the sun - Solar Elevation Angles -- Anomaly?

"The solar elevation angle is the altitude of the sun, the angle between the horizon and the centre of the sun's disc." - Wikipedia

The meridian is when the sun is directly facing earth and is at exactly 90° overhead at some point along the longitudinal line at a given location.

My question is, if the sun is 93 million miles away and travels overhead at 90° between the Tropic of Capricorn and the Tropic of Cancer, how is it that there are Meridian solar elevation angles reported from relatively nearby cities on Earth of less than 89.99°?

Let's take for instance Calgary, Canada as of today May, 04 2016 -- estimated ground distance from Calgary to where the Sun is at 90° on Calgary's longitudinal line is ~3000 miles. The solar elevation angle (aka "altitude") is reported by the timeanddate website to be 55°! -
TimeandDate

Reversing the angle, using right triangle math:

The base angle, our solar elevation angle, of a right angle triangle is:

    "solar elevation angle"= arctangent (h/a)

    89.998151749049 = arctangent (93000000/3000)

So using right triangle math alone with the presumption of the Sun being 93 million miles away and base distance of 3000 miles we get an expected solar elevation angle of 89.998151749049°.

That makes sense, since if something is 93 million miles away from us and you think of it as a triangle, then the distance between any place on earth and the location on earth where the sun is at exactly 90° overhead -- you should expect the other angle of the triangle to be at near 90°, connected by a relatively tiny sliver of distance in a much elongated triangle to the center point of the distant sun.

If we go the other direction and leave out the presumption that the sun is 93 million miles away. We take h as unknown, we know the reported solar elevation angle of 55°, we can again use right triangle math:

    Let a=3000, solar elevation angle=55°
    h = a * tangent ("solar elevation angle")
    4,284.4440202263 = 3000 * tangent(55°)

So according to my calculations, based on the advertised solar elevation angle of 55° at Calgary, Canada on May, 04 2016 at Meridian, at a ground distance of 3000 miles from where the Sun is at exactly 90° overhead --  the Sun is actually only 4,284.4440202263 miles high!

** Update **: The math would work out for a sphere Earth with a radius of 4000 miles, using the exact solar position and adding corrective angles due to a presumption of a spherical curvature:

    Calgary, Canada: 51°03′N 114°04′W
    Location of Sun at Calgary Meridian: 16°23′N 114°04′W
    Distance = 2395.386 miles
    Radius of Earth=4000 miles
    Arc distance = 4790.772 miles
    Width of arc = 4509.52059 miles
    Angle to center of the Earth=68.622754110849°
    68.622754110849°/2=34.31137705542450°

This math utilized calculations that may have had a bit less acurracy but to make it work you need to add a bandaid padding of approximately 34.31137705542450° to Calgary,Canada's advertised angles. If you do you arrive at 55°+34.31137705542450° = 89.31137705542450° which is close to what I would expect to see. If, we are truly living on a Flat Earth, then the calculations would be telling us that the Sun height is only 4,284.4440202263 miles high and we can use simple trignometry.
Arc Calculator
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If the math is difficult for you, you can check with an online triangle calculator such as: Ke!san

solar system - How did Jupiter form where it is?

It's at Jupiter's distance and beyond that ices were able to form out of the disk of material surrounding the early sun. Go much further in and there is too much energy from the sun for them to stay as solids (and will sublimate into gases); this is why asteroids are principally rocks and metals. So at this distance more of the materials of the planetary disk can form the base planetesimals.

At this point, your question is answered largely by a geometric consideration (or two) and one of Kepler's laws.

First, the geometric consideration. A circle of radius $r$ has area $pi r^2$. The bigger the radius, the more area. The material for Jupiter (or any other planet) came from an annulus: stuff outside one circle, but inside a slightly bigger circle. This annulus had a lot more area for the out planets than the inner planets, and so could contain a lot more mass.

Of course, that could makes us think that Jupiter shouldn't be the largest of the gas giants: it's the closest of them all to the sun, after all. The density of the disk needn't have been approximately constant throughout these regions, though. Quite possibly the density was such that Jupiter's region had more mass than the areas for the other planets. As HDE's answer (posted as I was finishing this) points out, these ices probably also helped stop materials from passing into the inner solar system, maintaining a higher density than you might otherwise expect in the inner solar system, as well as causing materials to sort of "dam up" right around Jupiter's orbit.

Now for the Kepler's law. The further you get from the sun, the slower your orbital period. Picking the correct units, we have $P^2=a^3$, where $P$ is the period measured in years, and $a$ is the semi-major axis of the orbit measured in AU. The further out you go, the slower you go around the sun; indeed, it's not just that it takes you a longer total time, but your actual velocity goes down. We can also see this as a Newton's law of gravity consequence. At Jupiter's furthest point from the Sun, the escape velocity is a little more than 18 km/s. At Saturn's maximum distance, the escape velocity drops to around 13.25 km/s. So things can go roughly 35% faster within Jupiter's orbit than they can closer to Saturn's, and they have less distance to travel to make a complete orbit.

What this means is that it takes longer for planetesimals to get close enough to each other to accrete together the further out you go, and there is a longer mean time between collisions.

Now, eventually, the Sun "turned on" and started blasting space with it's solar wind (before that, the heat came primarily from thermal radiation from the gravitational contraction of the sun). This ended up clearing out most of the unaccreted particles out of the solar system, stopping planetary growth (and removed portions of existing atmospheres; a very young Earth probably had a lot of H and He in its atmosphere, until the sun hit it with enough energy to knock any of it not locked up in rocks away).

So Jupiter was probably in a bit of a goldilocks situation. The average density of the region it formed in was probably higher than where the other giants formed in, it was at the perfect spot for lots of materials to start accreting early, and the accretion process would have been faster. So Jupiter is growing faster, and this gives it a competitive advantage: the bigger the growing planetesimals get the further their influence extends and the faster they can pull in more materials, and subsequently interfere with the growth of other planets (or planetesimals). Somewhere around a mass of 10-15 earth masses, the giants can start pulling in large quantities of the hydrogen and helium gasses. And, again, Jupiter likely hit this mass well before the other giants, and had more material to pull from, so it became much larger than the others could before the solar wind stopped the process.

expansion - Isn't the date of the Big Bang a bit bogus?

If I understand correctly, the date of the Big Bang is an extrapolation of acceleration of the universe's expansion through time based on the erroneous assumption that the universe is approximately 13 billion years old. The age of the universe, as far as I can tell is based on the speed of light and the fact that the furthest we could possibly see into the depths of the universe is approximately 13 billion light-years. Isn't that sort of a cat chasing his tail? It seems that there is no valid reason for believing that the universe is really of a radius of ~13 billion years. That is just as far as we can see. If this is wrong and there is a valid reason for believing that the universe is ~ 13 billion years old, then that fact makes for a very interesting possibility. That the expansion rate of the universe is equal to the speed of light, which leads to a very interesting suggestion that I pursued for quite awhile (until it became apparent that the suggested size of the universe is not what it seems), that the expansion rate of the universe is equal to the speed of light. Or, put another way, is it possible that the speed of light is based on the expansion rate of the universe?

orbit - Are trojans in L5 more likely than in L4?

There is a symmetry in the rotating gravitational field, which means that capture of an asteroid to the L4 is just as likely as to the L5 Lagrange point.

In the case of Mars the split is 1:6, and a simple binomial model suggests a probability of 0.0625 (the probability of a 1:6 or 0:7 split given the hypothesis that they are distributed randomly is 0.0625) This doesn't give a reason to suppose there is anything unusual happening, and as noted in the comments, there is no favouring of L5 when other planets are considered.

As noted by UserLTK: "Earth has an L4 asteroid, none in L5. Uranus also has an L4, no L5 and Neptune has more L4 than L5 objects."

The conclusion is that No, the greater number of L5 martian trojans is just a random effect.

How rare are earth-like solar eclipses?

We can certainly speculate.

There are dozens of moons in our own Solar System. I've done some preliminary calculations of their apparent size as seen from the planet vs. the apparent size of the Sun at that distance. The large moons (Jupiter's 4 Galilean satellites, Saturn's Titan, Neptune's Triton) are all substantially larger in the sky than the Sun is. I think there are some satellites that are fairly close to the apparent size of the Sun, but I haven't done all the calculations.

The apparent size of our Moon has changed over time, as the Moon has gradually moved farther away. We are coincidentally in a period of history in which it happens to be very nearly the same apparent size as the Sun.

The criterion for an exoplanet to have solar eclipses like the spectacular ones we have here on Earth is a moon that happens to have an angular size large enough to cover the photosphere, but not so large that it also hides the corona. For a larger moon, you'd have eclipses in which the corona is visible, but not all the way around the moon. For a smaller moon, you'd only have annular eclipses. (It could also get interesting if you consider eclipses as seen from other moons.)

So given the distribution of sizes and distances of moons in our Solar System, we can guess that situations where the apparent size of a moon and sun closely match is fairly rare, but since there are moons whose apparent size is smaller than the Sun and others whose apparent size is larger than the Sun, it probably happens sometimes.

The one piece of the puzzle that we're still missing, I think, is the typical sizes and distances of satellites of Earth-like planets. Moons the size of ours orbiting habitable planets might be common or very rare. If they're rare (say, if Mars-like moon systems are far more common), then Earth/Moon style eclipses might be very rare for habitable planets.

orbit - When will all eight planets in our solar system align?

Any estimate of the common period of more than two planets (i.e., after how much time do they approximately align in heliocentric longitude again?) depends very strongly on how much deviation from perfect alignment is acceptable.

If the period of planet $i$ is $P_i$, and if the acceptable deviation in time is $b$ (in the same units as $P_i$), then the combined period $P$ of all $n$ planets is approximately $$P approx frac{prod_i P_i}{b^{n-1}}$$ so reducing the acceptable deviation by a factor of 10 means increasing the common period by a factor of $10^{n-1}$, which for 8 planets is a factor of 10,000,000. So, it is meaningless to quote a common period if you don't also specify how much deviation was acceptable. When the acceptable deviation declines to 0 (to achieve "perfect alignment"), then the common period increases to infinity. This corresponds to several commenters' statements that there is no common period because the periods are not commensurate.

For the planets' periods listed by harogaston, $prod_i P_i approx 1.35times10^6$ when the $P_i$ are measured in Julian years of 365.25 days each, so the common period in years is approximately $$P approx frac{1.35times10^6}{b^7}$$ if $b$ is measured in years as well. If the periods are approximated to the nearest day, then $b approx 0.00274$ years and $P approx 1.2times10^{24}$ years. If the periods are approximated to the nearest 0.01 day, then $b approx 2.74times10^{-5}$ and $P approx 1.2times10^{38}$ years.

The derivation of the above formula is as follows:

Approximate the planets' periods by multiples of a base unit $b$: $P_i approx p_i b$ where $p_i$ is a whole number. Then the common period is at most equal to the product of all $p_i$. That product is still measured in units of $b$; we must multiply by $b$ to go back to the original units. So, the common period is approximately $$P approx b prod_i p_i approx b prod_i frac{P_i}{b} = b frac{prod_i P_i}{b^n} = frac{prod_i P_i}{b^{n-1}}$$

The above derivation doesn't take into account that the $p_i$ might have common factors so that the alignment occurs sooner than $prod_i p_i$ suggests. However, whether or not any two $p_i$ have common factors depends strongly on the chosen base period $b$, so it is effectively a random variable and does not affect the global dependence of $P$ on $b$.

If you express the acceptable deviation in terms of angle rather than time, then I expect you'll get answers that depend on the size of the acceptable deviation as strongly as for the above formula.

See http://aa.quae.nl/en/reken/periode.html for a graph of $P$ as a function of $b$ for all planets including Pluto.

EDIT:

Here is an estimate with acceptable deviation in terms of angle. We
want all planets to be within a range of longitude of width $δ$
centered on the longitude of the first planet; the longitude of the
first planet is free. We assume that all planets move in the same
direction in coplanar circular orbits around the Sun.

Because the planets' periods are not commensurate, all combinations of
longitudes of the planets occur with the same probability. The
probability $q_i$ that at some specific moment of time the longitude
of planet $i > 1$ is within the segment of width $δ$ centered on the
longitude of planet 1 is equal to $$q_i = frac{δ}{360°}$$

The probability $q$ that planets 2 through $n$ are all within that
same segment of longitude centered on planet 1 is then $$q =
prod_{i=2}^n q_i = left( frac{δ}{360°} right)^{n-1}$$

To translate that probability to an average period, we need to
estimate for how much time all planets are aligned (to within $δ$)
each time they are all aligned.

The first two planets to lose their mutual alignment are the fastest
and slowest of the planets. If their synodic period is $P_*$, then
they'll be in alignment for an interval $$A = P_* frac{δ}{360°}$$ and
then out of alignment for some time before coming into alignment
again. So, each alignment of all planets lasts about an interval $A$,
and all of those alignments together cover a fraction $q$ of all time.
If the average period after which another alignment of all planets
occurs is $P$, then we must have $qP = A$, so $$P = frac{A}{q} = P_*
left( frac{360°}{δ} right)^{n-2}$$

If there are only two planets, then $P = P_*$ regardless of $δ$, which is as expected.

If there are many planets, then the fastest planet is a lot faster
than the slowest one, so then $P_*$ is very nearly equal to the
orbital period of the fastest planet.

Here, too, the estimate for the average time between successive
alignments is very sensitive to the chosen deviation limit (if there
are more than two planets involved), so it is meaningless to quote
such a combined period if you don't also mention what deviation was
allowed.

It is also important to remember that (if there are more than two
planets) these (near-)alignments of all of them do not occur at
regular intervals.

Now let's plug in some numbers. If you want all 8 planets to be
aligned to within 1 degree of longitude, then the average time between
two such alignments is roughly equal to $P = 360^6 = 2.2×10^{15}$
orbits of the fastest planet. For the Solar System, Mercury is the
fastest planet, with a period of about 0.241 years, so then the
average time between two alignments of all 8 planets to within 1
degree of longitude is about $5×10^{14}$ years.

If you are satisfied already with an alignment to within 10 degrees
of longitude, then the average period between two such alignments is
roughly equal to $P = 36^6 = 2.2×10^9$ orbits of Mercury, which is
about 500 million years.

What is the best alignment that we can expect during the coming 1000
years? 1000 years are about 4150 orbits of Mercury, so $(360°/δ)^6
approx 4150$, so $δ approx 90°$. In an interval of 1000 years
chosen at random, there is on average one alignment of all 8 planets
to within a segment of 90°.

cosmology - How can cosmic inflation make an infinite universe homogeneous?

Inflation is used to explain why the observable universe is extremely homogeneous.

Without inflation, we can do the following crude calculation. The cosmic microwave background was formed about 300,000 years after the big bang, at a redshift of about 1100. Thus causally connected regions at the epoch of CMB formation would have a radius of $sim 300,000$ light years, which has now expanded by a factor of 1100 to be $3.3times 10^{8}$ light years in radius.

This can be compared with the radius of the observable universe, which is currently around 46 billion light years. This means that causally connected regions should only be $sim 4 times 10^{-7}$ of the observable universe, or equivalently, patches of CMB of $sim 2$ degrees radius on the sky are causally connected. This is clearly not the case as the variations in the CMB are no more than about 1 part in $10^{5}$ across the whole sky.

Inflation solves this by allowing previously causally connected regions to inflate to become larger than the entire observable universe.

You appear to understand this quite well, so I am not entirely clear what your question is. We cannot know whether the entire universe is homogeneous, since we cannot measure it. The cosmological principle is an assumption that appears to hold approximately true in the observable universe, but need not apply to the universe as a whole. Indeed it is not absolutely true in the observable universe otherwise it would be quite uninteresting, containing no galaxies, clusters, or other structure. I think the only requirement on inflation is that it blows up a patch of causally connected universe so that it becomes much bigger than the observable universe at the current epoch.

observational astronomy - Universal point for sky observation

With the axis of rotation of the earth matching the axis of rotation of the stars, and the equator of the earth matching the equator of the stars, any place along the equator would match what you have described. However due to closeness to the horizon, the polar-most stars wouldn't be visible. If you were able to get a hundred metres or so above the ground, with nothing in the surrounding landscape, you could see all of the stars in the sky over a year's time.

navigation - uses of coordinate systems

Wikipedia describes the five following celestial coordinate systems each with rectangular/spherical variants:

• horizontal


• equatorial


• ecliptic


• galactic


• supergalactic

What are the practical preferred uses of each, in relation to the others. If you were navigating within the solar system, which system would you use, or has been used by existing craft?

For example wikipedia describes the ecliptic system as "still useful for computing the apparent motions of the [celestial bodies]", which seems pretty bloody obsolete. Even though it seems to me as the most intuitively useful system.

Why are there no stars on New Horizons images of Pluto

I followed the New Horizons Mission a little, and saw among others this image of Pluto:

I wonder, why you can't see any stars on it. As far as my very basic knowledge in astronomy goes, I think you can even see stars during a "Sun-Moon-Eclipse" here on earth. So besides having here a spectacular "Sun-Pluto-Eclipse", I further more ask myself if that eclipse is even necessary to see other stars, as those pictures are already taken quit far away from sun, and outside an atmosphere.

Is that not enough to see stars? Or does NASA remove the stars from the images, before publishing them? (If so, why? Not to distract?)

Thanks for answers!

star - Conversion from Equatorial Coordinate to Horizon Coordinates

It depends from which celestial coordinates you are converting. I suppose you are working in equatorial coordinates. Then, you can convert as following:

$$tan A = {sin h over cos h sinphi_o - tandelta cosphi_o} $$ $$ begin{cases}
cos a sin A = cosdelta sin h \
cos a cos A = cosdelta cos h sinphi_o - sindelta cosphi_o
end{cases}$$ $$sin a = sinphi_o sindelta + cosphi_o cosdelta cos h$$

Where $A$ is azimuth, $a$ altitude, $alpha$ right ascension, $delta$ declination, $h$ hour angle and $phi_0$ the observer's latitude.

Source: Wikipedia

the sun - Why don't these Clouds show no depth perception in correspondence to the Sun

It may not seem related, but here's a detailed answer: http://photo.stackexchange.com/questions/27643/why-dont-cameras-capture-dynamic-range-as-our-eyes-do

The camera doesn't differentiate between different colors and brightness nearly as well as our eyes do so in the photos you linked, both the clouds in-front of the sun and sky that's kind of around but looks behind the sun appear the same color and to our brains, it appears the sun is rising in the middle of the clouds - like a tall building or mountain peak might appear in the middle of some clouds.

If you were there in person when the photo was taken, this illusion probably wouldn't be there. (nice photos though).