KNOWLEDGE TO GET FROM TODAY S CLASS MEETING Class Meeting #12, Monday, February 15 th 1) Taking a look at the 10-micrometer wavelength world.. 2) The decrease in RECEIVED RADIANT ENERGY as you move away from the source of the radiant energy, 1/distance 2 3) TELESCOPES (Chap 6).. how they help us compensate for the (1 / Distance 2 ) decline in received radiant energy as we attempt to view objects at greater and greater distances REFRACTOR-type telescopes REFLECTOR-type telescopes (these are BEST for astronomy science investigations)
EXAM #1 occurs 4 days from today, Friday, February 19 th An EXAM STUDY GUIDE was distributed Mon, Feb 8 th, and is available here in class today, and is also available online in the STUDY-GUIDES folder at the class web site
You WILL NOT need a scantron sheet for Exam #1 EXAM #1 will consist of multiple choice questions AND short written-answer questions (like quiz questions) Your QUIZZES and LABS are very good STUDY GUIDES, as are our In-Class exercises You WILL NOT need a scantron sheet for the exam!!
HOMEWORK #2 was distributed Monday, Feb 8 th With HW #2, you are obtaining OBSERVATIONS of the evening sky / MOON every-other evening for 10 more days Your set of OBSERVATIONS, and also your answers to QUESTIONS that will be distributed Wed (Feb 17 th ), are due to be handed in here in BX102 at the start of class on Friday, February 26 th HOMEWORK #2 is also available online within the class web page HOMEWORK folder You should already have 3 or 4 observations
THIS WEEK s Lab will occur in WH 232 You will conduct a lab focused upon PHASES of the MOON Lab available at: http://astronomy.nmsu.edu/murphy/astr105g-m040506-spring2016/lab-manual QUESTIONS?
PICTURE OF THE DAY VISIBLE Sunlight contains all colors in the VISIBLE portion of the Electromagnetic Spectrum As the light from setting the Sun passes through the Earth s atmosphere, the atmosphere acts like a PRISM and separates out the light in to its individual colors sometimes when this happens there is a GREEN FLASH during which the green light within the sunlight is VERY EVIDENT!
THE ELECTROMAGNETIC SPECTRUM All objects are attempting to emit radiant energy at all wavelengths, but hotter objects are more effective at emitting radiant energy at ALL wavelengths, especially at shorter wavelengths
The INTENSITY (watts per square meter) of RADIANT ENERGY that an object generates and emits is based upon that objects TEMPERATURE STEFAN-BOLTZMANN LAW QUANTITY = 5.67 x 10-8 x (TEMP) 4 Watts/m 2 The WAVELENGTH ( color ) at which an object emits the MOST RADIANT ENERGY: WIEN S LAW WAVELENGTH of MOST = (300 0 /TEMPERATURE) in units of micrometers (10-6 meters) We worked through an In-Class Exercise Friday.. any questions?
EARTH SUN SFC TEMPERATURE ~300 K ~5800 K (~60 o F) (~10,000 o F) Intensity of radiant 459 W/m 2 64,164,532 W/m 2 Energy Emitted (Watts per square meter) (the above numbers calculated using the Stefan-Boltzman Law) Wavelength of ~10 micrometers ~0.5 micrometers maximum radiant (infrared) yellow:green energy (your eyes and emitted mine DO NOT detect this infrared wavelength) (the above numbers calculated using Wien s Law) Let s look at the 10-micrometer infrared wavelength world which is showing us HOT and COLD surfaces / objects
Which stars in the image below are sending out the GREATEST intensity of Radiant Energy? Intensity = BRIGHTNESS If we assume that all of the stars are the same size, we would expect that the BLUEr stars, which Wien s Law tells us have hotter surfaces than the REDder stars, would also be the stars with a greater overall intensity of EMITTED RADIANT ENERGY, since the Stefan-Boltzman law tells us that the hotter surface emits away the greater intensity of radiant energy. BUT, some of the Red stars appear brighter than some blue stars. what might be going on???
The Sun emits > 64,000,000 Watts of radiant energy per square meter at its surface but you and I do not FEEL that much energy when we stand here on Earth in the sunlight we receive 1370 W/m 2 Why not???? Here on Earth, we are ~1.5 x 10 11 meters (150 billion meters = 150 million kilometers) from the Sun 1 AU from the Sun 64,000,000 Watts per square meter is equal to having 640,000 100-Watt light bulbs packed into an area that is smaller than the screen these words are being projected upon.
As the light that the Sun emits travels outward from the Sun, it becomes spread over a greater and greater area (think of an expanding sphere, like the surface of a balloon) This spreading of the light causes its local intensity (Watts per square meter received) to decrease as it moves away from the Sun Since the surface area of the expanding light sphere increases as: 4 times π times the radius squared= 4 π Radius 2, the radiant energy intensity must decrease at the same radius-dependent rate, ( 1 / Radius 2 )
As sunlight expands outward from the Sun, its local intensity declines with increasing distance from the Sun (with a 1 / distance 2 dependence), because the sunlight is being spread over a bigger and bigger spherical shape, just like an expanding balloon s material gets thinner-and-thinner as the balloon is inflated larger and larger but sunlight does not POP! Note that the expanding shells of light get dimmer as they move farther-andfarther from the Sun SUN
CHANGE in INTENSITY of RECEIVED SUNLIGHT with INCREASING DISTANCE DISTANCE RECEIVED RATIO FROM SUN SUNLIGHT INTENSITY Sfc (700,000 km) 64,000,000 W/m 2-1,000,000 km 31,830,980 W/m 2 1 2,000,000 km 7,957,747 W/m 2 1/4 4,000,000 km 1,989,436 W/m 2 1/16 8,000,000 km 497,359 W/m 2 1/64 16,000,000 km 124,340 W/m 2 1/256 32,000,000 km 31,085 W/m 2 1/1024 64,000,000 km (Mercury) 7,771 W/m 2 1/4096 105,000,000 km (Venus) 2,650 W/m 2 1 / 11,025 150,000,000 km (Earth) ~1,373 W/m 2 1 / 22,500
Since Pluto is on average 40 times farther from the Sun than Earth is: Sunlight intensity at Pluto = intensity @ Earth/40 2 (1370 W m -2 / 40 2 ) = 1370/1600 = 0.85 W/m 2 < 1 Watt per m 2!! Let s illustrate this one over distance squared received radiant energy dependence concept with a hands-on exercise.
An EXERCISE for us to perform 1) take the string and double loop it 2) loop the double loop string around the balloon and inflate the balloon to fill the loop 3) holding the inflated balloon end tightly, trace the outline of the paper square on to the balloon with the marker 4) unloop the string to its full length (twice its double looped length) 5) loop this full string around the balloon, and inflate the balloon to fill this new bigger loop 6) compare the size of the traced square on the balloon to the size of the paper square
WHAT HAS HAPPENED TO THE SIZE OF THE TRACED SQUARE?? Just as the square has gotten larger on this inflating balloon but the amount of balloon material within the square has remained the same, it is just stretched thinner, the area that a certain amount of light must fill also increases as the light travels outward from its source, so the magnitude of the light is thinner (less intense) with increasing distance
If the amount of total light (radiant energy) remains the same, BUT the total area that it must cover increases (as the ballon s surface expands), the INTENSITY of the light (units of Watts of energy per square meter) received within 1 square meter of viewing area must decline
In a perfect world, the larger inflated balloon square would be 4 times the size of the paper square, because the balloon has grown to twice its original radius, and surface area of a sphere increases as RADIUS squared (R 2 ), and : 2 2 = 2 times 2 = 4 times the surface area (so, the intensity of the twice-as-large balloon s energy per surface area is 1/4 of what it was after the initial inflation) But, we don t live in a perfect world, and the balloons are not exactly spherical, but hopefully this exercise gets the point across.. QUESTIONS?
SPECTRA how much energy is present at each wavelength of radiant energy that we are interested in Figures 5.12 5.13 5.14 5.15 5.16 and 5.17 and 5.18 and 5.19 in your text
Visible emitted radiant energy from Hydrogen atoms Note the very specific wavelengths at which a HYDROGEN atoms emits visible radiant energy.. These colors correspond to very specific electron orbital energy levels the electrons can jump between MICROMETERS 0.4 0.45 0.5 0.55 0.60 0.65 0.7 nanometers 0.6563 µm (6563 nm) Hydrogen gas being excited by an electronic current flowing through the gas.. this excitation causes electrons to jump around in their orbit energies
The Visible Emission Spectrum of Hydrogen Gas 6563 nanometers Nanometers (10-9 meters) the filled in spectrum above occurs for conditions where many, many atoms and molecules of various types are present and whacking in to each other
THE SUN This is an image of the SUN at a wavelength of 0.6563 micrometers (6.563 x 10-6 meters) This is a wavelength at which excited hydrogen atoms emit visible wavelength radiant energy.. compare to spectroscope color at ~650 nanometers(nm) x-axis position on previous slide) Any questions about: Total EMITTED INTENSITY of Radiant Energy? (Stefan Boltzman Law, Intensity = σ T 4 W / m 2 ) WAVELENGTH in micrometers of most Emission? (Wien s Law, Wavelength = 3000 / Temperature (in Kelvin units) Amount of RECEIVED RADIANT ENERGY? SPECTRA have us think about individual Wavelengths As Distance increases, Received Amount depends upon 1 / DISTANCE 2 (less received as distance increases)
An Exercise: RECEIVED RADIANT ENERGY What can you determine about two equal-sized sources (A & B) of radiant energy as indicated in the plot below? both curves peak at the RECEIVED INTENSITY (W/m 2 ) of RADIANT ENERGY same wavelength Source A Source B gamma ultraviolet visible infrared radio Wavelength How do the temperatures of the 2 sources compare? Why are the magnitudes (curve heights) different? WHAT HELP DOES WIEN S LAW PROVIDE?
RECEIVED INTENSITY (W/m 2 ) of RADIANT ENERGY Source A Source B gamma ultraviolet visible infrared radio Wavelength Source B (the lower curve source) is the SAME temperature as Source A [ using Wien s Law, since both curves ARE TALLEST at the same wavelength!!] BUT Source B is farther away from us so our received intensity from Source B is smaller (curve is shorter) due to the (1/distance 2 ) factor
If we want to study SOURCES A & B equally, we need a way of amplifying the radiant energy we receive from the objects, especially from B.. TELESCOPES provide this amplification
ANY QUESTIONS ABOUT RADIANT ENERGY STEFAN-BOLTZMANN Law? WIEN S LAW? TEMPERATURE? EMITTED ENERGY? WAVELENGTH? RECEIVED ENERGY?
TELESCOPES Telescopes help us compensate for the decrease of received light intensity with increasing distance The primary purpose of a telescope is to capture as much light as possible from an object (planet in our solar system, star in our galaxy, galaxy beyond our own, etc.) so that object will appear BRIGHTER (note: 'enlargement' is not the only effect a telescope can produce ) The more light we can capture from an object, the brighter the object will appear
We capture more light by having a large MIRROR (in a Reflecting telescope) or GLASS LENS (in a Refracting telescope) to do the COLLECTING (rather than just our small eye) Telescopes also allow us to make the object appear closer than it is (telescopes enlarge the image of the object) BUT if the telescope does not capture much light, a faint large object is still faint!!
How Big is your eye? A Human Eye s pupil is ~0.4 inch (1 centimeter) in diameter (0.5 centimeters in Radius) Its LIGHT COLLECTING AREA is: Π x Radius 2 = 0.8 cm 2 A LENS or MIRROR 12 inches (30.48 cm) in diameter (15.24 cm in radius) has a Light Collecting Area = Π x (15.24 cm) 2 = 729 cm 2 (900 times larger than your eye!!) IT IS THEIR LARGE LIGHT COLLECTING AREA THAT MAKES TELESCOPES USEFUL!
There are TWO types of telescopes used for observing visible wavelengths of radiant energy (and some nearby wavelengths): REFRACTOR: use a glass lens to focus the light glass lens Incoming incident white light white light (your eye here) Focus REFLECTOR: use a curved mirror to focus the light (your eye here) curved mirror incident white white light light Focus (at a another mirror to turn the light out the side)
The large, expensive telescopes used for astronomical research are all REFLECTORS? WHY? PROBLEM #1 with REFRACTORS If you shine light through a glass prism, what happens to the light? The light gets spread out in to its colors (its spectrum) like a RAINBOW, but it is difficult to then clearly see (in a realistic form) the object from which the light is coming
Also, it is difficult to make a thick piece of glass (such as for a BIG glass lens) that does not have some imperfections (bubbles, cracks, etc.), and these imperfections cause problems when trying to see faint objects (PROBLEM #2) Also, it is much easier to make large mirrors (to capture as much light as possible) that are near perfect (don t have blemishes and reflect all wavelengths of interest) AND We can make a very thin mirror but not necessarily a thin lens, so the REFLECTOR will be less massive and less likely to warp with time
So, for very practical reasons, large research telescopes are REFLECTORS Telescopes were originally (such as that used by Galileo Galilei) REFRACTOR type Eye Focus a lens here enlarges the object s apparent size Glass Lens incoming light refracts (bends) the travel direction of the light
The PRISM problem plus the difficulty of manufacturing large (meters in diameter) glass lenses that don t have imperfections (interior bubbles, un-smooth surfaces) has resulted in large modern telescopes all being REFLECTOR type: Eye Focus; second mirror here 3 rd mirror here Mirror #1; this is the primary mirror incoming light Some REFLECTORS use 2 mirrors, some use 3 mirrors..
Modern-day LARGE (5-10+ meters in diameter!!) telescopes (where the size indicates the diameter of the telescopes) are all REFLECTOR types because: 1) Mirrors do not suffer the prism-like effect of glass 2) It is not outrageously difficult to manufacture large mirrors that don t suffer from troublesome imperfections [use a lightweight, non-reflecting material to shape of the mirror, and then coat with silver, which is a very reflective material ] 3) a mirror can weigh much less than a glass prism of the same diameter
We have discussed the fact that as you move away from a source of RADIANT ENERGY, the intensity of the energy you receive from that source decreases in proportion to: Received Intensity = (1 / DISTANCE 2 ) Since the nearest star is 256,000 times as far from Earth as the Sun is, the nearest star is (1/256,000 2 )=: 1.5 x 10-11 times fainter to our eye than the Sun is AND most objects of interest in the Galaxy and Universe are much farther away than this!!!!! (very Faint!!!!!!)
If we want the nearest star (not the Sun) to provide as much collected light using our telescope as your eye collects from the SUN, than we need a telescope that is: Telescope size / Distance 2 ( (1 cm) 2 / 1 AU 2 ) = ((X cm) 2 / 256,000 AU 2 ) eye telescope X = telescope radius = 256,000 cm = 2.56 x 10 5 cm = 2.56 km radius = 5.12 km diameter (the current largest telescope is the 10-meter Keck telescope in Hawaii) 1 cm Your Eye 5.12 km Big Telescope Mirror
A PICTURE OF THE DAY Astrophysical Research Apache Point Observatory Consortium NMSU 1-meter Sloan Digital Sky Survey 2.5 meter 0.5 meter (visible spectra) 3.5 meter diameter telescope (visible and infrared wavelengths) Located ~15 miles south of Cloudcroft, NM
ANY QUESTIONS ABOUT TELESCOPES? Sections 6.1 & 6.2 & 6.3 (pages 165-178) in your text are most important for our purpose in this ASTR 105G course