* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download here
Gamma-ray burst wikipedia , lookup
Extraterrestrial life wikipedia , lookup
Tropical year wikipedia , lookup
International Ultraviolet Explorer wikipedia , lookup
Rare Earth hypothesis wikipedia , lookup
Hubble Deep Field wikipedia , lookup
Corvus (constellation) wikipedia , lookup
Expansion of the universe wikipedia , lookup
Extraterrestrial skies wikipedia , lookup
Geocentric model wikipedia , lookup
Aquarius (constellation) wikipedia , lookup
Perseus (constellation) wikipedia , lookup
Observable universe wikipedia , lookup
High-velocity cloud wikipedia , lookup
Andromeda Galaxy wikipedia , lookup
Observational astronomy wikipedia , lookup
Gravitational lens wikipedia , lookup
Dialogue Concerning the Two Chief World Systems wikipedia , lookup
Timeline of astronomy wikipedia , lookup
GEM2507 Physical Questions in Everyday Life Tutorial 1 Where are We?????? Name: Setiawan Office: S13-02-09, Phys. Departm. Phone: 6516-2988 Email: [email protected] Consultation: Wed. 10.00-12.00 1. What is a distance of 5 kilo parsec in lightyears? A minute of arc, arcminute, (MOA) = 1/60 of 1o An arcsecond = 1/60 * arcminute The parsec ("parallax of one arcsecond", symbol pc) = 30 trillion km or 3.3 light years. Defined as the distance from the sun at which two imaginary lines—one projected from the Earth, and one projected from the sun at a right angle to a third line connecting the Earth and the sun—intersect in space at an angle of 1 arcsecond Calculating the value of a parsec A B Calculating the value of a parsec S = Sun E = Earth D = point in space 1 parsec from the Sun 1 AU = 149,597,870,700 m 1 parsec ≈ 3.085 678×1016 m ≈ 3.261 564 light-years. Answer for question 1 5 kiloparsec = 5 * 3261.636 = 16308.181 light years Distances less than a parsec Distances measured in fractions of a parsec usually involve objects within a single star system. So, for example: one astronomical unit (AU) - the distance from the Sun to the Earth - is 4.85×10−6 pc. Parsecs and kiloparsecs Distances measured in parsecs include distances between nearby stars, such as those in the same spiral arm or globular cluster. For example Second Nearest star to the Earth (Proxima Centauri) 1.29 parsecs away. The center of the Milky Way is about 8 kpc from the Earth, and the Milky Way is about 30 kpc across. Andromeda Galaxy (M31), the most distant object visible to the naked eye, is a little under 800 kpc away from the Earth. Megaparsecs and gigaparsecs Astronomers typically measure the distances between neighboring galaxies and galaxy clusters in megaparsecs. For example: The Andromeda Galaxy is 0.77 Mpc away from the Earth. 2. Would it be a good idea to measure the distance to the closest galaxy outside our own by triangulation? No, as the parallax angle would be extremely small if it was used to measure the distance to the closest galaxy from our own galaxy. Indeed, the maximum distance measurable by triangulation is about 50 parsecs, only enough to explore our sun’s immediate neighborhood and far short of even being sufficient to map our own galaxy, the Milky Way Method used to measure distances in universe Method Distance (pc) Object Time(yr ) Hubble's Law 109-1010 Quasars 1011 Apparent luminosity - Galaxies 108 Virgo Cluster 108 106-107 Andromeda Cluster 107 105 Andromeda Cluster 105 Cepheid Variables - Type I 101-104 Center of Milky Way 103 Parallax 100 Alpha Centauri 101 Apparent luminosity - Super Giants Cepheid Variables - Type II 3. If we have a type-I Cepheid with the period of luminosity being approximately 140 hours and an apparent luminosity of 5 x 10-12 watt. What is it's distance from us in lightyears? From Figure 1.9 in the lecture note, we get the absolute luminosity (L) for period 140 hrs = 5.04 x 105 s, L = Ls = 5 * 4 x 1026 W = 2 x 1027 W The apparent luminosity is L LA 4d 2 L 2 x1027 18 d 5.63 x 10 m 182.62 pc 596.4lightyears 12 4 LA 4 (5 x10 ) 4. Joan (weighing 70kg) is running at a constant speed down a slope with an angle of 6 degrees with a speed of 10km/h. What is her acceleration? What is her momentum? (Use standard units!) y N f W x Constant speed = no acceleration Wy v = 10 km/h = 10/3.6 = 2.78 m/s p = m v = 70 x 2.78 = 194.6 kg m/s 5. Give the value of the Hubble constant in terms of seconds only (i.e. no other units) H0 = 7.2 x 104 m/s / Mpc = 7.2 x 104 m/s / 3.0857x 1022m = 2.33 x 10-18 /s Hubbles law V = H0 * d Question 6 6. Galileo was one of the first to systematically try to measure the speed of light. His attempt involved two observers positioned in two towers that were about 10km apart. The idea was that the first observer opens a shutter in a lantern and then as soon as the second observer sees the light from the first lantern, opens his shutter. Galileo would then measure the time it takes from opening the first shutter to seeing the light from the second lantern arrive at the first lantern. Unfortunately for Galileo, this experiment turned out to be inconclusive. Why is that so? Would it have been sensible to choose towers that are further apart? Answer for question 6 It is not possible because the velocity of light is tremendously large. As a matter of fact, light only takes 0.000005 seconds to travel one mile In order to measure the speed of light. It is extremely important to measure the time accurately which needs a sufficiently long distance and extremely negligible reaction time of the observer How to measure speed of light? http://www.colorado.edu/physics/2000/waves _particles/lightspeed_evidence.html