Wanton Waste and Pollution of the Air and Light and Soil

Can you figure out why is the northwestern corner of North Dakota lit up almost as bright as Manhattan?

I’m pretty sure I know.

And if you would like to see, in detail, what massive light pollution, air pollution, and increasing the CO2 content of the atmosphere looks like, you’ve come to the right place.

The map here is unique. I found it on  http://www.blue-marble.de/nightlights/2012 which allows you to see what the world looks like at night. As you might expect, big cities and their suburbs are all lit up, and remote, unpopulated places are mostly dark.

But there are some places way out in the boonies that are entirely toooo bright. Like northwestern North Dakota, as I hope you can see below.

waste and light pollution in north dakota

Part of that enormous blob of yellow in the center of the image is the super-bright lights on the oil rigs of the current North Dakota oil boom. The lighting, while probably rather cheaply and wastefully done, I at least understand. Drilling for oil in general, and fracking in particular, are dirty, dangerous and difficult jobs, and the work often goes on around the clock. Workers need to be able to see in order to be safe. However, I am sure that there are better lighting systems than ones that light up everything within 5 miles.

But that’s not the majority of that light.

Most of it is pure and simple waste.

Instead of bottling or piping out the natural gas (aka methane) that comes up along with the black,  oozing petroleum, they simply BURN OFF the gas.

It’s called “flaring”.

It’s a cold-blooded calculation by the corporate leadership: it is more profitable to them to burn up much of the gas than saving and selling it and using it later. So they light up enormous plumes that  light up the sky, literally 24/7, adding humongous amounts of carbon dioxide to the atmosphere, and warming up the planet both directly and indirectly. And turning that part of the Great Plains into something resembling Dante’s Inferno.

Oil companies say they are selflessly pursuing “energy independence” for the US.

Don’t you believe it. They are selfishly pursuing profits. If they were really interested in simply producing more energy for the good citizens of the USA or wherever, then all of that gas would be bottled up or saved to be used later in our stoves, heating systems, factories, or vehicles, where people need it.

Instead of burning it off for nothing.

What a waste.

That’s capitalism in a nutshell.

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Quoting from the NYT: (2011)

NEW TOWN, N.D. — Across western North Dakota, hundreds of fires rise above fields of wheat and sunflowers and bales of hay. At night, they illuminate the prairie skies like giant fireflies.

They are not wildfires caused by lightning strikes or other acts of nature, but the deliberate burning ofnatural gas by oil companies rushing to extract oil from the Bakken shale field and take advantage of the high price of crude. The gas bubbles up alongside the far more valuable oil, and with less economic incentive to capture it, the drillers treat the gas as waste and simply burn it.

Every day, more than 100 million cubic feet of natural gas is flared this way — enough energy to heat half a million homes for a day.

The flared gas also spews at least two million tons of carbon dioxide into the atmosphere every year, as much as 384,000 cars or a medium-size coal-fired power plant would emit, […]

All told, 30 percent of the natural gas produced in North Dakota is burned as waste. No other major domestic oil field currently flares close to that much, though the practice is still common in countries like Russia, Nigeria and Iran.

With few government regulations that limit the flaring, more burning is also taking place in the Eagle Ford shale field in Texas, and some environmentalists and industry executives say that it could happen in Oklahoma, Arkansas and Ohio, too, as drilling expands in new fields there unlocked by techniques like hydraulic fracturing and horizontal drilling.

“North Dakota is not as bad as Kazakhstan, but this is not what you would expect a civilized, efficient society to do: to flare off a perfectly good product just because it’s expensive to bring to market,” said Michael E. Webber, associate director of the Center for International Energy and Environmental Policy at the University of Texas at Austin.

If you’d like to see close up photos of flaring, look at NYT here.

Published in: on January 28, 2013 at 2:53 pm  Comments (5)  
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HATOVAUM (Hydrogen-Alpha Transit of Venus Astronomical Unit Measurement)

HATOVAUM

(Hydrogen-Alpha Transit of Venus Astronomical Unit Measurement)

Re-calculating the Astronomical Unit using the upcoming Transit of Venus — IN A NEW WAY

I think I have an original idea — perhaps for the first time in my life.

My proposal might allow us to measure, fairly accurately, one of life’s important questions: HOW BIG IS OUR UNIVERSE — using entirely equipment owned or built and operated by amateur astronomers.(*)

 
The general question to be solved is, how far it is from the Earth to the Sun. That distance, the Astronomical Unit, was first historically measured during previous Transits of Venus (which are very, very rare). Sadly, neither the accuracy nor the precision of results of these earlier measurements were much to write home about — and those were the days when “writing home” meant entrusting your letter to a ship that may or may not make it to its destination without being attacked, destroyed in a storm, or just sunk to the bottom of the ocean for reasons unknown.BTW: every single seafaring/colonialist nation or empire of the last three centuries has sent its very best scientists to tackle this problem, starting in the 1760s.

THIS IS NOT A TRIVIAL QUESTION.

The main difficulty is that the apparent parallax of Venus on the face of the sun was very small, even if your two observers were as far apart as possible. Plus, the Sun just appeared pretty much like a blank, featureless white or yellow disk with the occasional sunspot. Photographs were mostly used to help document contacts number 1 through 4. The famous “ink drop” problem made timing those contacts almost impossible, since different observers interpreted the phenomenon in different ways before. (Why did they get caught flat-footed like that? Simple: nobody had EVER done what they were doing before! The first time anyone tries something for the first time, there will be mistakes!)

My proposal is different. Forget about timing the contacts. Forget about the two parallel chords. None of that is necessary.

All we need is ten to twenty (or more) experienced solar imagers who know how to use a “hydrogen-alpha” filter (or dedicated solar telescope) to make sharp, crisp, in-focus and well-tuned images of the incredibly detailed surface of the Sun — at a large number of simultaneous events, for as long as they can. (The sun passing behind clouds, the computer runningout of disk space, the electricity cutting off, the sun going down, requirements for eating and household chorse, and so on are possible reasons for not continuing)

All these images must be time-stamped!!. Later on, all of these images from, hopefully, just about all over the world, get uploaded to some central repository. We take two images that were taken at the same instant from two observers, say, one in Australia, and one in British Columbia. We use some image processing software to align the two photos according to all of the easily visible h-alpha features on the surface of the sun — except for the two shadows of Venus. Those two shadows will block the view of slightly different locations on the face of the sun, which will be separated by roughly the diameter of Venus itself, if we calculated correctly, for those two locations. Then a few calculations give the angular parallax from this stereovision picture (’cause that’s what it really is!), and a few more calculations gives the distance from Earth to Venus, and then the diameters of the orbits of every single planet or comet or asteroid in our solar system, AND the sizes of the sun and everything around it, and so on and so on.

Having lots and lots of images means that we can beat down the errors to very low levels. How low, obviously, remains to be seen. Clearly some dry-run experiments involving some simultaneous images taken by as many observers as possible, spread over as many continents as possible, will be necessary.Again, this can be done strictly by amateur astrophotographers who have invested in their solar and astro equipment what an amateur or professional automobile mechanic might have in his/her tool set.Obviously I think this is a pretty cool project. But there’s not much time to get organized: the event is June 5/6, and it’s already May 9.Then again, I’m not asking these solar imagers to do anything they wouldn’t want to do already. I am sure that anybody who has an h-alpha scope is thinking very seriously about trying to take images of this event. After all, it won’t happen again in the lifetimes of anybody alive today. No one.

Exactly how the images should be made is up for discussion. I suspect that video recordings starting at each minute of viable observation, on the minute, stacked as each observer sees fit, and sent out in some sort of raw format (not compressed into a JPEG) will be advisable, but I am by no means any sort of an imaging expert. (Heck, I’m not even a duffer!)

So, if you know anybody who owns a hydrogen alpha filter or dedicated telescope, please put them it touch with me, so we can get this project going ASAP, because time really is of the essence. We already have committed observers in the USA, Pakistan, and the UK, but that’s just a start.

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(*) This measurement of the Astronomical Unit has been done before by scientists using a variety of methods. The best of them involved measuring parallaxes of asteroids and of Mars against either the Sun or the distant stars, as well as satellites and probes sent to the various planets. Most attempts using Venus were rather disappointing. I have not yet calculated how large the error bars will be in this proposed experiment, but I am hopeful that with lots of simultaneous images, we will be able to get those errors down to quite decent levels.

A Possible Origin for Dark Energy: Light!

An article by astronomer Tamara Davis of Australia in the July 2010 Scientific American got me to thinking about dark energy and redshifts.

The title of her article, “Is the Universe Leaking Energy?” is quite provocative. She points out that since photons from extremely distant galaxies are getting red-shifted, that means they are actually losing energy, since higher-frequency, shorter-wave-length (“bluer”) photons that are emitted by these galaxies have higher energies than the lower-frequency, longer-wave-length (“redder”) photons that we observe here on earth, billions of years later, with our eyes or telescopes. And where’s all that energy going?

She’s right: it’s a deep question.

Let me give some details. Suppose a star in some far-away galaxy emits some hard-UV (ultraviolet) light in our direction, at a wavelength of 100 nanometers. You can’t see this light with your eyes, but it can give you a nasty sunburn if you receive too much of it.  (100 nm = 100 x 10^-9 m, or just 10^-7 meters.)

I just looked it up and found that each of these light photons has an energy level of about 12.4 electron-volts.* (Google it yourself if you don’t believe me.)

Let’s also suppose that by the time these photons reach Earth, they have been red-shifted so far that they are now in the mid-infrared range, about 10 microns long, which you can’t see with your naked eye, but you can feel as heat on your skin. However, astronomers and others have built plenty of ingenious detector devices that can “see” IR photons quite well.  My same source says that each of these photons now has an energy of 0.124 electron volts, which is 100 times smaller.

(10 microns means 10 x 10 ^-6 meters, or just 10^-5 m)

In other words, since the decimal point in 12.4 got moved two places to the left in 0.124, the red-shifted photon only has ONE PERCENT of the energy that contained in the photon that was originally emitted. They do some really nifty detective work** with spectral ‘fingerprints’ of light to figure out exactly how much of a red shift occurs.

The red-shift in my hypothetical example is ordinarily assigned to the letter z (just like h often means “height” or the greek letter lambda (λ) is often used to denote the wavelength of something. Here is a formula to compute z:

wavelength observed (λo) minus wavelength emitted (λe) , all divided by the wavelength emitted (λe), gives you z.

Or z = (λo – λe) / λe.

So in our example, z would be (10^-5 – 10^-7)/10^-7.

There are lots of ways of working this, but I think I would prefer a simple method, which involves simplifying it by multiplying the top and bottom of the fraction by 10^7, aka ten million. If we do that and then simplify, we have (100 – 1)/1, which even I can do. I get a red shift of 99, which is rather big.

A different example: if a blue photon, with wavelength of 460 nanometers, gets red-shifted to a red wavelength of about 700 nanometers, it will have a redshift of z = (700 – 460) / 460 or about 0.525. *** In this case, the red light has an energy of about 1.77 electron-volts, and the blue light has an energy of 2.70 eV. Similar to finding the discount, we can calculate (2.70 – 1.77)/2.70 to calculate what percent of the energy was lost. I get that 34.4% of the energy is lost, much less than the first example.***

In any case, astronomers and others have been wondering where that 34% or that 99% of the original energy disappeared to.  And if you think about it for a while, you realize that this is an ENORMOUS amount of energy. Every second, our sun, and every other star, radiates astronomically huge amounts of energy, enough to keep us mostly fairly warm and comfortable here on Earth, and we are 93 MILLION miles away. (For comparison, my 2003 Subaru Forester has recently passed 93,000 miles; so if I continued at that rate, on a hypothetical highway to the Sun, it would take me about eight thousand years to drive all the way to dear old Sol. (Imagine the price for the fuel alone!)

And if 10% or 50% or 90% of that energy is simply disappearing, then astrophysics has a major problem, since that is a HUGE amount of energy. Our Sun, for example, emits roughly 3.84 x10^26 Joule/sec, which is about 9.2 x 10^25 calories, or umpty-gazillion times all of the nuclear explosives ever made or set off by all of humanity, all at once.

Now imagine, if you can, 50% or 99% of that incredible violent energy somehow disappearing into the dark reaches of space… That would be a theft along the lines of all those bricks and bales of hundred-dollar bills that were sent to Iraq and Afghanistan by the American occupation forces there, and which never, ever reached the people they were supposedly intended to help – but they did go somewhere. Where to? The logical conclusion is into the pockets of a handful of corrupt and well-placed officials….

But, as Davis writes: “Total energy must be conserved. Every student of physics learns this fundamental law… This principle, called conservation of energy, is one of our most cherished laws of physics …. The trouble is, it does not apply to the universe as a whole.”

Or so she says. Perhaps it really does apply?

There are lots of alternative explanations for the observed expansion and red shift and acceleration of the expansion of the universe. I will let you, dear reader, explore most of those other explanations on your own, but I’ll discuss a couple of them.

The most simple one is the red-shift is caused by the fact that the distant galaxy, and we ourselves, are rushing away from each other instead of towards each other. (This is an effect you notice every time a fast truck, ambulance, or car speeding by you: the musical pitch of the noises of the vehicle is higher when it is heading more-or-less towards you, and, assuming that the vehicle doesn’t hit you and keeps on going, the pitch is considerably lower when it is moving away. Sirens have sort of an “eeh-ahh” sound to them as they pass you.

Another possible explanation is that part of the theory of relativity indicates that it takes a certain amount of energy for a photon to “rise up” from a deep “energy well”. This is most evident near a black hole – any photon that enters the ‘event horizon’ will never be able to get out because gravity overwhelms it. But if a ray of light starts out at some point outside the event horizon and ends up escaping, it will have to lose energy to do so,  causing its wave length to get longer (or for it to get red-shifted).

However, it seems to me that this sort of gravitational-well red-shift would mostly apply to cases where photons are, in fact, escaping from the regions around black holes. And there just aren’t all that many black holes out there. Yes, all galaxies, and even most globular clusters have black holes at their very centers, but this effect only applies if the light source is right nearby – and that’s generally not the case.

Another problem is explaining the expansion of the universe in the first place, why the expansion was so fast at first that it got the name “inflation”, and why this expansion seems to be speeding up in the fast few billion years, caused by some increasing, mysterious force called “dark energy” – about which there is much debate and interest.

It occurred to me (though I was never a hard-science major in college, and am not a physics or astronomy professional, simply an interested amateur) that there might be  a relatively simple way to conjoin these two phenomena.

One is simply that this dark energy which causes this expansion, perhaps is nothing other than the energy “lost” when a red shift occurs.

In other words, every time a photon gets red-shifted and hence loses energy, that identical amount of energy goes to “pushing apart” the source and the target. In other words, it is part and parcel of the expansion of the universe itself. They are two sides of the same coin, so to speak.

However, this basically only applies to really, really distant objects. We can measure red- and blue-shifts of objects in our own galaxy, which is not noticeably expanding, and these relatively small shifts actually do correspond to the speed of the object relative to us. A week ago, I got to use a radio telescope dish for the very first time, with a group of other amateurs who attended a star party in West Virginia. We traveled to the US National Radio Observatory at Green Bank, WV and we saw red and blue shifts there as clouds of ordinary, atomic hydrogen were emitting their typical 21-cm radio lines and rushing this way and that. Our records consist of scrolls of paper with needle marks on them, looking a lot like what you get when you have an EKG at a hospital, or a seismograph recording earth tremors, and I got to take home a couple of them.

But how about the Doppler-effect at “small” distances like this inside the Milky Way, as my group was trying to measure?

I suspect that the loss of energy even in those relatively nearby photons (radio waves are photons, too; they just have very much longer wavelengths and lower frequencies than visible light) actually is the same as the one seen at the very edges of the observable universe. The energy “lost” in redshifting from, say 1420.4 megahertz to 1420.3 megahertz is not huge – only about 0.007%, I calculate, but it all adds up.

The bigger the redshift, the farther away the object is, the more the light itself, in losing its energy, is helping to push the universe apart.

This also might help to explain the twin phenomena of early-universe inflation and recent-times acceleration in a way that is consistent, perhaps, with the big bang hypothesis. When the universe was extremely young – much less than a second old – there was an enormous density of matter and energy concentrated, somehow, at a single point. If my theory has any validity whatsoever, then the initial mind-boggling inflation hypothesized by Alan Guth was instantly translated into, or was caused by, an absolutely phenomenal redshift from an amazingly high-frequency, dangerous x-ray universe to the point where this cosmic infrared background radiation now has a wavelength of about 2 millimeters (2×10^-3 m), lengthened from about 1 micron (10^-6 m or even much shorter), which means that virtually all of its energy was dissipated in the very inflation itself – and was inseparable from it.

In the relatively early universe, there were apparently not very many stars, so there were perhaps not too many photons to push the universe apart by getting red-shifted.

However, in the past 5 billion years, more and more stars and galaxies have been forming, as gravity has drawn ordinary clouds of hydrogen together, and thus there have been more and more photons. As they are emitted in greater and greater numbers, they get red-shifted, and this powers more and more expansion, which appears to be speeding up, according to measurements made by people who understand a lot more than I do. But it means that there is no need to hypothesize an additional repulsive particle that makes the dark energy that is expanding our universe. It’s simply the energy of light. Light energy. And it explains both inflation AND acceleration.

Here is the conclusion of Davis’ article, disagreeing with my conclusion:

“In the end … there is no mystery to the energy loss of photons: the energies are being measured by galaxies that are receding from each other, and the drop in energy is just a matter of perspective and relative motion.

“Still, when we tried to understand whether the universe as a whole conserves energy we faced a fundamental limitation, because there is no unique value we can ever attribute to something called the energy of the universe.

“Thus, the universe does not violate the conservation of energy; rather it lies outside that law’s jurisdiction.”

I don’t know if my hypothesis is reasonable, or whether it is nonsense. However, I’m pretty sure I don’t like Davis’ conclusion. Perhaps my explanation is better, but perhaps not.

Comments are welcome.

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* You may be wondering what on earth an electron volt is. It’s defined as the amount of energy needed to raise one electron’s electric potential by exactly one volt. It’s not a large quantity of energy, because electrons are so small and one volt is small, too. In fact, to make the energy represented by one single, ordinary calorie, you would need about 26,131,952,998,320,304,000 electron volts. – or what The New Yorker would fully spell out as twenty-six quintillion, one hundred thirty-one quadrillion, nine hundred fifty-two trillion, nine hundred ninety-eight billion, three hundred twenty million, three hundred four thousand electron-volts. Or somebody with less time to waste might write as being about 2.6 x 10 ^ 19 eV.

(By the way, I am referring to “calories” with a lower-case “c”, not a capitalized “C”. They are not the same, which is a bit confusing. A Calorie, the type of thing we typically associate with food, means exactly one thousand calories. Another word for Calorie is “kilocalorie” or “kcal”, which some of you may have encountered in some sort of science class. An ordinary lower-case calorie was originally defined as the amount of heat or other energy that is needed to raise exactly one cubic centimeter of water (aka one gram) by one degree Celsius (aka Centigrade).

** The details of that detective work are too complicated to go into here, but let me just say it’s a little bit similar how geologists reconstruct time lines in various strata of rocks, or tree experts figure out how old a given piece of timber is, based on the the lines representing climate data, left in the growth rings of trees.

*** Note that I left off all of the units! When you divide a unit by itself, say inches by inches or grams by grams, you get a pure, dimensionless ratio.

**** You can find a usable online calculator here for converting wavelengths of light and energies:

http://www.highpressurescience.com/onlinetools/conversion.html

Published in: on September 3, 2011 at 8:58 pm  Comments (4)  

How High the Moon? (Part Deux)

I have an experimental data point for the zenith angle of the moon last night.

We have a skylight in our bathroom. I was able to see the glow from right next to the moon, directly to the south, through the frosted glass. But not the moon itself, no matter where I stood. It appeared that I had lucked out and happened upon the moment when the moon was the highest in the sky. Or so I thought.

This morning I used a tape measure to measure the distance from the top of the skylight to where my head was (92 inches), and the horizontal distance that my head moved (40 inches). Assuming those form a right angle, and using the arc-tangent function on my calculator,  I get a zenith angle of about 23.5 degrees.

I don’t believe it!!!! It should be less!

Published in: on December 23, 2010 at 2:45 pm  Comments (2)  
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Why is the moon so high in the sky in winter?

And was it in fact directly overhead last night, near the beginning of the eclipse?

It sure looked like it was to me – though I didn’t take any measurements because I was only wearing my pajamas, my coat, and my slippers as I stood in the freezing cold on the snow-free but still-frozen concrete walk in front of our south-facing, Northeast DC  house.

[Yeah, I was being wimpy, only going out twice all night to look at the eclipse, but I was really tired, and I had to get up in the morning to give a full day’s guest lesson on astronomy to four, 70-minute middle school classes for a fellow teacher, so it was kind of  out of the question to stay up all night. (There is no way I could have followed through with the lessons if I had!)]

Maybe I’m just weird, but I have from time to time noticed, and marveled at, the fact that during the winter, the moon at times appears like it’s almost directly overhead. Let me emphasize that: to my unaided, subjective vision, without taking the trouble to measure it, during the winter, the moon sometimes appears to me to be directly overhead (at zenith).

However, everything I know about astronomy of the solar system tells me that this is probably impossible, simply because we do not live in the Tropics (with a capital T: the zone between the Tropic of Cancer and the Tropic of Capricorn). That’s the only part of our planet where the sun is ever directly overhead. (Don’t believe me? Use your internet resource skills and look it up. I’m not going to tell you just how to do that, because since you are reading this blog, you already know how.)

If you live in Washington,  the Sun will never appear directly over your house, no matter where you live in Washington, DC, and no matter how hot it may feel in the middle of summer.

And I figured that if the Sun and the Earth and the Moon were all aligned with each other, as in last night’s lunar eclipse, then the Moon would appear in our sky here in Washington as if it had simply traded places with the Sun for a while, and was at the same elevation. And that elevation just ain’t all that high.

Or so I thought.

Was I suffering from a version of the famous ‘moon effect’? (Which is a poorly-understood but almost-universal optical illusion about the apparent size of the moon,  a visual hallucination of sorts, caused by some internal human visual processing “bug” inside the various centers responsible for actually interpreting the photons and light waves that enter one’s eyes.)

Or is everybody else normal and it’s just me?

Or was the moon, in fact, at the zenith?

Or just very close to it, but within the theoretical and experimental range of error for this sort of thing?

I am going to try to settle this in two ways.

First of all, theoretically.

I used a rather widely-used piece of instructional geometry software called “Geometer’s Sketchpad” (version 5 in this case) and a couple of drawing and painting programs. I also used Google Earth to find out where on Earth are the places that are directly south of Washington and are on the Tropic of Cancer or on the Equator, as well as the spot on our planet that is diametrically opposite in position to Washington, DC.

I was rather surprised to find out where those places were. They weren’t really where I expected, and I of all people should have known better.

For example, I thought I remembered that Havana, Cuba, was just inside the Tropics, but Miami, Florida, was just north of the Cancerous Tropic. Or was that the Topic of Cancer? (Ha, ha, that was two intentional puns. If you don’t get them, or don’t think they are funny, that’s fine with me.) And I also remembered having been to some places in Florida that it was south of DC.  So I kinda figgered that the Tropic of Cancer would intersect our DC line of of longitude (about 77 degrees west) somewhere in the water between Havana and Miami.

Surprise: not very close. Just for fun, try guessing or figuring out the answer yourself. I’ll hide the answer at the end of this column, at (1).

And directly south of DC, on the equator? I always kinda figgered it would be somewhere in Brazil.

No surprise this time, I was wrong again. When I looked carefully, I discovered that 77 W and 0 degrees N or S is located… (2).

How about the point diametrically opposite to Washington, on the exact other side of the globe? Well, on this one I was fairly close. But calculating where this is, is a bit tricky. The latitude is OK. Any point at X degrees north is directly opposite some point that is X degrees south. So wherever it is, it’s at 39 degrees south. But the longitude is harder, because for most locations, Y degrees west is not opposite Y degrees east. What you have to do is change your latitude by exactly 180 degrees. Now here, you can either add or subtract. I would prefer to subtract, here. So 180 minus 77 gives us 103. (Of, if you prefer, 77 minus 180 gives -103.) And the way I interpret that 103, or -103, is to consider that as being 103 degrees east longitude.

Now knowing that DC’s literal antipode is roughly located at 39 degrees south and 103 degrees west, can you guess, or find, where that is? (3)

Here is the diagram that I made.

Bottom Line: if my diagram is correct, the full moon last night, at its greatest elevation or altitude last night, should have been about 15.5 degrees from the vertical (or 74.5 degrees from the horizontal). And that angular distance from the zenith should have been clearly and plainly obvious.

But it wasn’t. To me.

Now that’s just last night. Is it possible for the moon to be inclined a bit to the apparent orbit of the sun – that is – when the moon is not undergoing an eclipse? And can that cause the moon to be even higher in the sky than it was during last night’s eclipse?

Answer: YES. The moon ‘s orbit around the Earth is inclined by just about 5 degrees from the Sun’s apparent orbit. Thus, in different years and months, the details of which I will ignore right now ’cause it’s way too complicated for this here blog today, the moon might be as high as 10.5 from the vertical (79.5 degrees from the horizontal).

Next time: actual measurement

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Spoiler

answers below

or you could interpret my diagram..

(1) 77W and 23.5N turns out to be right next to Nassau, the capital of the Bahamas!

(2) it was about 100 or so miles east of Quito, Ecuador, along the banks of some huge jungle river that probably flows into the Amazon River, but doesn’t even seem to have a name. Nor any towns. Or roads.

(3) It’s located several hundred km, mi, or nm west-south-west of Perth, Australia, in the middle of the Indian Ocean. No land for hundreds of leagues in any direction, as the pirates or sailors or yarntellers of yore might say.

Total Lunar Eclipse Visible on the night of December 20-21

Lunar Eclipses are really amazing. The  one coming up on the date of the 2010 winter solstice starts rather late at night (around 1:30 AM on Tuesday the 21st of December if you live in the US eastern time zone), and takes about an hour to reach totality at roughly 2:41 AM. The moon will start leaving the earth’s shadow at about 3:53 AM.

The moon will actually enter the earth’s penumbra (partial shadow) about an hour or so earlier, but it’s hard to see.

Sometimes, the eclipsed moon is a bright coppery color. Other times it appears blue-ish, or brown. Other times it’s nearly totally invisible. Its exact appearance to the naked eye depends on how deeply it enters the earth’s total shadow, as well as on what’s going on in the earth’s atmosphere. Will our various levels of air bend enough light to illuminate the moon, or will our atmosphere be too full of dust and dirt to do so? You can only find out if you get up to watch it yourself.

A couple of web pages to peruse for further information:

http://science.nasa.gov/science-news/science-at-nasa/2010/17dec_solsticeeclipse/

and

http://shadowandsubstance.com/

and

http://eclipse.gsfc.nasa.gov/eclipse.html

Published in: on December 18, 2010 at 4:19 pm  Comments (1)  
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First Post

Hello, everybody,

This is my first attempt at putting up and maintaining a blog.

My main interests here will be education, public policy, math, astronomy, and science.

Recently retired from DC Public Schools, I have been doing some research on educational progress (or lack thereof) in that school system before and after the reign of our media-darling school Chancellor, Michelle Rhee. A lot of people have been throwing around a lot of rhetoric about how they are trying to put children first, but there has not been nearly enough in the way of actual facts. I have been trying to remedy this imbalance.

Some of my previous entries on this topic were sent as emails to various email list-serves made up of people interested in bettering the schools, but in a rather different way than Rhee. I will try to collect some of my previous posts and publish it here. One of those list-serves is http://groups.yahoo.com/group/concerned4DCPS/

I also posted on http://realeducationreformdc.blogspot.com/ which is a good resource as well.

Warning: since we have lots and lots of schools in DC, a good bit of my data will be in the form of huge spreadsheets comparing schools where Rhee replaced the previous administrations with the schools that did not experience such changes. The data take a while to wade through, and the arithmetic that I had to do was tedious in the extreme, but the results are quite informative. If you can compare, add, subtract, multiply and divide two decimals, and if you understand the meaning of percents and can read a graph, you should have no difficulty in understanding the results. (No algebra, geometry, or calculus is needed!)

I will have to learn how to post the spreadsheets and, in general, how to set up and maintain this blog.

I have another website that I put up about five years ago. Most of the things on it are still valid, but I imagine most of the links are probably broken. FWIW, here is the URL:

http://home.earthlink.net/~gfbranden/GFB_Home_Page.html

GFB

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