OUTLINE OF ASTRONOMY
CHAPTER PHYSICS FACTS ABOUT MASSES ABOUT GASES ABOUT LIGHT ABOUT DISTANCE THE SKY AND TELESCOPES COORDINATE SYSTEMS AND TIMESCALES INSTRUMENTATION TERRESTRIAL PLANETS FORMATION OF TERRESTRIAL PLANETS EVOLUTION OF THE TERRESTRIAL PLANETS MERCURY VENUS EARTH MOON MARS MOONS OF MARS JOVIAN PLANETS AND THEIR SATELLITES JUPITER SATURN URANUS NEPTUNE MOONS RINGS DEBRIS COMETS METEORITES ASTEROIDS PLUTO AND CHARON THE INTERSTELLAR MEDIUM AND STAR FORMATION THE INTERSTELLAR MEDIUM STAR FORMATION MAIN-SEQUENCE STARS AND THE SUN EQUILIBRIUM OF STARS OBSERVABLE PROPERTIES OF STARS THE SUN STELLAR EVOLUTION WHY DO STARS EVOLVE? HOW DO STARS EVOLVE? STARS<8MSUN STARS>8MSUN: SUPERNOVAE WHERE DO WE COME FROM? STELLAR REMNANTS (WHITE DWARFS, NEUTRON STARS, AND BLACK HOLES) DEGENERATE GAS PRESSURE WHITE DWARFS NEUTRON STARS BLACK HOLES GALAXIES AND CLUSTERS THE MILKY WAY NORMAL GALAXIES ACTIVE GALAXIES AND QUASARS COSMOLOGY HUBBLE’S LAW HUBBLE’S LAW AND THE EXPANSION OF THE UNIVERSE HUBBLE’S LAW AND THE AGE OF THE UNIVERSE HUBBLE’S LAW AND THE SIZE OF THE UNIVERSE THE BIG BANG LIFE IN THE UNIVERSE CONTENTS WHERE M AND M ARE THE MASSES OF THE TWO OBJECTS, D IS THE DISTANCE BETWEEN THEIR CENTERS, AND G IS THE GRAVITATIONAL CONSTANT: 6:67 1011 M3=KG=S2 THE UNIT OF FORCE IS THE NEWTON (N), WHICH IS EQUAL TO 1 KG M=S2 IF THE SIZES OF THE TWO OBJECTS ARE MUCH SMALLER THAN THEIR DISTANCE D, THEN THE ABOVE EQUATION IS VALID FOR ARBITRARY SHAPES AND ARBITRARY MASS DISTRIBUTIONS THE ELLIPSE THE PLANETS ORBIT THE SUN IN NEARLY CIRCULAR ELLIPTICAL ORBITS AN ELLIPSE IS DESCRIBED BY ITS MAJOR AXIS (LENGTH ¼ 2A) AND ITS MINOR AXIS (LENGTH ¼ 2B), AS SHOWN IN FIG 1-1 FOR EACH POINT A ON THE ELLIPSE, THE SUM OF THE DISTANCES TO THE FOCI AF AND AF 0 IS CONSTANT MORE SPECI?CALLY: AF Þ AF 0 ¼ 2A THE ECCENTRICITY OF THE ELLIPSE IS GIVEN BY E ¼ FF 0=2A IN TERMS OF THE SEMI-MAJOR AXIS, A, AND THE SEMI-MINOR AXIS, B, WE HAVE E ¼ ??????????????????? B A S THE SUN OCCUPIES ONE OF THE FOCI IN THE ELLIPSE DESCRIBED BY A PLANET ASSUME THAT THE SUN OCCUPIES THE FOCUS F WHEN THE PLANET IS ON THE MAJOR AXIS AND AT THE POINT NEAREST F, THEN THE PLANET IS AT PERIHELION ON THE FAR POINT ON THE MAJOR AXIS, THE PLANET IS AT APHELION BY DE?NITION, DP Þ DA ¼ 2A WHERE DP IS THE PLANET–SUN DISTANCE AT PERIHELION AND DA IS THE DISTANCE AT APHELION CHAPTER GÐM Þ MÞ WHERE M AND M ARE THE MASSES OF THE TWO BODIES THIS NEWTONIAN VERSION IS VERY USEFUL FOR DETERMINING THE MASSES OF OBJECTS OUTSIDE OF OUR SOLAR SYSTEM CHAPTER 1011 M3=KG=S2) ESCAPE VELOCITY AN OBJECT OF MASS M WILL REMAIN IN ORBIT IF ITS SPEED AT DISTANCE D DOES NOT EXCEED THE VALUE VE ¼ ??????????? 2GM D R THE SO-CALLED ESCAPE VELOCITY AGAIN, M IS THE MASS OF THE LARGER OBJECT THE ESCAPE VELOCITY IS INDEPENDENT OF THE MASS OF THE SMALLER OBJECT, M ANGULAR MOMENTUM ALL ORBITING OBJECTS HAVE A PROPERTY CALLED ANGULAR MOMENTUM ANGULAR MOMEN- TUM IS A CONSERVED QUANTITY CHANGING THE ANGULAR MOMENTUM OF A SYSTEM REQUIRES EXTERNAL ACTION (A NET TORQUE) THE ANGULAR MOMENTUM DEPENDS ON THE MASS, M, THE DISTANCE FROM THE OBJECT IT IS ORBITING, R, AND THE TRANSVERSE VELOCITY, V, OF THE ORBITING OBJECT, L ¼ M V R THE MASS OF PLANETS IS CONSTANT, SO CONSERVATION OF ANGULAR MOMENTUM REQUIRES THAT THE PRODUCT V R REMAINS CONSTANT THIS IS KEPLER’S SECOND LAW KINETIC ENERGY MOVING OBJECTS HAVE MORE ENERGY THAN STATIONARY ONES (AT THE SAME POTENTIAL ENERGY, FOR EXAMPLE, AT THE SAME HEIGHT O?THE GROUND) THE ENERGY OF THE MOTION IS CALLED THE KINETIC ENERGY, AND DEPENDS ON BOTH THE MASS OF THE OBJECT, M, AND ITS VELOCITY, V THE KINETIC ENERGY IS GIVEN BY KE ¼ KINETIC ENERGY (AND ENERGY IN GENERAL) IS MEASURED IN JOULES (J) THE MASS OF THE EARTH IS GIVEN IN APPENDIX 2, 5:97 1024 KG, AND THE RADIUS OF THE EARTH IS 6,378 KM (I E , 6,378,000 M OR 6:378 106 M) PLUGGING ALL OF THIS INTO THE EQUATION GIVES F ¼ 6:67 1011M3=KG=S2 65KG 5:97 1024 KG 6:378 106 M Ð Þ F ¼ 636 N THE GRAVITATIONAL FORCE ON YOU DUE TO THE EARTH IS 636 NEWTONS THIS IS ALSO THE GRAVITA- TIONAL FORCE THAT YOU EXERT ON THE EARTH (TRY THE CALCULATION THE OTHER WAY IF YOU DON’T BELIEVE THIS IS TRUE ) 1 2 WHAT IS THE MAXIMUM VALUE OF THE FORCE OF GRAVITY EXERTED ON YOU BY JUPITER? THE MAXIMUM VALUE OF THIS FORCE WILL OCCUR WHEN THE PLANETS ARE CLOSEST TOGETHER THIS WILL HAPPEN WHEN THEY ARE ON THE SAME SIDE OF THE SUN, IN A LINE, SO THAT THE DISTANCE BETWEEN THEM BECOMES D ¼ ÐDSUN TO JUPITERÞ ÐDSUN TO EARTHÞ D ¼ 5:2 AU 1 AU D ¼ 4:2 AU CHAPTER EARTH THE FACTORS OF G AND M CANCEL OUT, SO THAT THE EQUATION SIMPLI?ES TO CHAPTER FMARS FEARTH ¼ 2:6 S VC ¼ 7:72 103 M=S VC ¼ 7:72 KM=S SO THE CIRCULAR VELOCITY OF THE SPACE SHUTTLE IS 7 72 KM/S MULTIPLY BY 60 SECONDS PER MINUTE AND BY 60 MINUTES PER HOUR TO ?ND THAT THIS IS NEARLY 28,000 KM/H CHAPTER S VE ¼ 1:12 104 M=S VE ¼ 11:2 KM=S THIS MAY NOT SEEM VERY FAST, IF YOU ARE NOT USED TO THINKING IN KM/S CONVERT IT TO MILES PER HOUR BY MULTIPLYING BY 0 6214 MILES/KM, AND MULTIPLYING BY 3,600 SECONDS/HOUR NOW YOU SEE THAT THE ASTRONAUTS WERE TRAVELING AT 24,000 MILES/HOUR 1 7 WHAT IS THE DENSITY OF THE EARTH? HOW DOES THIS COMPARE TO THE DENSITY OF ROCKS (BETWEEN 2,000 AND 3,500 KG/M3)? WHAT DOES THIS MEAN? THE DENSITY IS THE MASS DIVIDED BY THE VOLUME IF WE ASSUME THE EARTH IS SPHERICAL, THE CALCULATION IS SIMPLI?ED ¼ M V ¼ MEARTH EARTH ¼ 5:97 1024 KG THE AVERAGE DENSITY OF THE EARTH IS HIGHER THAN THE DENSITY OF ROCK SINCE THE SURFACE OF THE EARTH IS MOSTLY ROCK, OR WATER, WHICH IS EVEN LESS DENSE, THIS MEANS THAT THE CORE MUST BE MADE OF MATERIAL THAT IS DENSER THAN THE SURFACE 1 8 THERE ARE ABOUT 7,000 ASTEROIDS IN OUR SOLAR SYSTEM ASSUME EACH ONE HAS A MASS OF 1017 KG WHAT IS THE TOTAL MASS OF ALL THE ASTEROIDS? IF THESE ASTEROIDS ARE ALL ROCKY, AND SO HAVE A DENSITY OF ABOUT 3,000 KG/M3, HOW LARGE A PLANET COULD BE FORMED FROM THEM? THE TOTAL MASS OF ALL THE ASTEROIDS IS JUST THE PRODUCT OF THE NUMBER OF ASTEROIDS AND THEIR INDIVIDUAL MASS: M ¼ N M M ¼ 7;000 1017 KG M ¼ 7 1020 KG THE VOLUME OF THE PLANET THAT COULD BE FORMED IS CHAPTER IF WE ASSUME THE PLANET IS SPHERICAL, THEN WE CAN ?ND THE RADIUS R ¼ ?????? 3V P ¼ ???? P P ¼ ????? 2A ¼ THE ECCENTRICITY OF THE ELLIPTICAL ORBIT IS 0 33 1 10 HALLEY’S COMET HAS AN ORBITAL PERIOD OF 76 YEARS, AND ITS FURTHEST DISTANCE FROM THE SUN IS 35 3 AU HOW CLOSE DOES HALLEY’S COMET COME TO THE SUN? HOW DOES THIS COMPARE TO THE EARTH’S DISTANCE FROM THE SUN? WHAT IS THE ORBIT’S ECCENTRI- CITY? SINCE HALLEY’S COMET ORBITS THE SUN, WE CAN USE THE SIMPLI?ED RELATION CHAPTER P2 ¼ PERIHELION Þ APHELION PERIHELION Þ APHELION ¼ 3P PERIHELION ¼ 3P APHELION PERIHELION ¼ ??????? PERIHELION ¼ THE FORCE BETWEEN THE TWO OBJECTS INCREASES BY A FACTOR OF FOUR WHEN THE PRODUCT OF THE MASSES INCREASES BY A FACTOR OF FOUR 1 12 HOW WOULD THE GRAVITATIONAL FORCE BETWEEN TWO BODIES CHANGE IF THE DISTANCE BETWEEN THEM INCREASED BY A FACTOR OF TWO? AGAIN, SET UP A RATIO SO THAT ALL THE UNCHANGED QUANTITIES CANCEL OUT (AS IN PROBLEM 1 4): CHAPTER M ¼ 2:0 1030 KG THIS IS STRIKINGLY CLOSE TO THE ACCEPTED VALUE FOR THE MASS OF THE SUN, 1:9891 1030 KG IT IS SO CLOSE THAT ANY DI?ERENCES MIGHT BE CAUSED BY A ROUND-O?ERROR IN OUR CALCULATORS PLUS THE ASSUMPTION THAT THE MASS OF THE EARTH IS NEGLIGIBLE 1 15 HOW FAST WOULD A SPACECRAFT IN SOLAR ORBIT HAVE TO BE MOVING AT THE DISTANCE OF NEPTUNE TO LEAVE THE SOLAR SYSTEM? THE ESCAPE VELOCITY IS GIVEN BY CHAPTER VE ¼ ??????????? 2GM D R VE ¼ ????????????????????????????????????????????????????????????????????????????? 2 6:67 1011 M3=KG=S2 2 1030 KG 4:5 1012 M S VE ¼ 7,700 M=S ¼ 7:7 KM=S IN ORDER FOR A SPACECRAFT TO ESCAPE THE SOLAR SYSTEM FROM THE ORBIT OF NEPTUNE, IT MUST BE TRAVELING AT LEAST 7 7 KM/S THIS IS NOT VERY MUCH LESS THAN THE ESCAPE VELOCITY OF A SPACE- CRAFT FROM THE EARTH (11 KM/S) EVEN THOUGH THE ORBIT OF NEPTUNE IS SO FAR AWAY, THE MASS OF THE SUN IS SO LARGE THAT OBJECTS ARE BOUND QUITE TIGHTLY TO THE SOLAR SYSTEM, AND MUST BE MOVING VERY QUICKLY TO ESCAPE 1 16 THE MOON ORBITS THE EARTH ONCE EVERY 27 3 DAYS (ON AVERAGE) HOW FAR AWAY IS THE MOON FROM THE EARTH? WE CANNOT USE THE SIMPLE RELATION BETWEEN P AND A FOR THIS PROBLEM, SINCE THE SUN IS NOT AT THE FOCUS OF THE ORBIT HOWEVER, WE CAN ASSUME THAT THE MOON IS MUCH LESS MASSIVE THAN THE EARTH FIRST, CONVERT 27 3 DAYS TO 2:36 106 SECONDS A3 ¼ P2 G ÐM Þ MÞ A3 ¼ 2:36 106 S 26:67 1011 M3=KG=S2Ð6 1024 KGÞ 4 ¼ ??