All About Astronomy

Archive for the ‘Material’ Category

Looking through a telescope at the stars there is very little information we can gain from them. To be sure, we know what color they are and we can see that some are more luminous than others. If we use a spectrograph we can tell what elements they are made up from. From these facts alone, it is difficult to tell just how much mass they contain.

By looking at pairs of stars that orbit one another we can try to answer the question, how much mass do the stars have?

Binary stars can be of two fundamental types:

  • Visual Binaries
  • Optical Doubles
Alberio (Visual Binary)
Visual Binaries are stars that are clearly gravitational associated with one another. They orbit each other around a common center called the barycenter. Visual binaries can be seen optically through a telescope. Only a small portion of binary stars are visual binaries. In order to see a visual binary, the stars must be separated by fairly wide distances, and the orbital periods are usually very long.

Optical Doubles are stars that appear to lie close together, but in fact do not, they only appear to us from our earthly observation to be close together. One of the stars in the pair is actually behind the first star and very far away. The stars of an optical double are not gravitationally bound.

William Herschel began looking for optical doubles in 1782 with the hope that he would find a measurable parallax, by comparing a close star to the more distant star in an optical double.
Herschel did not find any optical binaries, but he did catalog hundreds of visual binaries. In 1804 Herschel had so many measurements of visual binaries that he concluded that a pair of stars known as Castor were orbiting one another. This was an important discovery, because it was the first time observational evidence clearly showed two objects in orbit around each other outside of the influence of our own Sun and Solar System.

Spectroscopic Binary
It is also possible to detect binary stars using a spectroscope. If two stars are orbiting each other they will both produce a spectrum. If the stars are close to being the same brightness it is possible to see different spectral lines from both stars. These stars are of particular interest because it can be used to determine the radial velocity of the orbit of the two stars. Stars appear red shifted when receding away from the earth and blue shifted as they approach. This effect is caused by the Doppler effect which distorts arriving light waves from the stars depending on the direction if their motion. A Spectroscopic binary will alternate between blue and red shifted spectral lines.


Spectroscopic binaries are not detectable if we are seeing the star head on because no Doppler shifts would be present in the spectrum. If the Doppler shifts are present in a single line of the spectrum, we are seeing the light from only one star and we call this a single-line spectroscopic binary. If we can see the light from both stars the Doppler shifts will alternate, split and merge depending on the positions of the two stars in their orbits. This is called a double-line spectroscopic binary.

One very important detail, we do not know how the orbits of the two stars are inclined to earth. This inclination could be any angle, for that bit of information we have to go back to visual methods in order to see the individual stars to determine the inclination of their orbits relative to earth. Even so we can not for certain determine the true inclination of the orbit so our mass calculation is only a lower limit to the masses of the two stars.

Radial velocities permit astronomers to compute the total mass for the two stars, they do not provide the masses for the individual stars and other methods must be used to make that determination

Eclipsing Binary
Another type of binary called the Eclipsing binary can be studied. The information gathered can be used to calculate the individual stellar masses and the diameters of the individual stars. It is rare to find two stars in orbit around one another to have orbital inclination where the stars pass in front of one another to form one point of light as seen from earth.
When the orbital inclination if the eclipsing binary is edge on to earth, the stars will seem to pass in front of one another as they orbit, when the light from the brighter star is eclipsed we will see a deep decline in the amount of light received from the star (6/25/95 in Figure 1) we call this primary minimum, also when the light from the dimmer star is blocked by the brighter the light received declines again, but not so deep and we call this secondary minimum (see 6/9/95 in Figure 1) , otherwise we are able to collect some or all of the light from both stars.


The pattern of these light changes is called a light curve and the data for it gathered by the use of a photometer, making periodic measurements until the eclipsing binaries produce a complete orbital cycle.

We use the mass vs. luminosity relationship to determine what the difference is between the individual masses, then using the mass of the entire system calculated from the radial velocity information, we can determine what the individual masses of the two stars should be. The photometeric data removes some of the uncertainty in regard to the inclination because the shapes of the light curves will be different for a partial eclipse than for a total eclipse.
ALGOL is one of the best known and most studied eclipsing binary stars. ALGOL is normally about 2.3 magnitude, but every 10 hours or so it will dim to about 3.4 magnitude, in other words ALGOL becomes 68% dimmer. I suspect that humanity has known about ALGOL’s behavior for quite some time, since the Arabic name of ALGOL means “Demons Head”, and ALGOL is associated with the severed head of Medusa. ALGOL is often referred to as the winking eye of the demon.

An eclipsing binary occurs when the orbital plane of the binary system is exactly When one star passes directly in front of the other, as viewed from Earth, we seen an eclipsing binary perpendicular to the plane of the sky.

Dwarf Nova or Recurrent Nova


When an otherwise normal star is associated with a white dwarf companion, a type of binary called a recurrent nova, or dwarf nova may occur. The normal star transfers mass onto an accretion disk which forms around the white dwarf. As material falls onto the accretion disk some of the material may be transferred to the white dwarf by turbulence in the accretion disk, this causes a sudden brightening of the white dwarf as the hydrogen is converted into helium.
If enough material from the accretion disk falls onto the white dwarf the hydrogen gas will become compressed and will not immediately fuse until a substantial increase in temperature occurs; the material will suddenly and violently erupt fusing into a runaway fusion reaction and a violent eruption called a dwarf nova occurs which will blow the accretion disk away, but it will not disturb the normal star.


Mass transfer will quickly resume and a new accretion disk will form. The cycle will continue until enough mass is drawn off the normal star to halt the reaction.

Mass transfer in any type of binary system will affect the evolutionary cycle of the two stars. The normal star will burn its fuel more slowly as mass is removed and the star cools down due to less internal heating from gravitational forces. It will also accelerate the evolution of the star receiving the mass, for the same reasons, more mass, more internal heating and the hastening of the fusion process.

If the material transfers very quickly, the gravitational forces will prevent the hydrogen from fusing by compressing it even further until the hydrogen gas becomes degenerate matter. Degenerate matter does not expand due to the increases in temperature so the mass of the white dwarf increases until it exceeds the Chandrasekhar Limit. When this happens the white dwarf will collapse and a type I supernova will occur which may destroy the companion star and the white dwarf changes into a neutron star or a black hole.

Burster
A similar event can occur when a normal star is associated with a pulsar, the energy given off will be mostly X-rays however, and instead of being called a dwarf nova or recurrent nova, it is called an X-ray burster or more simply a burster. We think that as normal hydrogen falls onto the accretion disk it is quickly converted into helium, when the helium reaches a depth of 1 meter, it will explosively convert helium into carbon producing X-rays. The longer the delay in fusing carbon, the larger and more violent the burst will be. The main difference between the recurrent nova and the burster is that the accretion disk will be hotter in the burster because it is already fusing hydrogen into helium, also the burst will produce mostly X-rays instead of visible light.


When a black hole is associated with a normal star, it will produce the same events as an X-ray burster and the only way to be sure that the companion is a blackhole, is when the mass of the compact object is greater than 3 solar masses. This is far too much mass for the companion to be a neutron star. The gravitational forces would cause the collapse of the star beyond the point of the neutrons to support themselves against the force of gravity and the star would collapse to a zero radius creating a black hole.

Calculation of star’s properties with binary stars
Types of Binaries

  • Visual Binary: Can see both stars and follow their orbits over time.
  • Spectroscopic Binary: Cannot separate the two stars, but see their orbit motions as Doppler shifts in their spectral lines.
  • Eclipsing Binary: Can separate the stars, but see the total brightness drop when they periodically eclipse each other.

Visual Binaries –> Two stars orbiting about their center-of-mass.


Center of Mass
Two stars orbit about their center of mass.

  • Measure semi-major axis, a, from projected orbit & the distance.
  • Relative positions about the center give: M1/M2 = a2/a1

Measuring Masses
Newton’s Form of Kepler’s Third Law:
Procedure:
1. Measure the period, P, by following the orbit.
2. Measure semi-major axis, a, and the Mass Ratio, M1/M2, from the projected orbit on the sky.
3. Solve the equation above and separate Masses.

Problems
We need to follow an orbit long enough to trace it out in detail:

  • This can take decades
  • Need to work out the projection on the sky

Measurements depend on knowing the distance:

  • semi-major axis depends on d
  • derived mass depends on d^3

Small errors add up quickly (10% error in distance translates into a 30% error in the mass!).

Spectroscopic Binaries
Most binaries are too far away to be able to see both stars separately.
But, you can detect their orbital motions by the periodic Doppler shifts of the spectral lines:

• Determine the orbit period & size from the pattern of orbital velocities


Problems:
Cannot see the two stars separately:

  • Semi-major axis must be guessed from the orbit motions.
  • Can’t tell how the orbit is tilted on the sky

Everything depends critically on knowing the distance.
Again…

Eclipsing Binaries
Two stars orbiting nearly edge-on to our line-of-sight.

  • See a periodic drop in brightness as one star eclipses the other.
  • Combine with spectra which measure orbital speeds

With the best data, one can find the masses of the stars without having to know the distance!!!


Problems
Eclipsing Binary stars are very rare.
Measurement of the light curves is complicated by details:

  • Partial eclipses yield less accurate numbers.
  • The atmospheres of the stars soften the edges.
  • Close binaries can be tidally distorted.

However, the best masses are from eclipsing binaries.

Source : many different sites

Spectroscopy is a branch study in astronomy that focus on astronomical objects’ spectrum. From the spectrum, we can get informations, such as its temperatures, chemical compositions, movement speed, etc. That’s why spectroscopy can be considered as one of the fundamental field in astronomy. The spectrum of a star (or any other astronomical object) is acquired by using an instrument called spectrograph.

Fig 1. Spectrum

Fig 2. Spectrograph

One of the fundamental law in spectroscopy is Kirchoff Law (1859) which stated that:

  1. If a liquid or high pressure gas is ignited, they will emit energy in all wavelength which will produce a continuous spectrum.
  2. If a low temperature gas is ignited, it will only emit energy in certain range wavelength and produce spectrum which have a dark background and some bright lines. That kind of spectrum is called the emission spectrum. The wavelength of each bright lines are the precise indicator of what gas that produce them. So, the same gas will produce bright lines in certain exact wavelength.
  3. If a white light (which is a equal mixture of all colors) is passed through a cool low temperature gas, the gas will absorb energy at certain wavelength. The result spectrum will be continuous spectrum as the background with some dark lines in certain exact wavelength. The dark lines called absorption lines and that kind of spectrum is called the absorption spectrum. The wavelength of each dark lines are the precise indicator of what gas that produce them. So, the same gas will produce dark lines in certain exact wavelength.
Fig. 3 & 4. Continuous, emission and absorption spectrum (respectively)

Balmer Series
Switzerland scientist, Balmer, state a series equation to predict the wavelength of the absorption lines of hydrogen gas. The equation is widely known as Balmer series equation.

with : λ: the wavelength of the absorption lines [cm]
RH : Rydberg constant (= 109678 )

Fig. 5 : Emission spectrum of hydrogen that exhibit the first four emission lines in Balmer’s series

Planck’s Quantum Law

Planck postulates that light is radiated in the form of small discrete package called quantum. This theory is the foundation of the birth of a new field in physics called quantum physics.

Planck state that energy of each photon

Eo = h. f = hc//λ

h : Planck’s constant (h = 6,63 x 10^-34 J.s)
f : frequency of the photon [Hz]
c = speed of light (= 3.10^5 km/s)
λ = photon’s wavelength

Star’s spectrum
Star’s spectrum pattern is wide in variety. In 1863, an astronomer called Angelo Secchi classified star’s spectrum in 4 groups based on the similarities of its’ absorption lines.

Miss A. Maury from Harvard Observatory establish another way to classify star’s spectrum and it was revised by Miss Annie J. Cannon. Miss Cannon’s classification is the most widely adopted today.

Table 1 : Resume of the classification of star’s spectrum (to remember it use the donkey bridge : Oh Be A Fine Girl (or Guy), Kiss Me). (you can click the figure to get bigger and clearer version of the table above; .

Sub-classification of star’s spectrum
Star’s spectrum classification O, B, A, F, G, K, M is divided again to several sub-classes :
B0, B1, B2, B3, . . . . . . . . ., B9
A0, A1, A2, A3, . . . . . . . . ., A9
F0, F1, F2, F3, . . . . . . . . . ., F9

Bigger number represent lower temperature! The use of this sub-class is to narrow the specification’s range and become more precise.
(for further information, check this site.)

Fig 6. Star’s spectrum from different classes

M-K Classification (Star’s Luminosity Class)

Stars with same certain spectrum’s class is found to have different luminosities. In 1913, Adam dan Kohlscutter from Mount Wilson Observatory showed that the width of spectrum’s lines can be used to estimate star’s luminosity.
Based on these facts. in 1943 Morgan and Keenan from Yerkes Observatory divided stars to several luminosity class as shown in the table below.

Class1a

Very bright super giant star

Class 1b

Less bright super giant star

Class II

Bright giant star

Class III

Giant star

Class IV

Sub-giant star

Class V

Main sequence star

Table 2. Morgan Keenan’s Luminosity Class

Morgan Keenan’s Luminosity Class (M-K class) is sketched in a Hertzprung-Russell diagram (H-R diagram) below.

Fig 7. Star with different luminosity class in a H-R diagram

Now, star’s classifications use the combination of spectrum class and luminosity class. For example : A G2 V star is a main sequence star that belongs to spectrum class G2

Star’s motion
Contrary to widely beliefs that star isn’t moving in space, star DO move in space. However, the movement of stars is hard to track. Beause of its immense distance, the movement of star only produce extremely small apparent movement in sky. We have to wait several years (or decades!) to track star’s movement in sky. Warning : the star’s movement that is discussed above is not the apparent daily motion of the star !

The star’s angular motion of a star is called proper motion (μ). Proper motion is usually measured in arc-second per year. Star with biggest proper motion is Barnard Star with μ = 10”,25 per year (In 180 years, this star will (only) move in extent as full Moon’s disk).

Fig 8. Star’s motion

Relationship between tangential velocity (Vt) and the proper motion (μ):

Vt = 4,74 μ d

with :

Vt = tangential speed of the star [km/s]

μ = proper motion of the star [“/ year]

d = star’s distance [parsec]

the above equation also can be stated as :

Vt = 4,74 μ/p

with p is the parallax of the star (in arc second).

The proper motion is measured by two quantities: the position angle and the proper motion itself. The first quantity indicates the direction of the proper motion on the celestial sphere (with 0 degrees meaning the motion due north, 90 degrees due east, and so on), and the second quantity gives the motion’s magnitude, in seconds of arc per year.

Fig 9. Star’s proper motion

The equations used to find the quantity of star’s proper motion are :

μα cos δ = μ sin θ
μδ = μ cos θ

with :
μα = proper motion in right ascension
μδ = proper motion in declination
μα and μδ is measurable –> μ and θ can be determined.

Beside proper motion, information about star’s motion can be obtained from its radial motion, which is the component of star’s motion that lies parallel to our line of sight.
Radial velocity (Vr) can be measured by its spectrum lines that shift (Doppler shift). For star which radial velocity (Vr) is significant compared to the speed of light:

For Vr being much smaller compared to the speed of light (c), the equation can be simplified to:

Δλ/λo = Vr/c

with :
Δλ = the difference between static wavelength (λo) and observed wavelength (λ). [Å or nm]
λo = static wavelength. [Å or nm]
Vr = radial velocity [km/s]
c = speed of light (300.000 km/s )

Fig 10. Red shift and blue shift

Now, we are able to calculate Vt and Vr as discussed above and we will be able to calculate star’s true motion (linear motion):

V2 = (Vt)2 + (Vr)2

Reference:

  1. “Astrofisika I” lecture notes, by Dr. Djoni N. Dawanas
  2. Wikipedia
  3. Google images

For other pages that discuss this material, you are advised to visit these sites:
1. Spectroscopy
2. Astronomynotes.com

Spitzer, Chandra and Calar Alto Telescope

Tycho's Supernova Remnant. Credit : Spitzer, Chandra and Calar Alto Telescope

On November 11, 1572 Danish astronomer Tycho Brahe and other skywatchers observed what they thought was a new star. A bright object appeared in the constellation Cassiopeia, outshining even Venus, and it stayed there for several months until it faded from view. What Brahe actually saw was a supernova, a rare event where the violent death of a star sends out an extremely bright outburst of light and energy. The remains of this event can still be seen today as Tycho’s supernova remnant. Recently, a group of astronomers used the Subaru Telescope to attempt a type of time travel by observing the same light that Brahe saw back in the 16th century. They looked at ‘light echoes’ from the event in an effort to learn more about the ancient supernova.

A ‘light echo’ is light from the original supernova event that bounces off dust particles in surrounding interstellar clouds and reaches Earth many years after the direct light passes by; in this case, 436 years ago. This same team used similar methods to uncover the origin of supernova remnant Cassiopeia A in 2007. Lead project astronomer at Subaru, Dr. Tomonori Usuda, said “using light echoes in supernova remnants is time-traveling in a way, in that it allows us to go back hundreds of years to observe the first light from a supernova event. We got to relive a significant historical moment and see it as famed astronomer Tycho Brahe did hundreds of years ago. More importantly, we get to see how a supernova in our own galaxy behaves from its origin.”

Subaru Telescope

The view of the light echoes from Tycho’s supernova. Credit: Subaru Telescope

On September 24, 2008, using the Faint Object Camera and Spectrograph (FOCAS) instrument at Subaru, astronomers looked at the signatures of the light echoes to see the spectra that were present when Supernova 1572 exploded. They were able to obtain information about the nature of the original blast, and determine its origin and exact type, and relate that information to what we see from its remnant today. They also studied the explosion mechanism. What they discovered is that Supernova 1572 was very typical of a Type Ia supernova. In comparing this supernova with other Type Ia supernovae outside our galaxy, they were able to show that Tycho’s supernova belongs to the majority class of Normal Type Ia, and, therefore, is now the first confirmed and precisely classified supernova in our galaxy. This finding is significant because Type Ia supernovae are the primary source of heavy elements in the Universe, and play an important role as cosmological distance indicators, serving as ‘standard candles’ because the level of the luminosity is always the same for this type of supernova. For Type Ia supernovae, a white dwarf star in a close binary system is the typical source, and as the gas of the companion star accumulates onto the white dwarf, the white dwarf is progressively compressed, and eventually sets off a runaway nuclear reaction inside that eventually leads to a cataclysmic supernova outburst. However, as Type Ia supernovae with luminosity brighter/fainter than standard ones have been reported recently, the understanding of the supernova outburst mechanism has come under debate. In order to explain the diversity of the Type Ia supernovae, the Subaru team studied the outburst mechanisms in detail. This observational study at Subaru established how light echoes can be used in a spectroscopic manner to study supernovae outburst that occurred hundreds of years ago. The light echoes, when observed at different position angles from the source, enabled the team to look at the supernova in a three dimensional view. This study indicated Tycho’s supernova was an aspherical/nonsymmetrical explostion. For the future, this 3D aspect will accelerate the study of the outburst mechanism of supernova based on their spatial structure, which, to date, has been impossible with distant supernovae in galaxies outside the Milky Way.

The results of this study appear in the 4 December 2008 issue of the science journal Nature.

Source: Subaru Telescope

Cited from : universe today by Nancy Atkinson

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:

  1. when thermal equilibrium is achieved, the object’s temperature is a function of how many energy it absorbs per second
  2. 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 (λmaks) by a black body which temperature is T Kelvin is :

λmaks = 0,2898/ T …………………….. (eq. 1)

(λmaks expressed in cm and T in Kelvin)

The equation 1 is called Wien’s rule.

An example of implementing the Wien’s rule :

wien
(Warning : be clear that λmaks 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 T Kelvin is :

F = σT4 …………………….. (eq. 2)

(σ : Stefan-Boltzman constant: 5,67 x 10^-8 Watt/m2K4)

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 Luminosity. Its value (L) can be calculated by equation below:

L = A σT4 …………………….. (eq. 3)

For stars, we can assume it’s a perfect sphere. So, its surface area (A) is 4πR2 ; with R express star’s radius. So, a star’s luminosity (L) is equal to :

L = 4πR2 σT4 …………………….. (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/(4πd2) …………………….. (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 received energy flux or brightness. (Warning : differ between E and 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:

  1. From an observation result, we know that an area of 1 cm2 in Earth’s outer atmosphere received Sun’s energy with intensity of 1,37 x 106 erg/cm2/s. If we know that the distance between Sun and Earth is 150 million kilometres, determine the Sun’s luminosity.
  2. 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)?
  3. 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?
  4. Define the term Luminosity and Brightness using your own words
  5. Calculate the wavelength of maximum intensity radiation of a star which temperature is 10.000 Kelvin.
(source : Dr. Djoni N. Dawanas)
(translated from : belajar Astronomy).
ESO

A Pool of Distant Galaxies. Credit: ESO

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

angular diameter

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 dact, the angular diameter can be found with the formula:

for practical use, the distinction between d and dact 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.

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

trig parallax setup
take picture from opposite sides of orbit

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 very long, skinny trig parallax triangle

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 = 267,068.230220 AU = 1.2948 pc

the Earth–nearest star distance = 267,068.230222 AU = 1.2948 pc

Pluto–nearest star distance = 267,068.233146 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.

general parallax figure

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

  1. 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.
  2. 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 !
  3. 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:

  1. If a star has parallax 0″,711, determine its distance (in light years) from us!
  2. 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.
  3. 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.
  4. 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?
  5. 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.


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