## The Christmas Star

Posted December 13, 2008

on:

‘This is the season… And every year around this time people notice the brilliant ‘star’ to the west just after sunset. For astronomers, we know it’s the appearance of the planet Venus, but noticing it for the average person brings on questions about the holidays. Was the Christmas Star real?

Regardless of your faith, the story of the ‘Star of Bethlehem’ is one of the most powerful and enigmatic symbols of Christianity. For centuries, scientists, scholars and historians have debated about the nature of this biblical light that heralded an event. Was it purely a divine sign, created miraculously to mark a special birth? Or was it an astronomical event in its own right?

David Reneke, news editor of Australia’s Sky and Space Magazine, believes astronomers may have found the answer – or at least something that fits all the known facts – basing his research on the highly esteemed gospel according to Matthew, the first of the four gospels in the New Testament. It would appear to be the first written and this version places key players together in the same time period. “It’s generally accepted by most researchers that Christ was born between 3 BC and 1 AD.” says Dave. With the aid of modern astronomy software programs astronomers can reproduce the night sky exactly as it was, thousands of years ago. Humans are curious and so was Dave, so he turned back the hands of time and the stars to the time of that long ago Christmas…

“We found out something startling.” said Reneke, “It looks like the ‘Christmas star’ really did exist,”

Two thousand years ago, astronomy and astrology were considered one and the same. The motions of the heavenly bodies were used to determine the events of history, and the fate of people’s lives. Of the various groups of priests and prophets of this period, those which commanded the most respect were the Magi – whose origins are not entirely clear. Known as ‘wise men’ , we can only assume they were actually priests who relied on their knowledge of astronomy/astrology.

Armed with an approximate date, Dave assumed the ‘Star of Bethlehem’ was not just a localized event and could be observed by sky-watchers elsewhere in the world, not just by the Magi. Historical records and modern-day computer simulations indicate a rare series of planetary groupings, also known as conjunctions, during the years 3 BC and 2 BC In fact, this was one of the most remarkable periods in terms of celestial events in the last 3,000 years!

“Like the final pieces of a difficult jig-saw puzzle, our fabled biblical beacon is starting to reveal itself,” David said. “On 12 August, 3 BC, Jupiter and Venus appeared very close together just before sunrise, appearing as bright morning ‘stars.’ It would have been visible in the eastern dawn sky of the Middle East from about 3:45 to 5:20 a.m.”

But it didn’t stop there. The crowning touch came ten months later, on 17 June 2 BC, as Venus and Jupiter joined up again in the constellation Leo. This time the two planets were so close that, without the use of our modern optical aids, they would have looked like one single, brilliant star. According to Dave’s research, Jupiter was known as the “planet of Kings” and Saturn as the “Protector of the Jews”. This could easily have been interpreted as a sign that the Jewish Messiah had been, or was about to be, born. Also, Leo was thought to denote royalty and power. An interpretation? Perhaps. But, do not forget the times in which this occurred. Astronomy and astrology intermingled. This whole sequence of events could have been enough for at least three astrologers to see this as sign in the heavens and make their way Jerusalem.

“Now, this doesn’t mean that astrology works,” Reneke said. “We haven’t ruled out other possibilities for the Star of Bethlehem but it does make our search more rewarding to find a truly interesting astronomical event that happened during the most likely time for the Nativity.”

Whatever the Star of Bethlehem was, it has had more impact on humankind than any star before or since. It is also possible that the mystery of the Star will never be completely solved. For many of us though, it is the mystery itself that drives us to find the solution.

Source : *David Reneke*

David Reneke, one of Australia’s most well known and respected amateur astronomers and lecturers, has over 40 years experience in astronomy with links to some of the world’s leading astronomical institutions. David is also the News Editor for Australia’s Sky and Space Magazine, he teaches astronomy at college level, is an invited speaker at astronomy conventions throughout Australia, a feature writer for major Australian newspapers, and is a science correspondent for ABC and commercial radio stations. In these weekly radio interviews David regularly appears on about 60 networked stations around the nation with all the latest news and on general astronomy and space discovery issues.

Cited from : universe today

## Photometry (Part 1)

Posted December 7, 2008

on:- In: Material
**2**Comments

Photometry is a branch field of astrophysics which learned the quantity, quality, and the direction of the electromagnetic radiation from the sky’s objects. “Photo” in photometry which means “visual light” was used because the observation was used to limited in visual light.

Photometry was based on our knowledge about radiation law. We hypothesize that the astronomical objects have characteristics of a hypothetical black body.

The characteristics of the black body are:

- when thermal equilibrium is achieved, the object’s temperature is a function of how many energy it absorbs per second
- a black body doesn’t emit radiation in all wavelength in same intensity (some emits more radiation in blue region wavelength, and the way around. The wavelength which it emits most will determine its color).

The wavelength its emits most (**λ _{maks}**) by a black body which temperature is

**Kelvin is :**

*T***λ _{maks} = 0,2898/ T …………………….. (eq. 1)**

(**λ _{maks}** expressed in

*cm*and

**in**

*T**Kelvin*)

The equation 1 is called ** Wien’s rule**.

An example of implementing the Wien’s rule :

(Warning : be clear that λ_{maks} doesn’t mean the maximum wavelength but it means the wavelength that a body emits in the biggest intensity)

The total energy per time emitted by a black body per its surface area is called *emitted*** energy flux**. The value of the emitted energy flux from a black body with a surface temperatur

**Kelvin is :**

*T** F*** = σ T^{4} …………………….. (eq. 2)**

(σ : Stefan-Boltzman constant: 5,67 x 10^-8 Watt/m^{2}K^{4})

The total energy per unit time (= Power) that’s emitted by a black body with the surface temperature ** T** Kelvin and surface area A is known as

**. Its value (**

*Luminosity***) can be calculated by equation below:**

*L**L*** = A σT^{4} …………………….. (eq. 3)**

For stars, we can assume it’s a perfect sphere. So, its surface area (A) is 4π*R*^{2} ; with ** R** express star’s radius. So, a star’s luminosity (

**) is equal to :**

*L**L*** = 4π R^{2} σT^{4} …………………….. (eq. 4)**

he black body emits its radiation to all direction. We can assume the radiation pass through a sphere surface with a radius d in same** energy flux** (

**).**

*E**E*** = L/(4πd^{2}) …………………….. (eq. 5)**

This amount of flux is received by an observer from a distance ** d **from the black body. So, this flux is usually called

**or**

*received**energy flux***. (**

*brightness***Warning**: differ between

**and**

*E***).**

*F*The equation above is often termed as the inverse square law for brightness (E) because this equation shows that brightness is inversely proportional to the square of its distance (d). So, the farther the distance, the less bright it is.

**Review Questions:**

- From an observation result, we know that an area of 1 cm
^{2}in Earth’s outer atmosphere received Sun’s energy with intensity of 1,37 x 10^{6}erg/cm^{2}/s. If we know that the distance between Sun and Earth is 150 million kilometres, determine the Sun’s luminosity. - Calculate the intensity of Sun’s radiation received by Saturn’s surface if we know that the distance between Saturn and Sun is 9,5 Astronomical Units (use the information from number 1)?
- The luminosity of a star is 100 times larger than Sun is, but the temperature of the star is only half of Sun’s temperature. Calculate the radius of the star expressed in Sun’s radius unit?
- Define the term Luminosity and Brightness using your own words
- Calculate the wavelength of maximum intensity radiation of a star which temperature is 10.000 Kelvin.

- In: trivia
**2**Comments

Can you identify at least **TWO** constellations shown in the photograph above? Also, can you mention in what direction (north or south) this photograph was taken? You can submit the answer in the comment section.

Dive right in to this image that contains a sea of distant galaxies! The Very Large Telescope has obtained the deepest ground-based image in the ultraviolet band, and here, you can see this patch of the sky is almost completely covered by galaxies, each one, like our own Milky Way galaxy, and home of hundreds of billions of stars. A few notable things about this image: galaxies were detected that are a billion times fainter than the unaided eye can see, and also in colors not directly observable by the human eye. In this image, a large number of new galaxies were discovered that are so far away that they are seen as they were when the Universe was only 2 billion years old! Also…

This image contains more than 27 million pixels and is the result of 55 hours of observation, made primarily with the Visible Multi Object Spectrograph (VIMOS) instrument. To get the full glory of this image, here’s where you can download the full resolution version. It’s worth the wait while it downloads. Or click here to be able to zoom around the image.

In this sea of galaxies – or island universes as they are sometimes called – only a very few stars belonging to the Milky Way are seen. One of them is so close that it moves very fast on the sky. This “high proper motion star” is visible to the left of the second brightest star in the image. It appears as a funny elongated rainbow because the star moved while the data were being taken in the different filters over several years.

The VLT folks describe this image as a “uniquely beautiful patchwork image, with its myriad of brightly coloured galaxies.” It shows the Chandra Deep Field South (CDF-S), one of the most observed and best studied regions in the entire sky. The CDF-S is one of the two regions selected as part of the Great Observatories Origins Deep Survey (GOODS), an effort of the worldwide astronomical community that unites the deepest observations from ground- and space-based facilities at all wavelengths from X-ray to radio. Its primary purpose is to provide astronomers with the most sensitive census of the distant Universe to assist in their study of the formation and evolution of galaxies.

The image encompasses 40 hours of observations with the VLT, just staring at the same region of the sky. The VIMOS R-band image was obtained co-adding a large number of archival images totaling 15 hours of exposure.

Source: ESO

Cited from : universetoday

After we talk about parallax, now we will discuss about angular diameter.

**I. Definition**

The angle that the actual diameter of an object makes in the sky; also known as *angular size* or *apparent diameter*. The **angular diameter** of an object as seen from a given position is the “visual diameter” of the object measured as an angle. The visual diameter is the diameter of the perspective projection of the object on a plane through its center that is perpendicular to the viewing direction. Because of foreshortening, it may be quite different from the actual physical diameter for an object that is seen under an angle. For a disk-shaped object at a large distance, the visual and actual diameters are the same.The Moon, with an actual diameter of 3,476 kilometers, has an angular diameter of 29′ 21″ to 33′ 30″, depending on its distance from Earth. If both angular diameter and distance are known, *linear diameter* can be easily calculated.

The Sun and the Moon have angular diameters of about half a degree, as would a 10-centimeter (4-inch) diameter orange at a distance of 11.6 meters (38 feet). People with keen eyesight can distinguish objects that are about an arc minute in diameter, equivalent to distinguishing between two objects the size of a penny at a distance of 70 meters (226 feet). Modern telescopes allow astronomers to routinely distinguish objects one arc second in diameter, and less. The Hubble Space Telescope, for example, can distinguish objects as small as 0.1 arc seconds. For comparison, 1 arc second is the apparent size of a penny seen at a distance of 4 kilometers (2.5 miles).

The angular diameter is proportional to the actual diameter divided by its distance. If any two of these quantities are known, the third can be determined.

For example if an object is observed to have an apparent diameter of 1 arc second and is known to be at a distance of 5,000 light years, it can be determined that the actual diameter is 0.02 light years.

**II. Formulas**

The angular diameter of an object can be calculated using the formula:

in which δ is the angular diameter, and *d* and *D* are the visual diameter of and the distance to the object, expressed in the same units. When *D* is much larger than *d*, δ may be approximated by the formula δ = *d* / *D*, in which the result is in radians.

For a spherical object whose *actual* diameter equals *d*_{act}, the angular diameter can be found with the formula:

for practical use, the distinction between *d* and *d*_{act} only makes a difference for spherical objects that are relatively close.

**III. Estimating Angular Diameter**

This illustration shows how you can use your hand to make rough estimates of angular sizes. At arm’s length, your little finger is about 1 degree across, your fist is about 10 degrees across, etc. *Credit: NASA/CXC/M.Weiss*

**IV. Use in Astronomy**

In astronomy the sizes of objects in the sky are often given in terms of their angular diameter as seen from Earth, rather than their actual sizes.

The angular diameter of Earth’s orbit around the Sun, from a distance of one parsec, is 2″ (two arcseconds).

The angular diameter of the Sun, from a distance of one light-year, is 0.03″, and that of the Earth 0.0003″. The angular diameter 0.03″ of the Sun given above is approximately the same as that of a person at a distance of the diameter of the Earth.[1]

This table shows the angular sizes of noteworthy celestial bodies as seen from the Earth:

Sun | 31.6′ – 32.7′ |

Moon | 29.3′ – 34.1′ |

Venus | 10″ – 66″ |

Jupiter | 30″ – 49″ |

Saturn | 15″ – 20″ |

Mars | 4″ – 25″ |

Mercury | 5″ – 13″ |

Uranus | 3″ – 4″ |

Neptune | 2″ |

Ceres | 0.8″ |

Pluto | 0.1″ |

* Betelgeuse: 0.049″ – 0.060″

* Alpha Centauri A: ca. 0.007″

* Sirius: ca. 0.007″

This meaning the angular diameter of the Sun is ca. 250,000 that of Sirius (it has twice the diameter and the distance is 500,000 times as much; the Sun is 10,000,000,000 times as bright, corresponding to an angular diameter ratio of 100,000, so Sirius is roughly 6 times as bright per unit solid angle).

The angular diameter of the Sun is also ca. 250,000 that of Alpha Centauri A (it has the same diameter and the distance is 250,000 times as much; the Sun is 40,000,000,000 times as bright, corresponding to an angular diameter ratio of 200,000, so Alpha Centauri A is a little brighter per unit solid angle).

The angular diameter of the Sun is about the same as that of the Moon (the diameter is 400 times as large and the distance also; the Sun is 200,000-500,000 times as bright as the full Moon (figures vary), corresponding to an angular diameter ratio of 450-700, so a celestial body with a diameter of 2.5-4″ and the same brightness per unit solid angle would have the same brightness as the full Moon).

Even though Pluto is physically larger than Ceres, when viewed from Earth, e.g. through the Hubble Space Telescope, Ceres has a much larger apparent size.

While angular sizes measured in degrees are useful for larger patches of sky (in the constellation of Orion, for example, the three stars of the belt cover about 3 degrees of angular size), we need much finer units when talking about the angular size of galaxies, nebulae or other objects of the night sky.

Degrees, therefore, are subdivided as follows:

* 360 degrees (º) in a full circle

* 60 arc-minutes (′) in one degree

* 60 arc-seconds (′′) in one arc-minute

To put this in perspective, the full moon viewed from earth is about ½ degree, or 30 arc minutes (or 1800 arc-seconds). The moon’s motion across the sky can be measured in angular size: approximately 15 degrees every hour, or 15 arc-seconds per second. A one-mile-long line painted on the face of the moon would appear to us to be about one arc-second in length.

Source : Wikipedia and encyclopedia of science.

## Basic Astronomy (part 1)

Posted September 20, 2008

on:Before we learn further about astronomy, there are some basic knowledges that we must know and understand.

First, we will talk about measuring distance in astronomy.

Astronomical object lies in a very great distance from us. So far than our sense can perceive. That’s why our sense can’t have a 3-D visualization of the universe. Our sense can’t differ closer to farther objects. So, we need some trick to know how far an object from us. One of the simplest method used by astronomers to measure distance of some closest star is using the parallax effect.

Parallax is an optical effect seen when the observer seeing an object from two different positions. The object will be seen shifted relative to the farther background objects.

The parallax effect is one of those things you see everyday and think nothing of until it’s given some mysterious scientific-sounding name. There’s really no magic here. Consider the following simple situation.

You’re riding in a car on a highway out west. It’s a beautiful sunny day, and you can see for miles in every direction. Off to your left, in the distance, you see a snow-capped mountain. In front of that mountain, and much closer to the car, you see a lone ponderosa pine standing in a field next to the highway. I’ve diagramed this idyllic scene in the figure below:

As you drive by the field, you notice an interesting sight. When you’re in the position on the left side of the figure, the tree appears to be to the right of the mountain. You can see this in the figure by the fact that the line of sight to the tree (indicated by the green line) is rightward of the line of sight to the mountain (indicated by the blue line). A picture of what you see out the window of your car is shown below the car.

The interesting part is that as your drive on, you notice that the tree and mountain have switched positions; that is, by the time you reach the right hand position in the above figure, the tree appears to be to the left of the mountain. You can see this in the figure by noting that the line of sight to the tree (green line) is leftward of the line of sight to the mountain (blue line). A picture of what you see out the window of your car now is shown below the car.

What’s going on here? It’s pretty clear that the tree and mountain haven’t moved at all, yet the tree appears to have jumped from one side of the mountain to the other. By now, you’re probably saying *“Well, DUH, the tree is just closer to me than the mountain. What’s so remarkable about that?”* I would answer, *“There’s nothing at all remarkable about it. It’s just the effect of parallax.”* In fact, if you understand the above discussion, you already understand the parallax effect.

Now let’s talk about measuring the distance to the tree using this information. From the above information, you can see that it would be pretty easy to measure the angle between the direction to the tree and the direction to the mountain in both instances. Let’s call those angles **A** and **B**, respectively. Now, if the mountain is sufficiently distant so that the direction to the mountain from both viewpoints is the same, then the two blue lines in the figure below are parallel.

This helps a lot, because we can then show that the angle made by the two green lines (i.e., the difference in the direction to the pine tree from the two viewpoints) is equal to the sum of **A** and **B**. To see this, construct a line through the pine tree parallel to the two blue lines in the figure (this line is shown as a dotted line above). Then all of the blue lines are parallel, and each of the green lines crosses a pair of parallel lines. Reach deep back into your high school geometry (or equivalently, just stare at the above figure for a minute), and you’ll remember or realize that the angles at the pine tree labeled **A** and **B** have the same values as the angles **A** and **B** measured at the two car positions. Thus, the angle between the two green lines is the sum of **A** and **B**, which are angles we can measure from the comfort of our car.

Now, if we know the distance **D** we’ve traveled, then we have an Observer’s Triangle and we can solve for the distance to the tree using the Observer’s Triangle relation

**alpha/57.3 = D/R **where **alpha** is the angle at the tree (**A** + **B**), **D** is the distance we’ve traveled between views, and **R** is the distance from the road to the tree.** (source : Astronomy 101 Specials: Measuring Distance via the Parallax Effect).**

We will use the same method to measure the star’s distance. This method is called * trigonometric parallax* because we only use simple triangulation to find the distance. The only problem is star’s distance is so huge so the parallax effect will be so small (less than 1 arc second; 1 arc second = 1/3600 of a degree). So, that’s why this method can only measure accurately for several nearby stars. Farther star will need different, more complex, indirect method to derive its distance.

As explained before, the stars are so far away that observing a star from opposite sides of the Earth would produce a parallax angle much, much too small to detect (That’s why ancient people can’t detect this shifting to prove heliocentric view) . As a consequence, we must use large a baseline as possible. The largest one that can be easily used is the orbit of the Earth. In this case the baseline is the mean distance between the Earth and the Sun—an **astronomical unit** (AU) or 149.6 million kilometers! A picture of a nearby star is taken against the background of stars from opposite sides of the Earth’s orbit (six months apart). The parallax angle *p* is one-half of the total angular shift.

However, even with this large baseline, the distances to the stars in units of astronomical units are huge, so a more convenient unit of distance called a **parsec** is used (abbreviated with “pc”). A parsec is the distance of a star that has a parallax of one arc second using a baseline of 1 astronomical unit. Therefore, one parsec = 206,265 astronomical units. The nearest star is about 1.3 parsecs from the solar system. In order to convert parsecs into standard units like kilometers or meters, you must know the numerical value for the astronomical unit—it sets the scale for the rest of the universe. Its value was not know accurately until the early 20th century. In terms of light years, one parsec = 3.26 light years.

Using a parsec for the distance unit and an arc second for the angle, our simple angle formula above becomes extremely simple for measurements from Earth:

*p* = 1/*d*

Parallax angles as small as 1/50 arc second can be measured from the *surface* of the Earth. This means distances *from the ground* can be determined for stars that are up to 50 parsecs away. If a star is further away than that, its parallax angle *p* is too small to measure and you have to use more indirect methods to determine its distance. Stars are about a parsec apart from each other on average, so the method of trigonometric parallax works for just a few thousand nearby stars. The Hipparcos mission greatly extended the database of trigonometric parallax distances by getting above the blurring effect of the atmosphere. It measured the parallaxes of 118,000 stars to an astonishing precision of 1/1000 arc second (about 20 times better than from the ground)! It measured the parallaxes of 1 million other stars to a precision of about 1/20 arc seconds. Selecting the Hipparcos link will take you to the Hipparcos homepage and the catalogs.

The actual stellar parallax triangles are much longer and skinnier than the ones typically shown in astronomy textbooks. They are so long and skinny that you do not need to worry about which distance you actually determine: the distance between the Sun and the star or the distance between the Earth and the star. Taking a look at the skinny star parallax triangle above and realizing that the triangle should be over 4,500 times longer (!), you can see that it does not make any significant difference which distance you want to talk about. If Pluto’s entire orbit was fit within a quarter (2.4 centimeters across), the nearest star would be 80 meters away! But if you are stubborn, consider these figures for the planet-Sun-star star parallax triangle setup above (where the planet-star side is the hypotenuse of the triangle):

the Sun — nearest star distance = 267,068.23022*0* AU = 1.2948 pc

the Earth–nearest star distance = 267,068.23022*2* AU = 1.2948 pc

Pluto–nearest star distance = 267,068.23*3146* AU = 1.2948 pc !

If you are super-picky, then yes, there is a slight difference but no one would complain if you ignored the difference. For the more general case of parallaxes observed from any planet, the distance to the star in parsecs *d = ab/p*, where *p* is the parallax in arc seconds, and *ab* is the distance between the planet and the Sun in AU.

Formula (1) relates the planet-Sun baseline distance to the size of parallax measured. Formula (2) shows how the star-Sun distance *d* depends on the planet-Sun baseline and the parallax. In the case of Earth observations, the planet-Sun distance *ab* = 1 A.U. so *d = 1/p*. From Earth you simply flip the parallax angle over to get the distance! (Parallax of 1/2 arc seconds means a distance of 2 parsecs, parallax of 1/10 arc seconds means a distance of 10 parsecs, etc.)

A nice visualization of the parallax effect is the Distances to Nearby Stars and Their Motions lab (link will appear in a new window) created for the University of Washington’s introductory astronomy course. With this java-based lab, you can adjust the inclination of the star to the planet orbit, change the distance to the star, change the size of the planet orbit, and even add in the effect of proper motion. (**source : www.astronomynotes.com**)

**Units in Distance**

- Astronomical Unit (A.U). It is defined as the mean distance of the Sun from the Earth. Its value is about 149,6 million km. This unit is conveniently used to express distance to the object in solar system because we can directly compared the distance to Earth-Sun distance.
- One light year is defined as the distance that light has traveled in light years. Light has velocity about 300.000 km/s. So, one light year equals to 9,46 x 10^12 km. This unit is mostly used to express the distance of extragalactic object. Remember that light’s speed is finite so distant objects are seen as they are in the past. For example the Sun. The Sun that we see at this moment is the Sun as it was 8 minutes ago. Light needs about 8 minutes to travel the Earth-Sun distance. So, looking farther objects mean we’re looking even further to the past. That’s why light years is more commonly used to express distant object’s distance. When we say that a cluster’s distance is 8 billion light years, it means that the cluster that we seen right now is the way it looks 8 billion years ago !
- Parsec (
*Parallax second*). Star that have parallax 1 arc second have distance about 3,26 light years or 206.265 A.U (astronomical unit). Astronomer use this distance as a unit to express distance of the star. It is called a*parsec*. This unit is favorable to express star’s distance because it is closely related to star’s parallax (p). (remember that parallax = 1/distance, while the observer is on Earth, parallax is expressed in arc second and distance is expressed in parsec).

**So, for reviewing our understanding about the parallax, try to answer these questions:**

- If a star has parallax 0″,711, determine its distance (
*in light years*) from us! - Assume we can measure parallax from Mars (with the same technology that we used here on Earth). Assume that we can measure accurately using parallax method until 200 parsec from the Earth (distance limit). Determine the distance limit if we conduct the measurement of star’s distance using parallax method. Given that the distance of Mars from the Sun is about 1,52 AU.
- You observe an asteroid approaching the Earth. You have two observatories 3200 km apart, so you can measure the parallax shift of the incoming asteroid. You observe the parallax shift to be 0,022 degrees.Determine : (a) the parallax expressed in radians (b) the asteroid’s distance from Earth.
- If you measure the parallax of a star to be 0,1 arc seconds on Earth, how big would the parallax of the same star for an observer on Mars?
- If you measure the parallax of a star to be 0,5 arc seconds on Earth and an observer in a space station in the orbit around the Sun measures a parallax for the same star to be 1 arc seconds, how far is the space station from the Sun ?

You can share your solution of the above questions in the comment column.

## ANNOUNCEMENT

Posted September 20, 2008

on:Don’t forget to visit my main blog :

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