Posted on November 18th, 2010 No comments
No correct answers to this one. The floors of polar craters, where the Sun never shines, are the coldest places on Mercury.
It was long thought that Mercury is tidally locked to the Sun, as the Moon is to Earth, keeping one side always facing the Sun. There was once a hypothesis that the “dark side” of Mercury was among the coldest places in the Solar System. Now we know this is false.
In 1965, careful observations revealed that Mercury does indeed rotate with respect to the Sun. It has no “dark side.” All longitudes of Mercury are lit by the Sun at various times. Only the deep craters near the poles of Mercury are permanently shaded, and these are thought to contain vast quantities of water ice.
Mercury happens to have the smallest axial tilt of any planet in the Solar system. This ensures that the Sun never lights the polar craters. A crater at or near the poles of Earth, for example, would be lit by the Sun during that hemisphere’s summer, when that pole is tilted toward the Sun. Mercury’s tilt is so small that its polar craters are permanently in shadow.
More about Mercury: http://en.wikipedia.org/wiki/Mercury_(planet)
Posted on November 9th, 2010 No comments
There were several correct answers to yesterday’s question, but not all were as clear or complete as others. The first “complete” answer was from @jpwarren: Going to orbit of Mercury requires more fuel than leaving Solar System because must slow down. Quoting @jpwarren: “because you’re going deeper into the Sun’s gravity well and have to slow down.” Earlier correct but less clear/complete answers included:
- @jchaager: Entering orbit around mercury requires traveling faster than exiting the solar system due to the high gravity.
- @rozberk: I would say that it would take more power to fight increased gravitation forces than to exit from it.
And @rozberk had another answer about orbiting Mercury: “Because you’d just “have” to visit Vulcan while you’re there. :)”
As a spacecraft travels downsun, it gains kinetic energy (speed). To match Mercury’s orbital velocity, it must slow down tremendously. Thus, entering into orbit of Mercury with a starting point on Earth requires more rocket fuel than exiting the Solar System completely.
“Rocket fuel” is not necessarily the right term. The BepiColombo mission to Mercury will use solar ion drive to slow it down. This does not involve “rocket fuel” per se, but the spacecraft will still require a huge amount of propulsion to slow it down to match Mercury’s orbital velocity. The ion drive will do the job, but it will require a very long time to do so.
The current MESSENGER mission to Mercury uses conventional chemical thrusters, but it also makes extensive use of gravitational adjustments from planetary flybys to help it slow down to Mercury’s speed. That is why MESSENGER has already made three flybys of Earth since it was launched. Earth is helping to slow the spacecraft down (as are Venus, Mercury, and the Earth’s Moon).
BepiColombo is also making use of gravity assists from Earth, Venus, and Mercury to supplement the braking provided by its ion drive.
More about Mercury exploration: http://en.wikipedia.org/wiki/Exploration_of_Mercury and http://en.wikipedia.org/wiki/BepiColombo
Posted on October 15th, 2010 No comments
The first sort-of correct answer was from @rozberk: The “Hunter’s Moon” offers longer period of illumination than most full moons. But why? Nobody really explained that.
When the Earth is near its equinoxes, the Moon’s path (and the path of anything else that travels near the ecliptic) is nearly perpendicular to the horizon. Thus, it rises clear of the horizon more quickly than it does at other times of the year. This means the full Moon begins providing useful light soon after sunset in the months nearest the equinoxes. At other times of the year, the Moon lingers low on the horizon for much longer.
As some Tweeters pointed out, there are many myths regarding the “Hunter’s Moon” and “Harvest Moon” (the first full Moon after the September equinox). Most are false. The “Hunter’s Moon” is no nearer or brighter than any other full Moon. But it rises higher above the horizon more quickly than do most full Moons.
There is another subtle factor not mentioned in most sources that makes the Moon stay full longer in October than earlier in the year. I’ve never read this, but it is plain to see in any astronomical almanac.
After passing the September equinox, the Earth is accelerating, and moving closer to the Sun. The Moon has a hard time keeping up, so to speak. The interaction of Earth approaching perihelion and full Moon “keeping up” is very subtle, and varies depending on Moon’s position in its orbit relative to its own perigee and apogee. It could well happen, if the Moon is near perigee and near full in late October, it will move fast enough to cancel out the effect I’m describing here. However, in most years – and in every year that I’ve bothered to check – the Moon remains nearly full for longer in October and November than it does in the Earth’s slower half of its orbit.
Take the current “Hunter’s Moon” as an example. At “Full Moon” at 21:35 EDT on October 22, 2010, the Moon will be 99.8% illuminated (it’s never really 100% illuminated). However, it reaches 99.8% illumination at 14:40 EDT, and remains at 99.8% until 05:16 EDT on October 23. The Moon remains at “full” illumination for 14 hours and 36 minutes.
Going back 6 months, the Moon was full at 08:17 EDT on April 28, 2010. It reached 99.8% illumination at 02:21 EDT and remained at 99.8% until 14:57, a total of only 12 hours and 36 minutes.
Of course, the casual observer is not going to assign any significance to the fact that the Moon stays “full” for longer than 12 hours. The Sun will probably be up for part of that time, no matter where on Earth’s temperate zones you happen to be.
A more important effect of this phenomenon is the number of evenings when a casual observer would perceive the Moon as “pretty much full.” Astronomically speaking, the “Full Moon” is a particular instant in time when the Moon is exactly “downsun” from the Earth (though usually not in the same plane, which would be a lunar eclipse). It lasts only an instant. But in common parlance, one would be likely to call the Moon “full” if it were 90% or more illuminated.
Again, let’s use the current “Hunter’s Moon” as an example. The Moon will reach 90% illumination on October 19 at 18:27 EDT, and it will continue to be 90% or more illuminated until October 26 at 01:45 EDT. Thus, there will be seven evenings in a row when the farmer or the hunter might say there’s a Full Moon.
Around the April Full Moon this year, the Moon reached 90% illumination on April 25 at 18:11 EDT, and remained 90% or more illuminated until May 1 at 07:10 EDT. There were only 6 evenings of colloquially “Full Moon.”
Although I’ve never read this in any source, I really believe that this “lingering Full Moon” phenomenon may be at least as important a factor as the perpendicularity of the ecliptic to the horizon in the traditional attachment of special significance to the “Harvest Moon” and “Hunter’s Moon.”
More about “Hunter’s Moon”: http://en.wikipedia.org/wiki/Hunter%27s_moon#Variation_in_time_of_moonrise
Scientists speculate that an earth-size planet around Alpha Centauri A would be dry and airless. Why?Posted on October 5th, 2010 No comments
I can’t see any @ replies on Twitter since early August. Apologies to anyone who answered this question, correctly or otherwise, but Twitter is not 100% reliable. There’s even a hashtag for it: #twitterfail.
Scientists speculate that an earth-size planet at Alpha Centauri A would be dry and airless because, they speculate, there are no comets. There is a series of hypotheses and speculations behind this.
First, many scientists believe that Earth and other inner planets received their water and light elements via comet impacts over eons. I can’t find the source right now, but I heard one hypothesis that “microcomets” (approximately house-sized) continue to bombard the Earth every day, without being detected, and that such a bombardment from the time the Earth was formed would be sufficient to account for every drop of water in the oceans. Even without this hypothetical “microcomet” bombardment, it is widely accepted that much of our water and other volatile components of the atmosphere came from comets. It is not widely accepted that all of our water came from comets, but the hypothesis has its adherents.
Second, many scientists believe that Earth would have no comets if Jupiter did not deflect them toward inner solar system. Most scientist believe that early in the Earth’s formation, it was subject to bombardment by comets and asteroids for some hundreds of thousands of years, but that this period of heavy bombardment ended billions of years ago as the planets absorbed all of the smaller bodies. In effect, the Solar System ran out of bullets. Many scientists speculate that most of the comets that have struck the Earth in the last few billion years were deflected inward from the Kuiper Belt by gravitational perturbations from Jupiter and the other jovian planets. It is also widely beleived that at least some recent comets were perturbed out of the Oort Cloud by interactions with passing stars.
Third, we know for a fact that there are no jovian planets in the Alpha Centauri system. We would detect them if they were there. Our present exoplanet detection technology would surely see the indications of large planets in the Alpha Centauri system, and Alpha Centauri has been thoroughly scrutinized for many years. There are no large planets there.
So, if comets made Earth’s ocean and atmosphere, and Jupiter gave us comets, and the Alpha Centauri system has no jovian planets, an earth-like planet at Alpha Centauri A is dry and airless.
The question is a bit fuzzier in the case of Alpha Centarui B or Alpha Centauri C (Proxima Centauri). Some astronomers believe that Alpha Centauri A might be able to perturb the equivalent of the Kuiper Belt at Alpha Centauri B sufficiently that there may be quite a few comets in the inner parts of the Alpha Centauri B system. And Proxima Centauri might actually be within the equivalent of the Oort Clouds of either or both of the larger members of the Alpha Centauri system, so it might get plenty of comets.
In any case, all the members of Alpha Centauri are pretty much equally subject to perturbations of the equivalent of their Oort Clouds from passing stars as our own Solar System is. There would almost certainly be some comets near any of these stars. The question is, are there enough to enable planets near these stars to have oceans and atmospheres? Some scientists say no.
More about dry planets at Alpha Centauri A: http://en.wikipedia.org/wiki/Alpha_centauri#Theoretical_planets
Posted on September 20th, 2010 No comments
The best answer I received was from @rozberk, but I’m not quite sure it’s correct. He said the difference between a brown dwarf and gas giant is “density.” That’s close. More on why it’s not quite right later.
Like many such distinctions, some scientists disagree on the difference between a brown dwarf star and a gas giant planet. A brown dwarf has full convection from its interior to its surface. A gas giant planet is separated into layers, with denser material concentrated in the interior and less dense material in the atmosphere.
The evidence of full convection in a brown dwarf star would be heavy materials in its outer layers and atmosphere. However, that may depend on its composition, too. A gas giant planet that happened to be very rich in high-mass elements might look like a brown dwarf because of high-mass elements near its surface.
A better distinction is mass: a brown dwarf has mass greater than “X,” and a gas giant planet is less than “X” mass. (I’m not specifying the value of “X” because it’s a little fuzzy. Later.)
It might seem that this distinction by mass is arbitrary, but there is an important difference in behavior at masses greater than “X.” The best agreed-upon difference is that a brown dwarf has been hot enough to fuse deuterium (not normal hydrogen) at some point in its history, and in order to do so, it must have more than “X” mass. A brown dwarf may have used up all its deuterium and so no longer undergo fusion, although it once did.
If “X” is the mass necessary for deuterium fusion it’s not just an arbitrary number. It’s a real difference between a small “star” and a large planet.
The mass above which deuterium fusion can occur is about 13 times the mass of Jupiter, but it is affected by the composition of the “star.” The concentration of high-mass elements might either raise or lower the mass required for deuterium fusion. Thus, throwing composition into the mix, the distinction of brown dwarf/gas giant at 13 Jupiter-masses begins to seem arbitrary again. But it’s the best test we have.
Now, about that peculiarity related to @rozberk’s answer that “density” is the difference between a brown dwarf and a gas giant. There is a very odd thing about objects in the Jupiter-mass range and above: They’re all the same volume! More mass does not mean the object will have a bigger radius, but it will have greater density. For objects from about 1 to about 90 times the mass of Jupiter, density of a gas giant/brown dwarf is directly related to its mass.
A remarkable property of brown dwarfs is that they are all roughly the same radius as Jupiter. At the high end of their mass range (60–90 Jupiter masses), the volume of a brown dwarf is governed primarily by electron degeneracy pressure, as it is in white dwarfs; at the low end of the range (10 Jupiter masses), their volume is governed primarily by Coulomb pressure, as it is in planets. The net result is that the radii of brown dwarfs vary by only 10–15% over the range of possible masses. This can make distinguishing them from planets difficult.
In that sense, @rozberk was correct in answering “density.” More correctly, it is mass that distinguishes a brown dwarf star from a gas giant planet.
Of course, a planet is supposed to orbit a star. A Jupiter-like object on its own might best be called a brown dwarf. Or something else. Astronomers haven’t addressed this problem yet (as far as I know) because they have not yet detected such an object, and no current technology is likely to detect such an object.
More about brown dwarf stars: http://en.wikipedia.org/wiki/Brown_dwarf
Posted on September 8th, 2010 No comments
First correct answer was from @jchaager: Helium was first discovered on the sun, the only element first discovered in space.
The previously unknown element “helium” was proposed as an explanation for certain spectral emission lines found in the sun’s corona in 1868. Helium was discovered on earth (in natural gas deposits) in 1882, and confirmed to match the helium discovered on the sun.
The element “nebulium” was once proposed as an explanation of emissions from the Cat’s Eye Nebula, but they were later found to be emissions from doubly ionized oxygen.
More about helium: http://en.wikipedia.org/wiki/Helium
And about “nebulium”: http://en.wikipedia.org/wiki/Nebulium
What is the name of the most common explanation for why the night sky’s darkness proves the universe is finite?Posted on September 1st, 2010 No comments
The first correct answer was from @rozberk: Olbers’ Paradox says that in an infinite universe every point of the sky would be as bright as a star.
Since the time of Olbers (1823), the same argument has been used to demonstrate that there are a finite number of galaxies.
The mathematics and especially the physics behind Olbers’ Paradox get rather complex, but fundamentally, it’s almost intuitive. If there were an infinite number of stars, any straight line from your eye to “infinity” would end at a star. Even if there were intervening dust or gas absorbing the light of infinite stars, that material would become hot and would glow. If there were an infinite number of stars, the entire sky would be as bright as the surface of a star.
Introducing some of that complex physics for a moment, what if there were an infinite number of stars but that they came into being so recently that the light of some of them simply hasn’t had time to reach us? Well, it is notions like this that have led to refinements of Olbers’ Paradox that make it sound really complicated.
When really, really smart people really, really think about it, the blackness of the night sky proves that:
- The universe is not in a steady state that has persisted from an infinite past.
- The universe is not equally densely populated with stars.
- And a few other things.
Thus, the fact that the night sky is mostly black means that the universe had some beginning in time, and has some finite number of stars/galaxies in it.
The blackness of the sky, together with Olbers’ Paradox, supports the Big Bang Theory, but is not sufficient to prove it. (Not saying the Big Bang Theory is incorrect, just that this line of reasoning alone is not enough to prove it. There are other arguments, other sets of evidence, etc.)
More about Olbers’ Paradox: http://en.wikipedia.org/wiki/Olbers%27_paradox
Posted on June 23rd, 2010 No comments
First correct answer was from @mlv: Planets don’t twinkle because of their larger apparent size.
The twinkle phenomenon, technically “atmospheric scintillation,” is caused by shifting layers of atmosphere distorting light from a star. A star has infinitesimal cross-sectional area as seen from Earth. Therefore, even a slight change in atmosphere causes noticeable distortion. A planet has a finite cross-sectional area as seen from Earth. Atmospheric distortions within this small area tend to average out.
Picture the photons traveling from a star, through the Earth’s atmosphere to your eye. The beam of light that actually reaches your retina is so small in diameter that even a few extra molecules of air shifting in or out of it will make a slight difference in its position and brighness. And in the miles of atmosphere you’re looking through, there are uncountable billions of molecules of air even along that infinitesimally narrow beam of light. There are plenty of opportunities for that small cylinder of air to fluctuate by a few billions of molcules here and there. The result is that the star appears to twinkle.
In the case of a planet, the beam of light between the planet and your eye is of finite diamter. It’s pretty small, but not infinitesimal. Within each infinitesimal cross-section of that beam, the same sort of shifting of molecules and “twinkling” phenomenon occurs. However, there are enough of these tiny “twinkling” beams of light that, overall, they average out to a fairly steady position and brightness of the planet as seen from Earth.
The actual effect of the unsteadiness of the atmosphere on the light from a planet is that the details become fuzzy. There is a limit to how much detail you can see on a planet’s surface, no matter how powerful your telescope, as long as that telescope is inside the Earth’s atmosphere. Better to put your telescope into orbit. (Actually, most telescopes in Earth orbit are not used to observe planets, but more distant objects. But when they do point at planets, the images can be quite spectacular.)
Some sources say it is a myth that planets don’t twinkle. I say they’re splitting hairs. Yes, under extremely unstable seeing conditions but absent clouds, planets do twinkle. Yes, under extremely good seeing conditions, stars do not twinkle. But most of the time that the atmosphere is unstable enough to make planets twinkle, the sky is obscured by clouds. In the most common viewing conditions, if the atmosphere is clear enough to see stars, the stars twinkle and the planets don’t. In the most common viewing conditions, an otherwise untutored observer can tell stars from planets fairly reliably by their twinkling or not.
More about twinkling stars: http://en.wikipedia.org/wiki/Scintillation_(astronomy)
Some stars _do_ appear distinctly red or blue. If not due to redshift/blueshift, why are stars different colors?Posted on June 21st, 2010 No comments
First answer was from @rozberk: Stars appear different colors because of their different temperatures.
The colors of stars are related to the temperature of the star’s chromosphere, and therefore loosely related to the star’s overall temperature. Early astronomers did attribute the colors of stars to Doppler effect redshift/blueshift for a time, but this was soon proven to be incorrect.
Over the course of its lifetime, a star may exhibit a number of colors, but a main-sequence star will spend most of its life one color. The Sun is a class G “yellow” star, but most of the yellowish appearance is due to atmospheric scattering. In space, it’s pretty near white, but just a little yellowish.
If you don’t look closely, all stars appear pretty much white, and in fact, they are never far from white. But if you look closely, some stars are pale but unmistakably blue, yellow, orange, or red. Conveniently, we can see a pretty obvious blue star and a pretty obvious red star in one constellation. Orion’s heel is blue, and his hand is red. Rigel (ß Orionis), the heel, is a blue supergiant of magnitude 0.18. It is a pale, watery blue. Betelgeuse (α Orionis), whose name is thought to derive from “the hand of the giant,” is a red giant, varying between magnitude 1.2 and 0.3. It is plainly red.
To Nothern Hemisphere observers, when Orion is south of you, Betelgeuse is at upper left, Rigel at lower right. From the Southern Hemisphere, when Orion is north of you, Rigel is at upper left and Betelgeuse at lower right. If you’re on the equator, when Orion passes your meridian, it is right overhead, so it depends on which way you choose to face.
Gamma Orionis (Bellatrix, the Amazon) at upper right, is also a blue supergiant, but not so obviously blue as Rigel, in my opinion.
Perhaps the most obviously red star is Antares (α Scorpii), the “rival of Mars (Ares).” It does look like Mars, and is often brighter than Mars, but not always.
More about the colors of stars: http://en.wikipedia.org/wiki/Stellar_classification
Some sources say the star Sirius is red, but it’s white. What is the most accepted explanation for red Sirius?Posted on June 18th, 2010 No comments
First acceptable answer was from @rozberk: The most accepted explanation for some authors describing Sirius as red is “poetic license.”
I was actually looking for something else, but nobody mentioned it. On reflection, @rozberk’s answer is also correct.
The most accepted scientific explanation for “red Sirius” is that Sirius is subject to atmospheric scattering, just like the Sun and the Moon. When near the horizon, Sirius can appear reddened by its blue and green light being scattered away by Earth’s atmosphere. Most other stars are simply not visible close to the horizon, because they are too dim to shine through the thick atmosphere near the horizon.
Bear in mind that when you look straight up, you’re looking through only the depth of the atmosphere, about 60 miles of air, and most of it is very rarefied. When you look toward the horizon, you are looking obliquely through the atmosphere, through hundreds of miles of air, and through scores of miles of dense air near the surface of the Earth.
So, Sirius, being the brightest star in apparent magnitude in Earth’s sky, is visible through thick atmosphere close to horizon, but other stars are invisible when they are low enough that they would otherwise exhibit atmospheric scattering.
Some observers report flashes of red and blue from Sirius when it is close to the horizon, which is thought to be from atmospheric scattering.
Now, @microform mentioned J.K. Rowling and “Sirius Black.” I must assume that’s a reference to a character in books I haven’t read. I’m too old — my children are too old — to be Rowling fans. Most of my grandchildren are too young.
More about “red Sirius”: http://en.wikipedia.org/wiki/Sirius#Red_controversy