits in a spaceship and move away from Earth where r (distance between earth and my ship) is continuously increasing, can one keep going forever?

Yes. There are no edges of the universe that have been observed.

The standard cosmological models favoured by observations (open or flat universes) allow you to move outwards indefinitely. Closed universes are like the surface of a sphere, and you would eventually return to Earth unless the expansion was too fast; the same is true for the models with nontrivial topologies - but these have no evidence at present.


Has the Mars dust storm affected its naked-eye appearance? Mars has been undergoing a planet-wide dust storm for a couple of months now.

By my eye, while Mars currently looks a bright and handsome orange, there's been no change that I've been able to notice from this dust storm. Is it possible for an experienced naked-eye (or small telescope) astronomer to have noticed the change in appearance brought on by this dust storm?

Is it possible for an experienced naked-eye (or small telescope)
  astronomer to have noticed the change in appearance brought on by this
  dust storm?

Yes, it is definitely possible. Without dust storms it looks redder.

Source:

The opposition of Mars, 2001: Part I - Richard McKim


WOW! signal calculation of point source intensity The WOW! signal's measurement took the average signal strength over 12-second intervals. The average intensity's peak was just over 30-1 signal to noise, meaning that the 12 second interval had an average strength of 30-1. But evidence suggests the signal came from a point source. Has anyone ever done the math to estimate the point source strength? Is this even possible?

The plot is pretty straight forward.   6EQUJ5 puts the peak signal between Q and U, closer to U.   The letters represent two digit numbers, 1-9, then A-Z.  A being 10, Z being 35.   U was the highest signal strength received equivalent to a 30.

A plot fits rather neatly.

Wikipedia has a nice summary of the event.  The location is discussed here with some margin for error due to there being two listening devices and the design being, as I understand it, somewhat less than ideal.

While the signal remains unexplained, it was some 30 times stronger than baseline, but 3s and 4s were fairly common.  It was only 7-10 times stronger than signal strengths they received regularly.   I lean towards it being a natural but unexpected event, but there's not sufficient data to close the case, so to speak.


How to calculate the time for the solar disk to pass the horizon and transit's lines For a given date, longitude, and latitude:

Popular apps and websites give the time when the sun's center or edge is on the horizon (i.e., the time of sunrise and sunset) or its transit (i.e., noontime), but I want to ask how to calculate the time needed for the solar disk to pass the horizon's line and the transit's line (preferably in c# or java)?

The time it takes the sun to cross the horizon is given at:

Expression for length of sunrise/sunset as function of latitude and day-of-year
How long does a sunrise or sunset take?

Re the time it takes to cross the transit line, this doesn't really answer your question, but, to a good approximation, the time varies between 128 seconds at the equinoxes and 140 seconds at the solstices, regardless of latitude or longitude.

More specifically, the formula is $\frac{128}{\cos (\text{dec})}$ seconds, where $\text{dec}$ is the Sun's declination. You can calculate the Sun's declination itself using the formulas at https://en.wikipedia.org/wiki/Position_of_the_Sun

The calculation here is relatively simple:

The Sun travels $360 \cos (\text{dec})$ degrees in a 24-hour day, where $\text{dec}$ is the Sun's declination
When the Sun is transiting, the motion is perpendicular to the transit line (the Sun's motion is entirely in azimuth, not in altitude)
Therefore, all of the Sun's angular motion translates to motion across the transit line; in contrast, the Sun rises and sets a (non-perpendicular) angle (except at tropical latitudes on the two days where the Sun passes directly overhead), so sunsets and sunrises take longer than $\frac{128}{\cos (\text{dec})}$ seconds
Since the Sun has an angular diameter of 32 minutes or $\frac{8}{15}$ degrees, it takes $\frac{\frac{8}{15}}{360 \cos (\text{dec})}$ of a day for the Sun to cross the transit line
Since a day is 86400 seconds, this works out to $\frac{128}{\cos (\text{dec})}$ seconds.

Caveats and nitpicks:

I assume there are 86400 seconds between successive noons. This is incorrect for two reasons:

The time between noons is not exactly one day. The cumulative difference forms the Equation of Time but the day to day difference is quite small
The Earth's day is slightly longer than 86400 seconds, which is why we need leap seconds

The Sun's angular diameter actually varies based on Earth's distance from the Sun, but 32 minutes is a good approximation
I assume the Sun's altitude doesn't change while it's transiting. While this is a good approximation, the altitude does change slightly
I assume the Sun's declination doesn't change while it's transiting. The change in declination is very small, so this is a reasonable assumption
There are probably other assumptions I made implicitly that I am not noting here.
Refraction is not an issue, since the motion we are discussing is azimuthal and not in altitude.
I used the "one over cosine" form above to make things easier for non-mathematicians. The more compact form would use "secant".
The calculations I did for this problem are disorganized, but available at: https://github.com/barrycarter/bcapps/blob/master/STACK/bc-solar-transit.m


Is the moon's revolution speed increasing or decreasing I've come across multiple answers to this question. So, the moon is moving away from the Earth a couple of centimeters per year and the Earth's rotation is slowing down. Here it says the moon's revolution is speeding up, another place it says it is slowing down. So, due to tidal effects, is the moon's revolution slowing down or speeding up?

From the cited article,

  The slowing of the speed of rotation of Earth also results in the increase in speed of revolution of moon around the Earth.

I suspect that this is the statement that is causing the OP to be confused, and rightly so. This statement is incorrect. Whether "speed of rotation" means orbital velocity or angular velocity, both are decreasing as the Moon slowly retreats from the Earth.

The Moon is retreating very slowly from the Earth, currently about 3.78 cm per year. Dividing by 385000 km (the Earth-Moon mean orbital radius) yields about 10-10 parts per year. That qualifies as "very slowly". For simplicity, I'll assume the Moon's orbit is essentially circular, plus this very slow spiraling out. For a circular orbit, the relation between orbital radius and angular velocity is $$r^3\omega^2 = G(M+m)$$
This essentially is Kepler's third law. Assuming $G$, $M$, and $m$ are constant, differentiating both sides with respect to time yields
$$r^2\omega (3\dot r \omega + 2 r \dot \omega) = 0$$
meaning
$$r \dot \omega = -\frac32 \dot r \omega$$
Suppose some quantity $q$ is of the form to $q=r^n\omega^m$. Differentiating with respect to time yields $\dot q = r^{n-1}\omega^{m-1}\left(n\dot r\omega + m r \dot\omega\right)$. Since $r\dot\omega = -\frac32 \dot r \omega$, this can be rewritten as $$\dot q = \left(n - \frac 3 2 m\right) \dot r r^{n-1}\omega^m$$
Quantities of this form include angular velocity ($q=\omega$, so $n=0, m=1$ in this case), orbital velocity $(q=r\omega$, so $n=m=1$), and angular momentum (q$=r^2\omega$, so $n=2, m=1$).  Note that $(n-\frac 32m)$ is negative for angular and orbital velocity but positive for orbit angular momentum. Orbit angular momentum thus has the same sign as does $\dot r$ while angular velocity and orbital velocity have the opposite sign as does $\dot r$.

As a sanity check, look to centripetal acceleration. This is given by $a = r\omega^2$. In terms of the above, $n=1$ and $m=2$. The term $n-\frac32m$ is clearly negative in this case, so centripetal acceleration decreases as the Moon retreats from the Earth. Centripetal acceleration is also give by Newton's law of gravitation: It is proportional to the inverse of the square of the distance between the Earth and the Moon, so this does indeed decrease as the Moon retreats from the Earth.


How high were the tides back when the Moon was much closer to Earth? In this article it is stated that the Moon was originally much closer to Earth at a distance of 22,500 km rather than the 384,400 km it is today. What was the height of the tides back then in meters? Surely they would've been orders of magnitude higher?

To expand on Keith's comment, the equation for the tidal force is based upon the difference between the acceleration due to the orbiting object, and the gravity of the body itself, i.e. the difference between how hard the Moon pulls on you compared to how hard the Earth pulls.  This results in an overall equation of:
$$|a_{tidal}| = 2\Delta r G \frac{M}{R^3} $$
where $\Delta r$ is the radius of the Earth (6371 km), G is the gravitational constant (6.674 x 10$^{-11} m^3kg^{-1}s^{-2}$, M is the mass of the Moon (7.342 x 10$^{22}$kg), and R is the distance between the two (here either 22,500km or 384,400km). To check our equation we can calculate the current tidal acceleration: using 384,400km, we arrive at a value of 1.099 μm/s$^2$, which is close to the accepted value of 1.10 μm/s$^2$ (2).  Calculating the tidal force using the distance of 22,500km, we arrive at a value of 0.0055 m/s$^2$, or about 5,000 times the current tidal force.  That could make for some massive tides, however, were there oceans?
Current research suggests that the Moon formed as the result of a major impact, around 4.5 billion years ago (3), while the first oceans formed around 4.4 billion years ago (4).  This leaves a gap of around 100 million years between that 22,500km distance and the formation of the oceans, but it is close enough that the tidal force is likely not that different.  However, translating the height of the tides is difficult.  As the average tide is about 0.6m (5), a tide 5,000 times higher would be almost 3km high, which seems impossible.  Additionally, the height of the tide appears to be more strongly affected by geography than by the tidal force, as the current range of tides is from 0-15m in height.  The earliest evidence we have of tides is about 2.5 billion years ago (6), but unfortunately, the geologic record only records the pattern of the tides, not the height.  We can tell that the Earth was spinning faster then, with a day 4.5 billion years ago being about 4-5 hours long, but it may have been too cold to melt the oceans.
So, to sum up:  The tidal force with the Moon at that distance would be 5,000 times greater than it is today, but there may not have been oceans, and the tide height is affected more by the shape of the ocean than the tidal force itself.  However, something like the water planet in Interstellar, with massive, 3,000km-high tides sweeping over the Earth every couple of hours is possible, although a boring ice planet is possible too.


What is the acceleration for the expansion of the Universe? I've found that the speed of expansion is about $70\frac{km}{s\cdot 10^{6}pc}$.

But what is the acceleration of expansion in terms of $\frac{km}{s^{2}\cdot 10^{6}pc}$ ?

I want to know how fast this value of "70" is changing, ie what is $\mathrm{d}H_0/\mathrm{d}t$?

The figure below shows the evolution of the Hubble parameter $H$ from 10 billion years (Gyr) ago, to 10 Gyr into the future:

As you can see, the change in $H$ is modest nowadays, compared to the past. The "acceleration" of the expansion at any point in time is given by the tangent to the curve at that time. Today, the change is $dH/dt \simeq -1.2\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}\,\mathrm{Gyr}^{-1}$, or roughly $-10^{-17}\,\mathrm{km}\,\mathrm{s}^{-2}\,\mathrm{Mpc}^{-1}$.

In other words, it will take roughly a billion years before the expansion rate has decreased by $1\,\mathrm{km}\,\mathrm{s}^{-1}$, and it asymptotically approaches some $57\,\mathrm{km}\,\mathrm{s}^{-1}$.

Since $H$ is defined as $(da/dt)\,/\,a$, where $a$ is the scale factor (the "size" of the Universe), a constant $H$ implies that $a\propto e^{Ht}$; that is, the size of the Universe increases exponentially.


How to convert horizontal coordinates using NOVAS? I'm using NOVAS 3.1. I know that I can convert equatorial coordinates to horizontal coordinates using the equ2hor function.

Is it possible to make NOVAS do the inverse transformation: from horizontal to equatorial? There is no hor2equ, but maybe some other function implements this functionality?

This is the reply I got from the US Naval Observatory itself:

  No, we do not have that transformation in NOVAS.  Generally, altitude and azimuth are not determined to very high accuracy, so using them to obtain RA and Dec, which are much more precisely defined and measured, does not make a lot of sense.


Need help with the calculations/conversion of a celestial object I'm developing a telescope controller open-source application. I started this project with a very little knowledge of Astronomy.
Basically, the app is going to send data to telescope over a wireless connection.

Data of the celestial object is as follow:

["SAO#": 308,
"HD": 8890,
"Con": "Alpha Ursae Minoris",
"StarName": "Polaris",
"RAH": 2,
"RAM": 31.812,
"DED": 89,
"DEM": 15.85,
"Mag": 2.02,
"PRA": 0.038,
"PDec": -0.015]

RAH = RA hours, RAM = RA minutes,
DED = Dec degrees, DEM = Dec minutes

(PRA and PDec are the drift of the star in arc-seconds per year (from the catalog Epoch J2000.0). I don't need to use these in the app, that I'm told.)

The controller server (data receiver) accepts in this format:
Set target RA- HH:MM:SS,
Set target Dec- DD:MM:SS,
Set target Azm- DDD:MM:SS,
Set target Alt- DD:MM:SS

I would want to convert the given celestial object data to Ra, Dec, Azm, Alt.

I'm told that:

RA = ((RAH + (RAM / 60.0)) * 15.0); //in degrees, RA is an earthly longitude projected onto the sky
DEC = (DED + (DEM / 60.0)); //in degrees, Dec is an earthly latitude projected onto the sky

For a star or other celestial object from the catalogs they have a fixed Epoch (J2000.)  I need to apply a correction for precession/nutation (wobble of the Earth's axis of rotation) to get a half decent estimate of the star's RA/Dec "now".

I'm using a code library that seems to have support for this and does the calculation on it own.
I would want to know the procedure of the data conversion from equatorial coordinates to horizontal including.
For an instance this is how the code works:

    let jd = Date().julianDay
    let RAH = 2.0
    let RAM = 31.812
    let DED = 89.0
    let DEM = 15.85

    let eqCoor = EquatorialCoordinates.init(rightAscension: Hour((RAH+(RAM/60.0))*15.0), declination: (Degree(DED+(DEM/60.0))), epoch: Epoch.J2000, equinox: Equinox.standardJ2000)
    let annualAbb = eqCoor.correctedForAnnualAberration(julianDay: jd, highPrecision: true)
    let initEQ = AstronomicalObject.init(name: "Polaris", coordinates: annualAbb, julianDay: jd, highPrecision: true)
    print("EquatorialCoordinates -> RA(α):", initEQ.equatorialCoordinates.alpha, "DEC(δ):", initEQ.equatorialCoordinates.delta)

    let userLocation = GeographicCoordinates(positivelyWestwardLongitude: Degree(.plus, 75, 51, 13.65), latitude: Degree(.plus, 30, 54, 43.55), altitude: 256)
    print("User Location -> Latitude:", userLocation.latitude, "Longitude:", userLocation.longitude)

    let preccCoor = annualAbb.precessedCoordinates(to: Equinox.standardJ2000)
    let horizontalCoor = preccCoor.makeHorizontalCoordinates(for: userLocation, at: jd)
    print("Horizontal Coordinates -> Azimuth:", horizontalCoor.azimuth.inHours, "Altitude:", horizontalCoor.altitude.inRadians.inDegrees)

This is the data that code returns:

EquatorialCoordinates -> RA(α): +37h56m06.009s DEC(δ): +89°16'05.165"
User Location -> Latitude: +30°54'43.550" Longitude: +75°51'13.650"
Horizontal Coordinates -> Azimuth: +12h1m07.396s Altitude: +30°13'16.650"

Here's the data from Stellarium according to the same location as my code.

The data calculated from my code is not even close to the one from Stellarium except apparent altitude.

I couldn’t really figure out the way to get RA Dec from the data mentioned. I ended up using a catalog that already has RA Dec of objects.


What is a singularity? What is at the center of a black hole? Specifically regarding space-time So because I can only really think of space-time in 2-dimensions like a sheet of something, my assumptions might be wrong to begin with. I was watching a YouTube video on black-holes and there was a quip regarding black holes and their warping of space time, to the point where the narrator said something along the lines of “there is nothing at the center of the black hole but gravity has warped space time so much that it's this ’pinhole’ like affect that causes the blackhole behavior”.

My questions are:

When a star dies and collapses into a black hole, what is at its center? The star's mass compacted into the size of the plank length, or something similarly small? Is there really nothing at the center of a black hole? Surely, the core collapsed into something really small right?
What is meant by singularity? Is it just the warping of space-time that makes it this way?

This is more of a question for the Physics stack, but I'll give it a shot, since it's fairly basic.

You need to understand something before we begin. The theoretical framework we have to gauge and answer this sort of thing is called General Relativity, which was proposed by Einstein in 1915. It describes things such as gravity, black holes, or just about any phenomena where large densities of mass or energy are involved.

There's another chapter in Physics called Quantum Mechanics. This describes, usually, what happens at very small scales - things that are super-tiny.

Both GR and QM are fine in their own way. Both are tested against reality and work very well. But they are not compatible with each other. Meaning: you cannot describe a phenomenon from a GR and a QM perspective, both at once. Or meaning: we don't have a coherent set of equations that we could write down, and then "extract" out of them either a GR-like view of reality, or a QM-like view.

The problem is, the center of a black hole is both very high mass density and very high gravity (and therefore right in the field of GR), and very small (and therefore "quantum-like"). To properly deal with it, we'd have to reconcile GR and QM and work with both at once. This is not possible with current physics.

We pretty much have to stick to GR only for now, when talking about black holes. This basically means that anything we say about the center of a black hole is probably incomplete, and subject to further revision.

  A star dies, collapses into a black hole, what is at the center? The
  star's mass compacted into the size of the plank length or something
  similarly small? Is there really nothing at the center of a black
  hole?, surely the core collapsed into something, just really small
  right?

According to General Relativity, it collapses all the way down to nothing. Not just "very small", but smaller and smaller until it's exactly zero in size. Density becomes infinite.

You can't say "Plank length" because, remember, we can't combine GR and QM, we just don't know how. All we have here is GR, and GR says it goes all the way down.

It is quite possible that the singularity is not physical, but just mathematical - in other words, whatever is at the center is not actually zero size. Quantum mechanics in particular would be offended by zero-size things. But we can't say for sure because our knowledge here is incomplete.

I'm using words such as "size" (which implies space) and "becomes" (which implies time). But both space and time in the context of a black hole are very seriously warped. The "becoming" of a black hole all the way down to the zero-size dot is a reality only for the unlucky observer that gets caught in it. But for a distant, external observer, this process is slowed and extended all the way to plus infinity (it's only complete after an infinitely long time). Both observers are correct, BTW.

EDIT:

So, when we are saying "density is infinite and size is zero at the singularity", this language applies to the unfortunate observer being dragged down in the middle of the initial collapse of the star.

But from the perspective of the distant observer, a black hole is still a chunk of mass (the original star) in a non-zero volume (the event horizon of the BH). To this observer, the density of that object is finite, and its size is definitely not zero. From this perspective, anything falling into the BH never quite finishes falling, but just slows down more and more.

Both observers are correct. So, keep in mind, when I talk about "infinite density", that's the inside observer point of view.

  What is a singularity? Is it just the warping of space time that makes
  it this way?

You get a singularity whenever there's a division by zero in the equations, or when the equations misbehave somehow at that point. There are many different kinds of singularities in science.

http://en.wikipedia.org/wiki/Mathematical_singularity

In the context of a black hole, the center is said to be a gravitational singularity, because density and gravity are suggested to become infinite, according to the GR equations.

GR says: when you have a lump of matter that's big enough, it starts to collapse into itself so hard, there's nothing to stop it. It keeps falling and falling into itself, with no limit whatsoever. Extrapolate this process, and it's easy to see that the size of it tends to zero, and density tends to an infinite value.

EDIT:

Put another way - if density becomes large enough, gravity is so huge, no other force is strong enough to resist it. It just crushes all barriers that matter raises to oppose further crushing. That lump of matter simply crushes itself, its own gravity pulls it together smaller and smaller... and smaller... and so on. According to current theories, there's nothing to stop it (QM might stop it, but we cannot prove it, because we don't have the math). So it just spirals down in a vicious cycle of ever-increasing gravity that increases itself.

Space and time are really pathologic inside the event horizon. If you are already inside, there's no way out. This is not because you can't move out fast enough, but because there's really no way out. No matter which way you turn, you're looking towards the central singularity - in both space and time. There is no conceivable trajectory that you could draw, starting from the inside of the event horizon, that leads outside. All trajectories point at the singularity. All your possible futures, if you're inside the event horizon, end at the central singularity.

So, why the center of a black hole is called a "singularity"? Because all sorts of discontinuities and divisions by zero jump out of the equations, when you push math to the limit, trying to describe the very center of a black hole, within a GR frame.

http://en.wikipedia.org/wiki/Gravitational_singularity

Speaking in general, physicists don't like singularities. In most cases, this is an indication that the mathematical apparatus has broken down, and some other calculations are necessary at that point. Or it might indicate that new physics are taking place there, superseding the old physics.

One last thing: just because we don't have a combined GR/QM theory to fully describe the center of black holes, that doesn't mean a pure GR research in this area is "wrong" or "useless". It doesn't mean one could imagine some arbitrary fantasy taking place inside a black hole.

Astronomers these days are starting to observe cosmic objects that are very much like black holes, and their observed properties are in very close accord with what GR predicts for such things. So research in this field must continue, because it's clearly on the right track, at least in the ways we can verify today in astronomy.


Any record of the Earth passing through the tail (not trail) of a comet? Discussion below the Space SE question How hard is it to fly through the tail of a comet? Has it been done? has led me to ask if there is any record of the Earth passing through the tail of a comet.
That's tail not trail. The annual meteor showers we talk about, even the unicorns are the Earth passing through the trail of breadcrumbs distributed long ago. This question is about the tail(s) of comets, those things we see in photos and sometimes by eye.

Earth passed through the tail of Halley's Comet in 1910. It caused a bit of a panic due to claims that cyanogen detected in the tail would cause harm: https://en.wikipedia.org/wiki/Halley%27s_Comet#1910
Linked there is https://timesmachine.nytimes.com/timesmachine/1910/02/08/104920328.pdf
 (click for larger)


What is the most distant observable White Dwarf known? I attended a talk last week about Type 1a supernovae with the focus being on double-degenerate systems as potential progenitors for them. We were shown a graph of White Dwarf (WD) mass distribution like the one below (source), and it got me thinking, because even though they are very faint objects they are also very numerous in our galaxy.
So this led me to wonder what is the furthest known WD from Earth given that they are so faint but so numerous. I have looked for an answer online but the closest I got was a Type 1a progenitor at redshift 2 called "SN UDS10Wil" but this is not what I am looking for because this is not observable from Earth, it is just inferred from the supernova classification.

There is the unusual white dwarf RX J0439.8−6809.  It seems to be a carbon-oxygen white dwarf with a temperature of around 250,000 K!  It is 9200 pc away.  You might want to look for other PG 1159 stars to see how far they can be detected.


What is the star whose western traditional name is the longest? My definition of traditional name is that name that appears in the stellar catalogs made in classical antiquity by Greeks and Romans and which were later translated and compiled by the Arabs to be later used again by Europeans from the Renaissance. I do not look for traditional Chinese or Mayan names, for example.

Most stars visible without a telescope were given names in Antiquity. These Greek or Babylonian names were transmitted and modified during the Middle Ages by Arabian compilers and many of them are still popular, such as Sirius or Vega.

Other names are longer and more "exotic", like Betelgeuse. I am thinking, for example, of really long and exotic names like Zubenelgenubi (Alpha Librae) or Zubeneschamali (β Librae). Are there any stars whose "ancient" or "medieval" name are longer?

Here is a list of arabian star names, later modified by europeans, but it may be incomplete, because greek catalogues included about 1.000 stars and this list only includes 200.

According to Google, the longest name for any star is “Shurnarkabtishashutu”, the Arabic for “under the southern horn of the bull”. The star is known as Zeta Tauri. You could also scroll thru Wikipedia's list of star names for different countries.


How to get the longitude of the perihelion to find the Heliocentric Equatorial coordinates of a planet I'm writing a program to randomly generate star systems and I have been having a difficult time figuring this out. I've found the equations I need to get the heliocentric equatorial coordinates, but I haven't been able to figure out how to determine the longitude of perihelion. Is it just the opposite of the longitude of the ascending node or is it a completely different variable?

It might just me, but the names for this stuff is definitely making it more confusing than it should be.

Here is all my notes:

** = power

P = planet's period

a = semi-major axis

b = semi-minor axis

M = mean anomaly

E = Eccentric anomaly

e = eccentricity of orbit

v = true anomaly

r = radial distance

A = longitude of asending node

w = longitude of perihelion

i = inclination

t = time

(r, v) = polar coordinates

(x, y, z) = Heliocentric Ecliptic coordinates

(X, Y, Z) = Heliocentric Equalorial coordinates

eccentricity

e = sqrt(1 - (b ** 2 / a ** 2))

Period

P = sqrt(a ** 3)

Kepler's Equation

M = E - e * sin(E)

Mean Anomaly
use as starting point for eccentric anomaly

M = (2 * pi * t) / P

True Anomaly

v = 2 * atan(sqrt((1 + e) / (1 - e)) * tan(E / 2))

Radial Distance

r = (a(1 - e ** 2)) / (1 + e * cos(v))

Heliocentric Ecliptic coordinates

x = r(cos(A) * cos(w + v) - sin(A) * sin(w + v) * cos(i)

y = r(sin(A) * cos(w + v) + cos(A) * sin(w + v) * cos(i)

z = r * sin(w + v) * sin(i)

Heliocentric Equalorial coordinates

X = x

Y = y * cos(i) - z * sin(i)

Z = y * sin(i) + z * cos(i)

As opposed to what @user21 said, the longitude of periastron does depend on the longitude of ascending node (I was surprised by that myself).
In orbital dynamics, there exist the argument of periastron (symbol ω) as well as the longitude of periastron (symbol π [or ϖ but may lead to confusion) ), which is equal to the argument of periastron (ω) plus the longitude of ascending node (Ω). In other words:
π = ω + Ω
Now, the argument of periastron is equal to:
ω = arccos $\frac{\mathbf{n}\ ·\ \mathbf{e}} {|\mathbf{n}||\mathbf{e}|}$
If ez < 0 then ω → 2π − ω.
where:
$\mathbf{n}$ is a vector pointing towards the ascending node (i.e. the z-component of $\mathbf{n}$ is zero), and
$\mathbf{e}$ is the eccentricity vector (a vector pointing towards the periastron).
This formula can be found at https://en.wikipedia.org/wiki/Argument_of_periapsis (I call it instead the “argument of periastron,” as Meeus says: “The word ‘periapse’, used by some authors, is incorrect. The word perihelion means the point of the orbit that is closest to the Sun [from the Greek peri = near + helios = Sun]. Similarly, perigee is the point closest to the Earth [ge = Earth]. Therefore, ‘periapse’ would really mean the point closest to the apse; but this is ridiculous, because what is meant is the apse itself!”)
So, using the above formula, you should be able to figure out the longitude of the ascending node for each of your hypothetical bodies.
Have fun!


The dark side of the moon I encountered this statement that the dark side of the moon is hidden from our sights.
Could someone please explain?

Yes, the dark side of the moon is mostly hidden from earthbound observers. This is due to tidal locking:

  A tidally locked body takes just as long to rotate around its own axis as it does to revolve around its partner. This causes one hemisphere constantly to face the partner body.

In other words, one moon-year is just as long as one moon-day.

The left-hand side shows the actual configuration of earth and moon: The orbit is locked and we can only observe one side of the moon. On the (fictional) right-hand side, the moon does not rotate around itself and we can observe different areas of the moon at different times.

I should note that we can actually observe not 50%, but around 59% of the moon's surface because of libration and parallax effects that cause the moon to slightly "wobble":

You can find all those images and further explanation in the linked wikipedia articles on tidal locking and libration.


Data for spectrum of Wolf-Rayet stars I'm looking for a way to access the data used to produce these plots for instance. These plots show the spectrum of various Wolf-Rayet stars. I found a few websites where you have some informations on those stars here for instance, but I'm not able to find and download a file containing the spectrum data of the stars.

If you know how to reduce raw spectrum data, you may look at the official archives of big observatories (ESO, MAST, CADC). But a quicker and much easier way could be to use the following portal:

http://archive.eso.org/scienceportal/home

Typing the name of some Wolf Rayet stars will lead you to reduced data and thus spectra.


Why does the solar wind switch to flowing straight by the time it reaches Earth? Between 02:01 and 02:29 the new NASA Goddard video 5 New Discoveries from NASA's Parker Solar Probe explains

finding # 4, A Breaking Point:
Before parker, scientists knew that the corona rotates with the visible surface below it. But they didn’t know how — or where — the solar wind switched to flowing straight by the time it reaches Earth.
Parker has finally spotted signs of this transition — and the changeover happens significantly farther out than expected.

Naively I'd apply conservation of angular momentum and assume that the particles are in individual ballistic orbits, but the solar wind contains charged particles as well as neutrals and so even though of low density and long mean free paths they can interact collectively.
But I don't understand why the solar wind would suddenly stop rotating beyond some radius from the Sun well inside Earth's orbit, rather than beyond the heliopause.
Why does this happen?

When plasma moves in a magnetic field then the charged particles follow helical paths around the field lines, due to the $q\vec{v}\times \vec{B}$ Lorentz force. If the radius of gyration is small, then the plasma is effectively tied (or frozen) to the field lines.

However to decide whether it is the motion of the magnetic field that enforces motion of the plasma or vice versa, we must compare the magnetic pressure with the gas pressure (or equivalently, compare the magnetic and kinetic energy densities).

The magnetic energy density scales with magnetic field squared and the field drops further from the Sun (e.g. a dipole field falls as $r^{-3}$), whereas the plasma kinetic energy density (which depends on the plasma density and temperature) could remain relatively high (the solar wind density drops only as $r^{-2}$ and the temperature is roughly constant).  There is a transition between the magnetic field dominating the plasma close to the Sun, to where the plasma dominates and effectively carries the field with it. This transition point is known as the Alfven radius and is usually at a few tenths of an au from the Sun.

Below this radius, the plasma (sort-of) co-rotates with the field, which is anchored to the photosphere. Beyond that the plasma is (sort-of) free to move radially outwards. This how the Sun loses angular momentum.


How was the time of equinox measured in ancient times? In the book "The Crime of Claudius Ptolemy" by Robert Newton, on page 81, it reads as follows:

  When we measure the time of a solstice, we measure the time when the
  meridian elevation angle of the sun has an extreme value, a maximum in
  the summer, and a minimum in the winter.... When we measure the time
  of an equinox, on the other hand, we measure the time when the
  meridian elevation takes on a specified value.

What does he mean by a "specified value". What specified value?

Also, how would the ancients have determined the time of the maximum, or the specified value? Time devices in ancient Greece, like water clocks, I suppose would have significant errors. Would they just start the water clock when the sun rose, then marked its level when the "specified value" occurred? That method would seem to have a lot of error.

The "specified value" would be the latitude of the observer, and would be the average of the maximum and minimum values observed at the solstice.

The altitude of the sun above the horizon is measured when it is due South. This is what "meridian elevation" means. So no clock is required. However this method can at best determine the date of the equinox. Precise observations over several years could be combined to obtain an time of equinox.

The position of the sun above the horizon can be simply determined by measuring the length of a shadow. However Ptolemy had access to astrolabes and quadrants for more accurate measurement of the sun's position.


Where exactly does the Moon flip, given that it appears the other way up in the other hemisphere? I've read in multiple sources that the Moon appears the other way up in Northern Hemisphere vs Southern Hemisphere. Here is one that explains why: Does the Moon look different in the northern and southern hemispheres?

My question is, where is the line drawn in this case where the moon 'flips'? Is it exactly the equator? I'm not sure if it is, the simple 'thought experiment' in the above link assumes (for simplicity's sake) that the orbit of the Moon is exactly above the equator. According to Wikipedia, the orbit of the Moon is slightly tilted, and our own earth also tilts from time to time. Where does the moon flip exactly? Does it change throughout the year as our earth change its tilt? If so, how? Which are the areas that always see the northern look or the southern one throughout the year?

The moon doesn't flip. Is us who flip.

Those in the antipodes have their heads pointing in the opposite direction. So the moon appears inverted. But there is no sudden point where it "flips". We don't live on a discworld.

You can see this yourself. Because as the Earth spins you will change the direction in which your head is pointing. Look at the full moon over the course of one night. As it appears to travel from East to West in the sky, it's orientation will slowly change. There is no mystery here. You are being rotated by the Earth.


How can a naked singularity be possible? How can we have a singularity, without a black hole around it? How would one form? What does it do?

Nobody likes the idea of naked singularities, as they would have a toxic effect on causality. If a singularity existed that was not separated from us by an event horizon, then not only would the future not be predictable, but the past would not be fixed. Like the grandfather paradox, it wouldn't make sense, so it cant exist. The trouble is that GR doesn't implicitly rule them out.

If a black hole is spinning fast enough, or has enough charge then it seems a naked singularity could form. Naively you may think that the centrifugal force in the first case, and the electromagnetic force in the second are sufficient to overcome gravity.

I particular if $G^2M^4/c^2 < J^2$, where J is angular momentum, M is the mass, G is the gravitation constant, and c is the speed of light. The there will be a naked singularity. For a small, stellar mass black hole would need an angular momentum of $10^{42} \mathrm{kg m^2 s^{-1}}$ to lose its event horizon


Do blue giants have a habitable zone? Or would the planet(s) need to be too far away for life to exist?

Just because a planet is in the "Habitable zone" doesn't mean it is habitable. A planet is said to be in the habitable zone if it is possible for liquid water to exist on the surface. A planet may be in the habitable zone, and yet not be habitable if, for instance, it's atmospheric pressure is too low (like Mars) or too high (like Venus), both of which are in the habitable zone of the Sun.

A star may be unsuitable for life for other reasons: it may be prone to extreme flares, for example. A blue giant is a star that burns bright and dies young, only lasting a few million years. Planets around such a star would have only just formed (they may still have molten surfaces). Blue giants tend to be unstable, being variable over a range of timescales, and ejecting large amounts of matter into space.

So while there would be a region which is "habitable" in the sense that the surface temperature of a planet could be in the range 0-100, you wouldn't expect such a planet to actually be suitable for life.


Why do GPS satellites move in relation to earth's surface? Wouldn't it be easier with geostationary satellites? Has is to do with the monitoring of gamma-ray bursts?

Geostationary orbits (GEO) are necessarily over the equator, and satellites in GEO are not visible at all poleward of about 80 degrees -- they are always below the horizon (There's a good explanation on physics.stackexchange.com).

This is a problem because for GPS to work, you need at least four satellites to be visible and they need to be well-separated in the sky.  Because it's rare for the whole sky to be "visible" all the time -- parts are blocked by buildings, metal vehicles, hills and suchlike -- you in fact need more than four in the sky as seen from any point on Earth to have a minimum of four actually visible all the time.

So even at lower latitudes where GEO satellites are visible, they still all are in the southern (or all in the northern) sky and thus won't give a good GPS fix.  And the military (who were paying for this) wanted to be able to use GPS in the north, not just in the equatorial regions.

Also, possible GEO satellite locations trace a gently curved line across the sky, and even if there were a dozen visible and even at low latitude, they'd give very poor accuracy perpendicular to that line.

The solution was to put the GPS satellites into many different orbits which were strongly inclined to the equator. They wanted them to be fairly slow-moving as viewed from Earth, so LEO wasn't viable.  (LEO also required too many satellites to get adequate coverage.)   They needed to be above the Van Allen belts for longevity. But they also needed them to be as low as possible to maximize signal strength. 10,000 miles up turned out to be the right engineering compromise.

Gamma ray bursts were lagniappe.


what planet is the most radioactive? So as I understand it the source of most geothermal energy on Earth is due to radioactive decay. Jupiter is the most massive planet but if the bulk of it's mass comes from gaseous elements instead of heavier ones like uranium or whatever than it being the most massive might not necessarily mean it has the most radioactive material.

There's a few parts to this question, I'll give it a shot.  Also, since you're asking about the geothermal energy of the earth, I'm going to ignore radiation from space and radioactive van allen belts (Jupiter has a very radioactive van Allen belt).   Also, as voyager discovered when it left the solar system, there's more radiation in the form of cosmic rays out there, that the sun effectively shields most of the planets from.

But, because you mention the Earth's geothermal heat from radiation, so I'll focus on that.

Here's a very well answered similar question about radioactive elements inside the earth.  what percentage of the Earth's core is Uranium

  So as I understand it the source of most geothermal energy on Earth is
  due to radioactive decay.

A healthy share of Earth's core heat is from the planets formation.  This is a pretty good article on that subject - http://physicsworld.com/cws/article/news/2011/jul/19/radioactive-decay-accounts-for-half-of-earths-heat, it says that of the current heat, about 50% is from the energy left from the formation of the planet and about 50% is from radiation - so the radiation is very important, but it's probably not the cause of "most" of the heat in the earth.

As far as I know, no similar studies have been done on any of the other planets.

  Jupiter is the most massive planet but if the bulk of it's mass comes
  from gaseous elements instead of heavier ones like uranium or whatever
  than it being the most massive might not necessarily mean it has the
  most radioactive material.

That I pretty much agree with.  The two primary differences between the 4 gas giant planets (or, two gas giant, two ice giants if you like) and the 4 inner rocky planets is that the gas giants formed outside the frost line and key difference #2, the 4 outer planets are much larger, so they were more able to hold onto more hydrogen and helium, which are the 2 most abundant elements in the solar system.

Obviously, the 4 outer planets have a lower percentage of heavy elements overall.    As to their percentage of Uranium - well, your guess is as good as mine, but, probably a lower percentage of radioactive elements than the inner planets seems reasonable to assume.

Also, any radioactive elements in the 4 outer planets would likely be well below their thick gaseous surfaces, so it would be difficult to detect, outside of Neutrino detection I suppose, and more trouble than it's worth to try to collect.

The 4 inner planets (and lets throw the moon in there as well, and asteroids if you like) are more interesting I'm assuming David Hammen's answer is correct in the linked question above, and that most of the Uranium on earth is in the crust or mantle, the highest percentage in the crust, and so that's probably true for any other rocky planet that formed and was, at one time, molten enough to circulate, so all the inner planets probably have similarly small amounts of uranium and other radioactive elements as Earth.

Mercury may have had much of it's crust and some of it's mantle blown off in a giant collision, and if that's the case, Mercury might have lost a portion of it's radioactive elements.  Brief summary of that here, point #6:  http://messenger.jhuapl.edu/faq/faq_science.html

unrelated footnote, but if Mercury's crust was blown off, it might be a good planet to mine for rare earth elements, or maybe asteroids will be easier, who knows, but I've always been curious of the mining possibilities on Mercury.  Abundant solar energy too, but it's a hard planet to fly to because of the sun's gravity, so, maybe asteroids are the way to go for space mining.

The moon was also formed by a giant impact, so, the moon was formed by material from Earth's but it has a much higher percentage of mantle than the Earth, so it could have a higher percentage of Uranium and other radioactive elements than the earth.  How well the Moon's material was circulated in it's cooling process is another question.   Being smaller it probably cooled more quickly.

Venus and Mars, I see no reason why they shouldn't have similar percentages of radioactive elements than Earth.

 Source:  https://en.wikipedia.org/wiki/Internal_structure_of_the_Moon

One way to look at this question is to investigate how quickly the other inner planets cooled.  Far from exact, but that probably tells us something, since heat generated by radioactive decay inside the planet would probably lead to volcanism at some point.  Even the Moon appears to have had volcanic eruptions as recently as 33 million years ago, so some ongoing internal heating seems likely.

Volcanoes on the moon

Size is also a key factor in what keeps the inside of a planet hot, and in some cases, tidal heating (Io, Enceladus), but that's not really relevant to this question.

Neither Mars nor Venus are volcanically active like the earth, but Olympus Mons may have erupted as recently as 25 million years ago, and there is evidence of large lava flows on Venus.   As to how big a role radioactivity plays in heating up the cores of those 2 planets, I don't know precisely how much.

Overall, I think this is a good question on planet cores and radioactivity, as the inside of the planet is just as interesting as the surface, well, to me, at least.   I know this doesn't answer your question specifically as to which is the most radioactive, cause I don't know, but I like the question.


Would Hubble Space Telescope improve black hole image observed by EHT if it joined array of telesopes? My question is related to black hole image released in April. As far as I understand idea of EHT, it joins observations from multiple locations to work like one telescope with radius that is equal to the distance between the farthest telescopes in array. Moreover Earth atmosphere affects observation, e.g. by introducing delays. Would it beneficial to add Hubble Space Telescope to EHT and if yes, why HST was not included in observation array?

No, it would not, because it operates in the visible spectrum and the EHT is an array of radio telescopes. For the "very long baseline interferometry" technique to work, all the telescopes have to be operating at the same wavelength, because combining the signals involves measuring exactly how well the peaks and troughs of the radio waves from the different telescopes line up.
You can do VLBI in the visible spectrum, but you have to match up the waves even more precisely since light has shorter wavelengths than radio waves. The EHT collected all the data first, and then spent a lot of time combining it by computer, but that required using very precise clocks to sync the data. We don't have clocks precise enough to do that for light, so a direct optical connection is required between the multiple telescopes. So there isn't a good way to do a planetwide VLBI optical telescope yet.
Addendum inspired by the comments, especially from TazAstroSpacial: To get the same resolution with optical light, you can use a much smaller telescope array. I didn't mention that fact because I was thinking about keeping the size of the array the same and getting better resolution. But in any case the problem is that the challenges of doing the interferometry at smaller wavelengths more than make up for the advantages of needing a smaller aperture. At least at the current state of the art.