ASTR 101 Notes / Homeworks / Exams

Topics: Earth, Light, Moon Pages: 11 (3861 words) Published: December 7, 2013
LESSON ONE:
Speed of light: C= 3 x 10^8 m/s, C=3x10^5 km/s
Light year= dist= speed x time
Earth rotates 360 deg every SIDEREAL DAY
1 Sidereal day= 24 sidereal hrs= 23:56 solar hours
Earth revolves 360 around the sun ever 365.24 days= tropical year Earth’s rotation axis precesses every 26,000 yrs
Moon revolves 360 deg ever 27.3 days= Sidereal mon
Moon’s cyc repeats every 29.5 days= Syndonic mon
Saros cycle: one eclipse yr is only 346 days. 19 eclipse years= 223 syndonic months. LESSON ONE HOMEWORK:
1. You send a message to Buzz Aldrin on the moon, 384,000 km and he sends you an immediate reply. Both messages travel at the speed of light. How long do you have to wait between sending your message and receiving his? Over 2 sec.  (time= dist/speed= 384,000km/ 3x10^5)=1.28 Sec. (then X2 for there and back) 2. You watch a star rotate 120 deg around the n. celestial pole. How long have you been watching? 8 hrs.  (120x 24hr/360deg). 3. You watch Orion rise above the horizon at 2am. What time will you have to go out to watch it rise two months earlier? 10 PM (2mons x (360deg/12mos)=60 deg. 60degx(24hr/360deg)= 4hr. 2am-4hr= 10PM) 4. It is hotter in summer and colder in winter b/c:? Your hemisphere of earth is tipped toward the sun. 5. The sun at midday is the lowest in the sky on the:? Winter solstice. 6. The day is 12 hrs long and the night is 12 hrs long:? On the vernal and autumnal equinoxes. 7. In 26,000 yrs:? Polaris will be the pole star (precession, n. celestial pole traces circle every 26,000 yrs) 8. When lunar eclipses occur, the moon is full.  (lunar eclipses occur when earth is between sun and moon, =moon is full. Since the earth-moon plane is tilted with respect to the earth-sun plane, the line of nodes must point towards sun) 9. There are apx 19 eclipse seasons every 9 tropical yrs. 10. Every Saros cycle: ? All lunar eclipses reoccur.

LESSON TWO:
2) baseline= dist btwn different observing points
If you know the baseline and ang shift in degrees use:
Dist = (360 deg/ 2pi) x (baseline/ ang shift)
If you know the baseline and dist can calc the ang shift = (360 deg/ 2pi) x (baseline/ dist) For Objects orbiting SUN: P= (a/1AU)^3/2 or
A= (p/1 yr) ^2/3 AU
For Objects orbiting EARTH: P= (a/1 earth-moon dist)^3/2 lunar mos. A= (p/ 1 lunar month)^2/3 EM Newton’s 2nd= F= MA (force = mass of obj x acc)
LESSON TWO HOMEWORK:
1. Principal astronomical alignment at Stonehenge:? Summer solstice sunrise. 2. Perfected the geocentric model of the universe:? Ptolemy. 3. First to argue for the heliocentric model of the universe:? Aristarchus. 4. Which of Galileo’s discoveries is evidence for the heliocentric model? Venusian phases 5. One person stands at the north pole and one person stands at the south pole. When the moon is directly over the equator, they both take a picture of the moon and the background stars. Afterword, they compare their pictures. Between the pictures, the moon appears to be shifted with respect to the background stars by how many lunar diameters? (diameter of Earth = 12,756 km; circumference = 2π × radius; π = 3.14; distance to moon = 384,400 km; 1 lunar diameter = 0.50°)?: 3.8 lunar diameters.  ((angular shift/ 360deg)= (baseline/circm), (angular shift/ 360deg)= (12,756km/(2x3.14x384,400km)=.00528. ANG SHIFT= .00528x360=1.9 deg. 1.9x(1 lunar diameter/ .50deg)=3.8) 6. Every year, Proxima Centauri, the closest star to the sun, is observed to move 1.5 arcseconds with respect to the background stars for six months, followed by a return to its original position over the course of the next six months. How far away is Proxima Centauri? [diameter of Earth's orbit = 2 AU (astronomical units)].? 280,000 AU.  (ANG SHIFT TO DEG..1.5 arcsecx (1arcmin/60arcsec)x(1deg/60 arcmin)=.000417deg. DIST.. (baseline/2pi)x(360deg/.000417deg) =275,000AU. 7. new planet disocered is an approximately circular orbit beyond plut. It moves at a rate of approx 1 degree/yr. It’s semi-major axis is probz:? 50AU....
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