Wednesday 31 July 2013

earth - What planets and moons have we seeded with life (if any)?

Which planets and moons have we accidentally (or purposely) seeded with life? I would expect this to be any planet or moon we have sent an unsterilized (or not fully sterilized) lander to, or crashed anything into.



I am speaking largely of bacteria and other microorganisms that might have hitched a ride on our spacecraft, or any life that was deliberately taken to and left on another world (which I am not aware). I know we went to the moon, we at the very least "seeded the moon" temporarily with life (humans) and probably bacteria on the space suits / lander / equipment we sent.



I wondering where we have "seeded" / brought life (even if it doesn't / didn't survive there).

Tuesday 30 July 2013

asteroids - How often do meteorites hit earth

There is definitely not a catalogue of all meteorites hitting Earth. For instance, the ones falling in desert areas and in the ocean aren't found, and even the ones falling in more populous regions are easily mistaken for normal rocks.



Meteorite rate



Meteorites come in all sizes, from sand grains to dinosaur-annihilating rocks.
The size distribution of these rocks (as well as most other things in space) follow an approximate power law, meaning that the smaller the meteorites you consider, the more often they hit Earth.



Estimating the total number of meteorites hitting Earth is difficult, but one study (Bland et al. 1996), looking at the composite of the earth on three arid site and considering the weathering over time, finds that between 36 and 116 meteorites over 10 g hit Earth every year per $10^6,mathrm{km}^2$. Multiplying by the surface area of Earth $A_oplus$, that's roughly 20,000 to 60,000 meteorites hitting Earth per year.



Meteor rate



UPDATE: You want to know the rate of incoming meteors, i.e. all rocks that hit Earth's atmosphere. By definition, a meteorite is a meteor that lands on the surface of Earth. Before it enters the atmosphere and becomes a meteor, you can call it asteroid, meteoroid, rock, pebble, or dust grain, depending on its size and your mood.



Taylor et al. (1996) have monitored incoming meteors with the AMOR radar in New Zealand and detected in one year 350,000 trails from incoming meteors of size 10–100 $mu$m vaporizing in ~100 km's height. I don't know exactly how large an area AMOR monitors, but from simple trigonometry, and as an order-of-magnitude estimate, if the events happen in $h sim 100,mathrm{km}$'s height, the radar can "see" a distance $d=sqrt{h^2+2R_oplus h} sim 1100,mathrm{km}$, and thus covers an area $Asim4times10^6,mathrm{km}^2$. Thus, again multiplying by $A_oplus$, this amounts to the order of 40 to 50 million meteors per year.



Killer meteor rate and the Sudan asteroid



The meteors discussed above is probably not what you had in mind. As mentioned, the bulk of these "rocks" are actually more like sand grains. So what is the rate of devastating killer-rocks from space? Brown et al. (2002) fit a power law to several different data sets (LINEAR, Spacewatch, NEAT and others), and find that the number $N$ of meteors above a diameter $D$ in meters is roughly
$$
N(>D) simeq 37 D^{-2.7}.
$$
The following plot is from their paper, with the long black line showing their fit, and the short black lines indicating the uncertainty:



meteors



You mention the asteroid 2008 TC3, which had a diameter of 4.1 m.
According to the fit, such an event happens roughly $1/(37[4.1,mathrm{m}]^{-2.7})sim$ once per year.



You can also see that the largest meteor hitting Earth in a century would be $(37times100,mathrm{years})^{1/2.7}sim20,mathrm{m}$, equivalent to $sim500,mathrm{kiloton,TNT}$.
Similarly, hundred-meter-sized objects ($sim1,mathrm{megaton,TNT}$) should hit us roughly once per 10,000 years. By extrapolation of their fit beyond observational data, objects of $D>1,mathrm{km}$ — or 100 gigaton TNT — should hit Earth on average once every ~3.4 million years.

stellar evolution - Gyrochronology? - Astronomy

Please point us to the peer-reviewed reference! As far as I'm aware the problems that plague gyrochronology still persist. These are (in no particular order).



  1. A lack of calibrating clusters for stars with ages older than the Sun and for stars older than about 1 Gyr and mass much less than a solar mass.


  2. It just doesn't work for stars younger than about 200 Myr at a solar mass and younger than 1 Gyr at half a solar mass, because rotation rates have not converged to a single-valued function of age.


  3. You need rotation periods to use it. You cannot get around the unknown inclination angle so if you only have a projected equatorial velocity $v sin i$, then this is considerably less useful.


  4. Measuring the rotation periods for stars takes a lot of effort. For young stars they potentially have a large photometric modulation caused by starspots that can be measured through monitoring over timescales of days to weeks. For older stars like the Sun, the spot modulation is so small that you need something like Kepler in order to be able to measure it. It is not something that could be routinely done over huge swathes of the sky.


  5. The precision is not that great. Aside from the calibration issue there is the problem that there is a dispersion in rotation rates at a given age that leads to an uncertainty in the estimated age for a given rotation rate. This amounts to at least 10-20% for stars younger than a Gyr but probably decreases for older stars. However, this is counterbalanced by the fact that older stars appear to suffer more differential rotation with latitude. That means if you measure the rotation period on one occasion it is quite possible it will be different the next time you measure it, if spots are at a different latitude. This would give rise to errors approaching 20% in a star like the Sun.


Work to overcome these issue will be slow and incremental. Asteroseismology is certainly helping for solar-type stars because for these stars, you can estimate the age in an independent (though model-dependent) way, which offers an alternative calibration route. The original Kepler project has also observed a number of older clusters and there are a few more which might prove to be decent calibrators in the K2 fields. However, my bet would be that the problems of calibration for K and M dwarfs will still be there even after all this data has come in.



You can find a discussion of the above and a comparison with other age estimation techniques in Jeffries (2014). Estimates of age based on metallicity would be very crude indeed, and subject to catastrophic errors. For instance, the metallicity of young, star-forming clusters in the solar vicinity is almost identical to the Sun. What you can probably say, is that if a star has a metallicity less than about 0.5 solar then it is likely to be older than 5 Gyr; anything less than 0.1 solar is likely to be older than 10 Gyr. But that's about it.



EDIT: So, the paper that triggered your question is Meibom et al. (2015) "A spin-down clock for cool stars from observations of a 2.5-billion-year-old cluster", which talks about Kepler observations of the open cluster NGC 6819, in which the rotation periods have been found for 30 cool stars with masses between 0.85 and 1.4 solar masses. These data are indeed an extremely useful calibration set which fills in the gap for solar type stars between the ages of the Sun and the well-studies clusters like the Hyades and Praesepe below 1 Gyr.



However, all the problems I mentioned above still exist. The range of masses probed here is rather small, but what they are able to establish is that the 20 or so stars below about 1.2 solar masses have a dispersion around a mean period-colour relationship with a standard deviation of only 5%. This means that the solar mass stars have converged to a fairly tight relationship between rotation period, mass and age and that errors in measuring the periods are also relatively small. Because $t propto P^{2}$, this leads to the
claim that ages could be determined with a precision of 10% at this age, if the age-rotation-mass space were fully calibrated. Note though that these rotation periods were measured by Kepler; that the median peak-to-peak amplitude was only 4 milli-mag and that in most cases the period is a mean from several independent measurements of the same star. Such excellent data will not be forthcoming for very large samples of field stars anytime soon.

the sun - Finding Latitude and Longitude of a person or of where a picture was taken

EDIT: It turns out I answered this question earlier (in a slightly different format):



Cancelling out earth rotation speed, Altazimuth mount



Imagine you are at 35N,106W at 5h local sidereal time.



The celestial north pole is at (0,90) in the celestial frame and
(0,35) in your frame.



The point (5h,0) is (75,0) in the celestial sphere (1h = 15 degrees),
and (180,55) in your frame



The point (11h,0) is (165,0) in the celestial sphere and (90,0)
(setting in the east) in your frame.



Converting to rectangular coordinates, we need the matrix that does this:



{0,0,1} -> {Cos[35 Degree], 0, Sin[35 Degree]}

{Cos[75*Degree],Sin[75*Degree],0} -> {-Sin[35 Degree], 0, Cos[35 Degree]}

{Sin[-75*Degree],Cos[75*Degree],0} -> {0,1,0}


or, generalizing a bit (lst = local sidereal time as an angle, lat =
latitude, notice that lon is irrelevant since we're using local
sidereal time)



{0,0,1} -> {Cos[lat], 0, Sin[lat]}

{Cos[lst],Sin[lst],0} -> {-Sin[lat], 0, Cos[lat]}

{-Sin[lst],Cos[lst],0} -> {0,1,0}


Solving (Mathematica):



m0 = Table[a[i,j],{i,1,3},{j,1,3}]

(* Some simplifying assumptions, intentionally avoiding corner cases *)

conds = {-Pi/2 < dec < Pi/2, -Pi/2 < lat < Pi/2, 0 < ra < 2*Pi, -Pi < lst < Pi}

m = Simplify[m0 /. Solve[{
m0.{0,0,1} == {Cos[lat], 0, Sin[lat]},
m0.{Cos[lst],Sin[lst],0} == {-Sin[lat], 0, Cos[lat]},
m0.{-Sin[lst],Cos[lst],0} == {0,1,0}
},Flatten[m0]],conds]

m = m[[1]]


This gives us



$
left(
begin{array}{ccc}
-cos (text{lst}) sin (text{lat}) & -sin (text{lat}) sin (text{lst})
& cos (text{lat}) \
-sin (text{lst}) & cos (text{lst}) & 0 \
cos (text{lat}) cos (text{lst}) & cos (text{lat}) sin (text{lst}) &
sin (text{lat}) \
end{array}
right)
$



The inverse of this matrix:



$
left(
begin{array}{ccc}
-cos (text{lst}) sin (text{lat}) & -sin (text{lst}) & cos
(text{lat}) cos (text{lst}) \
-sin (text{lat}) sin (text{lst}) & cos (text{lst}) & cos (text{lat})
sin (text{lst}) \
cos (text{lat}) & 0 & sin (text{lat}) \
end{array}
right)
$



will transform rectangular azimuth and elevation back to right
ascension and declination and right ascension. This would be an
interesting alternative approach to this problem, but I won't be using
it.



OK, if we convert right ascension and declination to rectangular
coordinates, multiply by the first matrix above and convert the result
back to spherical coordinates, we will have azimuth and
elevation. Let's do that...



The spherical coordinates for (ra,dec) are:



{Cos[dec] Cos[ra], Cos[dec] Sin[ra], Sin[dec]}


Multiplying by the matrix, we get



$
{sin (text{dec}) cos (text{lat})-cos (text{dec}) sin (text{lat})
cos (text{lst}-text{ra}),-cos (text{dec}) sin
(text{lst}-text{ra}),cos (text{dec}) cos (text{lat}) cos
(text{lst}-text{ra})+sin (text{dec}) sin (text{lat})}
$



(note that lst-ra is sometimes called the "hour angle", which would
simplify the above)



Converting back to spherical coordinates and simplifying:



{x,y,z} = Simplify[m.{Cos[dec] Cos[ra], Cos[dec] Sin[ra], Sin[dec]},conds]

(* NOTE: the use of single argument arctangent here introduces
ambiguity, and the final answer must compensate for this *)

az = Simplify[ArcTan[y/x],conds]

el = Simplify[ArcSin[z],conds]


yielding:



$
text{azimuth}=-tan ^{-1}left(frac{cos (text{dec}) sin
(text{lst}-text{ra})}{sin (text{dec}) cos (text{lat})-cos
(text{dec}) sin (text{lat}) cos (text{lst}-text{ra})}right)
$



$
text{elevation}=sin ^{-1}(cos (text{dec}) cos (text{lat}) cos
(text{lst}-text{ra})+sin (text{dec}) sin (text{lat}))
$



Assuming I haven't made any mistakes, the next steps would be:



  • Converting local time to local sidereal time


  • Inverting these equations to get latitude and longitude from
    azimuth, elevation, and time.


I may or may not edit this answer to add those steps, but this
hopefully provides a start.

Monday 29 July 2013

Formation of our Universe - Astronomy

I didn't read the SciAm article, but: Any cosmological theory compatible with "the" mainstream big bang theory doesn't assume material to have been contained in "Our Universe" (needs to be defined) from the very beginning.
Instead a very hot, unified state, is assumed, which then underwent several symmetry breakings (https://en.wikipedia.org/wiki/Symmetry_breaking, https://en.wikipedia.org/wiki/Higgs_mechanism), before "material" came into existence, as we observe it by now.



So, the very hypothetical, presumed four-dimensional star just needs to trigger an initial state causing the big bang. Whether "Our Universe" is a brane universe embedded in a higher-dimensional universe, is another mere hypothesis. Whatever "star" or "black hole" in a higher-dimensional universe would mean, the physics would be very different from the physics we can observe. For instance, inverse square laws wouldn't hold due to the additional dimension(s), hence orbital dynamics, electromagnetism needed for atoms, etc., all would either fail, or work rather different. Presuming the existence of stars in such a hypothetical universe is very daring, already.

Monday 22 July 2013

light pollution - Do Noctilucent clouds cause problems with telescopes?

Noctilucent clouds are not a problem for space telescopes because their orbits are always more than 85 km. The Hubble Telescope orbits at about 570 km. Noctilucent clouds are a problem for ground-based telescopes (although only at high latitude sites,> 50$^{circ}$), especially if you are trying to get accurate photometric brightnesses. However, typically you can see them reflecting city lights or moonlight, or you can tell from the large variance in your photometry counts that they are there and you should try again another night.

We are sure that there are no brown dwarfs closer than ______

I think the answer is roughly 3-5 light years.



The recent WISE survey in the near-infrared should have been capable of detecting even a very cool nearby brown dwarf (and indeed it has detected some very cool brown dwarfs - e.g. a 250K brown dwarf only 6 light years away Luhman et al. 2014). Since WISE was an all-sky survey and there is great interest in finding these ultracool brown dwarfs, then it is reasonable to assume that any similar, object within a few light years would already have been found. The caveat to this is that the point spread function of the WISE survey s not that great an so near the galactic plane it could be that source confusion might have hidden the brown dwarf. On the other hand, nearby objects usually have high proper motion, so it would probably have been uncovered in combination with the earlier 2MASS survey, since as I show below, such an object may have been detected in both.



Some details: As an example let's take a very low mass brown dwarf - say 20 Jupiter masses and 5 billion years old - and place it 4 light years from the Sun. According to the evolutionary models of Saumon & Marley (2008), such an object has an intrinsic luminosity of $10^{-7}$ times (1 ten millionth) that of the Sun and a temperature of 400 Kelvin and would appear to have a spectral type of late T or early Y. This is similar to the coolest brown dwarfs yet found.



From the calibration of absolute magnitudes versus spectral type for cool brown dwarfs in Marsh et al. (2013) we know that at 4 light years, such a brown dwarf would have magnitudes of $H=16$ and $W2=10$. The former is just about bright enough to have been seen in the 2MASS all-sky survey and the latter is easily detected by WISE. The combination of data would also easily have revealed the large parallax of such an object. We can conclude that an object would have to have much lower mass than this to remain undetected and thus would arguaby not be called a brown dwarf, but would really be a "free-floating planet".



Now you might ask, what about a cooler brown dwarf? Well the theoretical predictions of magnitudes and colours become much more uncertain, but I would simply refer to the Luhman discovery. This brown dwarf is about as cool as a brown dwarf could be given the age of the Galaxy, may only be 10 Jupiter masses, yet was still easily detected in the WISE survey. Therefore I think this sets a fairly firm empirical limit.

Tuesday 16 July 2013

Could gravity be the 4th dimension?

According to Michio Kaku, in his book Hyperspace, he said that in this Hyperspace, there exist a total of 10 dimensions that govern the laws of physics. Is it possible that we too live in a Hyperspace, and if so, would gravity be the 4th dimension that would govern the laws of physics we are yet to discover?

What percentage of galaxies rotate with trailing arms?

I am looking for the percentage of galaxies that rotate with trailing arms. I know that only a small percentage of galaxies rotate with leading arms and would like to know whether there are any results that determine the percentage of spiral galaxies that rotate with tailing arms.



I am interested in publications in this topic.



Note:
So far I have found the following:



Do you know of any newer publication on this question? More recent than Pasha & Smirnov (1982) where the sample is quite small?

Saturday 13 July 2013

Does natural satellite(s) of a planet affects its orbital velocity around a star?

One way to look at this question is to not think of the Moon orbiting the Earth but look at it as two bodies orbiting each other, then it becomes obvious that orbital velocities of two objects around each other does affect the orbital velocity around the star.



The Moon and Earth both orbit the barycenter between them, which is inside the earth, but in the direction of the moon.



The moon orbits the barycenter about 3,640 km/hr (slightly slower than it orbits the Earth I think), so the corresponding earth's movement for 81 times the mass is an ellipse 1/81st the size, giving the Earth an elliptical orbital velocity of about 45 km/hr, which, being in an elliptical motion is a vector addition of velocity and so it's only about plus or minus 45 km/hr at full or no moon, when the motion is relatively parallel with, or 180 degrees opposite to the Earth's motion around the Sun.



see picture:



enter image description here



This velocity variation has a period of about 27.3 days (sidereal not synodic) with an average diameter of 1/81st the Moon's orbital diameter, or about 9,400 KM, which is about the same distance the Earth orbits around the sun in just over 5 minutes, so, the effect is tiny, and is probably tiny for every planet-satellite system, but feel free to calculate Jupiter-Ganymede or Pluto-Charon if you like.



Using the 30,000 km/s approximation, we get 30,045 km/hr at no moon and 29,955 km/hr at full moon or about 3/10ths of 1%, peak to troth, every 13.65 days or so.



For comparison, The Earth's perihelion is about 3.28% closer to the sun than it's aphelion, which using Kepler's equal areas law and an approximation of area = 1/2 base x height, the Earth's orbital velocity is about 3.28% greater at perihelion, or about 11 times as much variation in every 182.62 days of elliptical orbit.



For the most part, the Earth's orbital velocity isn't that relevant. The Earth's position, not it's velocity, determines which stars or planets you see at night and if the Earth's spans slightly more or slightly less of it's orbit than usual, over a 24 hour period because of the Moon, 5 minutes of fluctuation over 13 days is only relevant to the most rigorous of astronomers.

Wednesday 10 July 2013

Is the Earth going to evolve towards Mars' fate or Venus' fate?

Either or neither. It's impossible to tell from the present.



If runaway climate change occurs, then yes, the conditions on Venus could be a potential analogue for the kind of environment on Earth due to the greenhouse effect.



Mars's atmosphere is assumed to have been much thicker in the past, otherwise it could not have sustained liquid water on the surface; it would have evaporated away. This water is required to explain the gorges and riverbeds we see on the Martian surface today. Researchers are still unsure where the atmosphere went, but the prime candidate is through top-loss to outer space, rather than sequestration through minerals on the surface (see here).



The likelihood of such an atmospheric escape process becoming dominant on Earth is small due to the Earth's greater mass and strong relative magnetic field, which prevents ion escape. In fact, the dominant loss process on the earth is sequestration. Some estimates put the reservoirs of sequestered carbon from original CO$_2$ at 250000 times the size of the existing atmosphere.



The alternative is that the Earth remains in its current steady state, or close to it, far in to the future. There are many mechanisms to enable such an equilibirum, such as oceanic sequestration of C0$_2$ highlighted above. There are also more contested theories such as the Gaia hypothesis. This proposes that the global biological ecosystem on the Earth is self regulating, helping to maintain life enabling conditions on the Earth. It's pretty much a dead-cert that life doesn't exist on either the Martian or Venusian surfaces, so we can accept that if the Gaia hypoethesis is true, the Earth will never reach these conditions.

Tuesday 9 July 2013

planetary formation - How can you determine the initial volume of a planet's atmosphere?

Since the surface pressure of a planet is determined by the mass of the column of gasses above it one would surmise that to determine the pressure you must know the volume and mass of the atmosphere.



Knowing the composition and volume of the gasses in the atmosphere would allow you to calculate their mass but how do we determine what that volume is?



Assuming we are only concerned with initial conditions when the primary atmosphere (created through accretion) was formed would the volume be arbitrary?



I hope this isn't too broad of a question. I'm trying to pin down a possible initial state that would be created when a new planet is formed and the conditions stabilize.

How to calculate a planet's current position within the solar system

(DISCLAIMER) I am not an expert on this subject, and I only hope that my answer will attract better, and more complete ones.



The Meeus book that @Gerard Ashton mentioned is a great resource for amateurs, but I am worried whether it is accurate enough for, let's say, eclipse or transit predictions. Also I glanced at the site linked in the question and saw something like "A live view of Plutonian system". To be honest I don't think that the Meeus book can enable you to do that kind of thing.



As far as I know, serious calculations of planetary positions are done today with the help of dynamical ephemerides developed and maintained by specialized agencies, such as VSOP87 by the Bureau des Longitudes in France, and DE431 by Jet propulsion laboratory of NASA. The NASA Horizons system is an online interface from which you can get the planets' positions. (It's powered by some version of the DE ephemeris, I think.) It is well documented, so feel free to check out their documentation if you want to have a go.

Sunday 7 July 2013

How do you calculate the lookback time distance to a given galaxy?

No you can't. Other information is required.



For low redshifts - let's say smaller than 0.1 - and by that I mean the wavelength increases by 10 percent, you might get away with using Hubble's law to estimate the distance and then get the look back time by dividing by the speed of light.
$$ t simeq frac{lambda - lambda_0}{H_0 lambda_0},$$
where $H_0$ is the present day Hubble parameter of about 70 km/s per Mpc, $lambda$ is the measured wavelength and $lambda_0$ is the rest wavelength.



So far so good, you just need to know $H_0$. However for larger redshifts it gets horribly complicated because the Hubble parameter changes with time in a way that depends on the curvature of the Universe and hence on the cosmological parameters defining the matter density and dark energy density.



There is indeed a highly complicated, non- linear formula involving integrals that I will look up a reference for. But possibly the better way to proceed is to use a simple look-up table produced from such calculations with certain assumed values for ratios of matter and dark energy densities with respect to the critical density; a.k.a. $Omega_M$ and $Omega_{Lambda}$.



The plot below is an example taken from http://www.astro.caltech.edu/~eran/MATLAB/Cosmology.html which shows look back time versus redshift for two different cosmologial models (but with the same value of $H_0$). The curves are very different at high redshifts, but converge at small redshifts.



Here is a cosmological calculator that can do the job for you. Enter the redshift and your assumptions about the Hubble parameter and other cosmological parameter and it will tell you the age of the universe at that redshift as well as the lookback time.



Lookback time versus redshift

ceres - Subterranean Oceans On Other Planets/Planetoids: How Do Astronomers Deduce This

How they have concluded that there could b.e a subterranean ocean on Ceres is by Spectralscopy.



Spectral Signature can be summarised to like this:




Different elements emit different emission spectra when they are
excited because each type of element has a unique energy shell or
energy level system. Each element has a different set of emission
colors because they have different energy level spacings. We will see
the emission spectra or pattern of wavelengths (atomic spectra)
emitted by six different elements in this lab. We will then identify
an unknown element by comparing the color of the unknown with the
flame color of our knowns.




And another:




When something is hot enough to glow (like a star), it gives you information about what it is made of, because different substances give off a different spectrum of light when they vaporize. Each substance produces a unique spectrum, almost like a fingerprint.




So how the scientists would have concluded that there was water because of the water vapours in the atmosphere. It gives off a certain wavelength that could be compared the closest element that gives off the same wavelength which is water.



Then the scientists would have used Galileo's magnetometer(an instrument which measures the strength and direction of magnetic fields) to conclude that there could be an ocean on the dwarf planet. The strength and response of the induced field would tell the scientists a rough estimate of the dwarf planet's sub surface which in return they could deduce the fact the there is a large amount of water present which equals an ocean.

the sun - Source of high energy cosmic particles

Earth receives shower of both solar and cosmic radiations every seconds. The Sun put up a heliosphere around the solar system and thus foreign high energy charged particles could not penetrate easily. Is there any other source most probably that lies within our solar system?



Note:



  1. Very high energy cosmic particle beyond 5.7 x 10^19 eV were observed
    by Pierre Auger Observatory in Argentina. (rare but the GZK limit
    seems to rule out distance sources)

  2. GZK limit roughly capped the upper energy level of cosmic ray as
    they are slowed due to CMB over vast distance

Saturday 6 July 2013

atmosphere - Oldest Reference to Astronomical Seeing

I'm writing a paper on astronomical seeing. Sir Isaac Newton identified both the phenomenon and origin of astronomical seeing in his Opticks. He writes:



If the Theory of making Telescopes could at
length be fully brought into Practice, yet there would
be certain Bounds beyond which Telescopes could
not perform. For the Air through which we look
upon the Stars, is in a perpetual Tremor; as may be
seen by the tremulous Motion of Shadows cast from
high Towers, and by the twinkling of the fix'd Stars.


is anyone aware of any older reference from Gilbert, Galileo, Harvey, Ptolemy, Copernicus or Kepler?

positional astronomy - Moving-Cluster method for determination of the distance of Hyades. A starter problem

I am currently following a class of observational astronomy lab. I will present a brief description of the method in first and then proceed to the question, so anyone is welcomed to read the entire post or move directly to the problem. The current question is about the Moving-Cluster method, for which I quote from Wikipedia




In astrometry, the moving cluster method and the closely related convergent point method are means, primarily of historical interest, for determining the distance to star clusters. They were used on several nearby clusters in the first half of the 1900s to determine distance. The method is now largely superseded by other, usually more accurate distance measures.*




An image for the Convergence point of a Cluster



One can find more here http://pages.uoregon.edu/soper/Stars/movingcluster.html . A brief description of the method is like that:



From photographs taken, say, 10 years apart, we can see that a star of the cluster has moved. (It has proper motion.) Let us suppose that the proper motion is an angle μ. Then the angle θ is the angle between the convergent point and the star or equally the angle between the radial and tangential speed of the moving star



enter image description here



If one knows the angle θ, he/she will be able to find the movement of the star and in the end, calculate it' s distance, through deduced relations beginning with $ v_t =v_r tan θ.$



Question
I know the declination and the right ascension of both the star and the convergent point.



My problem as is to find the angle θ. In the exercise I have it is requested to find the angle between the star and the convergent point, that is θ , using spherical trigonometry. My mind goes on to using the cosine relationship:



$$ cos a = cos b cos c + sin b sin c cos A ,$$ for the triangle



enter image description here



But, although it may be simple, I cannot understand where this triangle is to be implemented. What are it' s angles in my problem. Wherever I have looked, the angle θ is taken as granted.



So, should I use something like the triangle? If yes, what the triangle should be? Should the Sun be on one angle as the image above with the convergent point? Should I define another triangle? In the end, for whatever triangle I have thought, how will I calculate the needed sides? Note that I know the right ascension and the declination of both the convergent point and the star's.

Thursday 4 July 2013

light - How does gravity interact with a photon?

It is simply not true that gravity can only interact with mass. Rather, any long-range spin-2 force interacts with all energy-momentum equally, and it source is the stress-energy-momentum tensor. That is one way to state the equivalence principle.



Note that a massive object in its own rest frame has an associated energy $E = mc^2$, which under ordinary conditions is usually much higher than any stress terms (including pressure) or momentum (outside the rest frame). Thus one can usually pretend that gravity couples to mass, but it isn't so--rather, the gravitational charge is energy, and the entire stress-energy-momentum tensor couples to the gravitational field. This is analogous to how for electromagnetism, there is electric charge, but the electric currents also make a difference.



In the weak-field limit of general relativity, one can consider the fact that light is gravitationally deflected twice as much as a naive Newtonian prediction would be (for an object at the speed of light under Newtonian mechanics, that is) as the fault of light putting a pressure in the direction of its propagation equal to its energy density. But in general, the motion of a test particle is determined by its four-velocity and the geometry of spacetime only, and the light would just be a special case of having a 'lightlike' four-velocity.

Tuesday 2 July 2013

gravity - Do all the objects in the universe exert force on all other objects?

No. It's impossible for every object to interact with every other object, due to the assertion by general relativity, that the universe can, and is, expanding faster than the speed of light.



I then assume that the universe initially was expanding at, or close to the speed of light, and that it immediately after the big bang was expanding faster than the speed of light.



Some of the particles/forms of energy that would have reached us are also bound to have been "held back or deflected", even in the young stages of the big bang, and are now at a distance at which they can never reach us. They could have been held back by for example a black hole.



Potentially, if the expansion of the universe at one point was so slow that gravity from every particle had time to propagate to every other particle, then yes - every particle and energy in the universe affects every other particle.