Introduction to Needed Tools - Lecture Notes | ASTR 1020, Study notes of Astronomy

Download Introduction to Needed Tools - Lecture Notes | ASTR 1020 and more Study notes Astronomy in PDF only on Docsity! ASTR-1020: Astronomy II Course Lecture Notes Section I Dr. Donald G. Luttermoser East Tennessee State University Edition 4.0 Abstract These class notes are designed for use of the instructor and students of the course ASTR-1020: Astronomy II at East Tennessee State University. Donald G. Luttermoser, ETSU I–3 d) Light Year (ly) = the distance light travels in one year = 9.46 × 1015 m = 63,240 AU. e) Parsec (pc) = 3.09 × 1016 m = 206,265 AU = 3.26 ly. C. Scientific Notation 1. Powers of Ten. 1,000,000 = 106 1 = 100 100,000 = 105 0.1 = 10−1 10,000 = 104 0.01 = 10−2 1,000 = 103 0.001 = 10−3 100 = 102 0.0001 = 10−4 10 = 101 0.00001 = 10−5 1 = 100 0.000001 = 10−6 (exponent = # of zeros (exponent = # of decimal after the “1”) point moves to the right to just get past the “1”) 2. Notation: m × 10n 276 = 2.67 × 102 −3126.25 = −3.12625 × 103 .00276 = 2.76 × 10−3 −0.0000000312625 = −3.12625 × 10−8 a) Rule #1: m is a positive or negative real number and 1 ≤ m < 10. b) Rule #2: n must be an integer (−∞, ..., -3, -2, -1, 0, 1, 2, 3, ..., ∞). I–4 ASTR-1020: Astronomy II 3. Arithmetic with Scientific Notation. a) Multiplication: (−4.6 × 1016)(2.0 × 102) = (−4.6 × 2.0)(1016+2) = −9.2 × 1018 (5.0 × 108)(6.0 × 10−10) = (5.0 × 6.0)(108−10) = 30. × 10−2 = 3.0 × 10−1 = 0.3 b) Division: 6.3 × 108 3.0 × 104 = 6.3 3.0 × 108−4 = 2.1 × 104 −6.3 × 108 3.0 × 10−4 = −6.3 3.0 × 108−(−4) = −2.1 × 108+4 = −2.1 × 1012 c) Roots and Powers: (200)2 = (2 × 102)2 = 22 × (102)2 = 4 × 104 (1600)1/2 = (16 × 102)1/2 = (16)1/2 × 102/2 = (42)1/2 × 10 = 42/2 × 10 = 4 × 10 = 40 Donald G. Luttermoser, ETSU I–5 d) Significant Figures: Never write down a ton of digits after a decimal point. Your answer can have no more significant digits than the input number with the least amount of significant digits. (3.0379624 × 10−24) · (2.6 × 10−2) = (3.0379624 × 2.6) · (10−24+(−2)) = 7.898702 × 10−26 (incorrect) = 7.9 × 10−26 (correct) 29 0000︸ ︷︷ ︸ n.s. . 0. 000︸ ︷︷ ︸ n.s. 139 The underlined digits are significant in the above num- bers. Leading zeros are never significant and trailing ze- ros sometimes are not significant (in this example they are labeled with a n.s. for not significant). D. Angular Measure 1. Astronomers measure positions on the sky in terms of angles. a) A full circle is divided into 360◦ (◦ = degree). b) A right angle is 90◦. c) 1◦ = 60 arcmin = 60’. d) 1’ = 60 arcsec = 60”. 2. You can use common everyday objects to measure angles on the sky. a) Width of your finger at arm’s length = 1◦. b) Width of your fist at arm’s length = 10◦. I–8 ASTR-1020: Astronomy II b) Law 2: A line joining a planet and the Sun sweeps out equal areas in equal amounts of time (law of equal areas) =⇒ this means that a planet moves faster when it is near perihelion than at aphelion. i) Perihelion: (rp) Point on an orbit when a planet is closest to the Sun. rp = a (1 − e). (I-3) ii) Aphelion: (ra) Point on an orbit when a planet is farthest from the Sun. ra = a (1 + e). (I-4) iii) Note: rp + ra = 2a. (I-5) Sun A 1 A 2 t1 t2 if A 1 = A 2 (area), then t1 = t2 (time) rp ra Donald G. Luttermoser, ETSU I–9 c) Law 3: The square of the sidereal period of a planet is proportional to the cube of the semimajor axis of a planet’s orbit about the Sun (harmonic law). ( P P⊕ )2 = ( a a⊕ )3 (I-6) ( P 1 yr )2 = ( a 1 AU )3 (I-7) OR P 2yr = a 3 AU (I-8) Example I–2. A comet has a semimajor axis of 100 AU and a perihelion distance of 5 AU. What is the period, aphelion distance, and eccentricity of its orbit? ( P 1 yr )2 = ( 100 AU 1 AU )3 = (102)3 = 106 P 1 yr = (106)1/2 = 106/2 = 103 P = 1000 × 1 yr = 1000 yrs rp + ra = 2a ra = 2a − rp = 2(100 AU) − 5 AU = 195 AU rp = a(1 − e) rp a = 1 − e e = 1 − rp a = 1 − 5 AU 100 AU = 1 − 0.05 = 0.95 2. Newton’s laws of motion: a) The First Law: A body remains at rest, or moves in a straight line at a constant speed, unless acted upon by an external force (law of inertia). I–10 ASTR-1020: Astronomy II i) Force (F ): Something that produces a change in the state of motion of an object. ii) Inertia: The tendency of an object to remain in uniform motion. b) The Second Law: F = ma, (I-9) where m = mass of object, a = object’s acceleration. Force is measured in newtons (N = kg m/s2). F = ma is the most important equation in all of science! c) The Third Law: Whenever one body exerts a force on second body, the second body exerts an equal and opposite force on the first body. This law is the reason why rockets work. 3. Newton’s Universal Law of Gravity. a) Gravity is a force that every object with mass possesses. b) The universal law of gravity states: Two bodies attract each other with a force that is directly proportional to the product of their masses, m1 & m2, and inversely pro- portional to the square of the distance, r, between them: F = G ( m1 m2 r2 ) . (I-10) i) Mass measures how much material a body pos- sesses, while weight measures the amount of grav- itational force on an object. ii) Weight changes depending on the gravitational field the object is in. Donald G. Luttermoser, ETSU I–13 b) Room temperature ≈ 300 K. 3. An object at thermal equilibrium emits a thermal spectrum and is called a blackbody radiator. a) A blackbody does not reflect any light, it absorbs all radiation falling on it. b) All radiation it does emit results from its temperature. c) A blackbody spectrum is represented by a Planck curve: F lu x λλmax Planck Curve d) The energy flux (F ) is the amount of energy emitted from each square meter of an objects surface per second. The flux of a blackbody is a function only of its temper- ature and is given by the Stefan-Boltzmann law: F = σ T 4, (I-13) where T is the temperature and σ = 5.67×10−8 W m−2 K−4 is the Stefan-Boltzmann constant. I–14 ASTR-1020: Astronomy II e) The total brightness, or luminosity (L), of a blackbody is just the flux integrated over all of the surface of the object. For a spherical object, the surface area is 4π R2, so L = 4π R2 F = 4π σ R2 T 4. (I-14) Note that we can rewrite this equation by making a ra- tio equation with respect to the Sun (as shown in the example below): L L = ( R R )2 ( T T )4 . (I-15) f) The hotter a blackbody, the bluer its peak emission of light =⇒ the cooler, the redder its light. The wavelength of peak brightness for a blackbody is given by Wien’s law: λmax = 0.0029 m K T . (I-16) Example I–3. A star has a temperature of 10,000 K and a radius of 20 R , what is its flux and wavelength of maximum flux? What is its luminosity with respect to the Sun? (Note that R = 6.96 × 108 m and T = 5800 K.) F = (5.67 × 10−8 W m−2K−4) (10, 000 K)4 = (5.67 × 10−8 W m−2K−4) (104 K)4 = (5.67 × 10−8 W m−2K−4) (1016 K4) = 5.67 × 108 W m−2 λmax = 0.0029 m K 10, 000 K = 2.9 × 10−7 m = 2900 Å =⇒ UV light! L = 4π σ R2 T 4 Donald G. Luttermoser, ETSU I–15 L L = 4π σ R2 T 4 4π σ R2 T 4 = ( R R )2 ( T T )4 = ( 20 R R )2 (10, 000 K 5800 K )4 = (400) (1.72)4 = (400) (8.84) = 3500 L = 3500 L 4. The energy of a single photon is given by: E = h ν = h c λ , (I-17) where h = 6.625 × 10−34 J s is Planck’s constant and c is the speed of light.