In all cases assume the observer is located at the latitude of Charlottesville (38 degrees north).

Transcription

1 1. Recalling that azimuth is measured around the sky from North (North is 0 degrees, East is 90 degrees, South is 180 degrees, and West is 270 degrees) estimate (do not calculate precisely) the azimuth of a) the rising Sun at the winter solstice: the Sun is at -23 declination. A star at that declination rises south of due East (90) so an answer around azimuth = 110 is good b) the setting Sun at the spring equinox: this one is easy. The setting sun sets due West on the equinox because it is on the celestial equator azimuth = 270 exactly. c) a rising star that has a declination of -40 degrees: This is a star that will reach an altitude of only 12 degrees looking due South (180). It must rise just East of due South so lets guess azimuth = 160 d) a setting star that has a declination of 50 degrees: This is a star that is practically circumpolar. It will set very close to due north but on the west side so roughly azimuth = 340 In all cases assume the observer is located at the latitude of Charlottesville (38 degrees north). 2. Describe the experiment of either Eratosthenes or Aristarchus. Be sure to include a sketch, and possibly an equation, to make the mathematical aspects of the experiment apparent. Eratosthenes Circumference of Earth Measuring shadow lengths at two locations of known separation at Noon on the same day. As seen from the center of the Earth these two locations subtend an angle theta directly measured by the different shadow lengths (actually the difference in angle between the zenith and Sun seen from two observing sites). Theta / 2p = (distance between the two cities)/(circumference of the Earth). Aristarchus Relative distance of the Moon vs. Sun. Find the instant the Moon is half illuminated and thus establish a right triangle between the Sun, Moon, and Earth with the right angle at the Moon. The angle between the Sun and Moon as seen from Earth at that time is established from knowing how long it has been since New Moon vs. the synodic period of the Moon. The closer that angle is to 90 degrees the farther away the Sun must be. 3. Explain how the concepts of the Roche Limit and Hill Sphere are conceptually related? Both quantify the influence of tidal forces on a pair of self gravitating particles and establish the boundary between tidal dominance and self-gravitational dominance formation of rings vs. satellites for Roche, keeping vs. losing a satellite for Hill.

2 4. Explain why a point on Earth will experience two high tides a day. Include a sketch to support your explanation. The Earth rotates within the ocean tidal envelope raised by the Moon (because the Moon orbits much more slowly than the Earth rotates). Twice a day a location on Earth finds itself first aligned with the Moon and then roughly 12.5 hours later, opposite the Moon. Both are times of high tide. 5. Why are there just over 200,000 Astronomical Units in a Parsec? Because a parsec is the distance at which 1 AU subtends an angle of one arcsecond. There are arcseconds in a radian. 6. Why is the sidereal year different from the tropical year? The location of the vernal equinox shifts slowly along the celestial equator completing a circuit every 26,000 years as the Earth s rotation axis precesses. The drift occurs steadily so as the Earth completes a sidereal year the vernal equinox location shifts and as a result the Sun reaches the celestial equator (the instant of Spring ) 1/26,000th of a year early each orbit relative to the true sidereal orbital period of Earth. 7. Jupiter has a mass of one thousandth the mass of the Sun (a milli-sun!). How far must a particle orbit from Jupiter to have an orbital period of one year? 1 MP 2 = a 3 so a=m P 2 P = 1 (year) and M = 1/1000 since the units are solar masses and 1/10 is the cube root of 1/1000. In this formulation a is in AU so the particle orbits 1/10 AU from Jupiter

3 8. Consider the diagram below representing the local observing conditions for an observer. Label the horizon, celestial equator, and meridian. What is the approximate hour angle and declination of the indicated star? What is the approximate latitude of the observer? If the right ascension of the star is 12 hours and it is midnight, what is the approximate date (approximate meaning within a few weeks)? The last question part translates to, in what month is the sidereal time 16 hours at midnight?... because the hour angle shows the star at 12 hours transited the meridian 4 hours ago, so the sidereal time (the time on The Meridian) must be 16 hours. We know that the time on the meridian at midnight is 12 hours on the spring equinox (because the Sun is at 0 hours). To get 16 hours on the meridian 2 months must have passed since the Spring equinox so it is very close to May 21.

4 9. The sidereal rotation period of the Earth is solar days. The sidereal orbital period of the Moon is days. Consider someone standing on the surface of the Earth to represent one planet and the Moon another one. Calculate the synodic orbital period for the Moon (the time between Full Moons and thus the time from opposition to opposition for the Moon). As outlined in class, this question is messed up. It is phrased so that you are actually calculating the time between meridian transits of the Moon (related to the time between high tides), but ends talking about the Synodic month. Calculating either yielded full credit. The solutions to both are outlined in the Week 6 class notes. 10. Consider an asteroid in orbit around the Sun with eccentricity=0.4 and a semi-minor axis of 4 AU. What is the semi-major axis and the aphelion distances for this orbit measured in AU? What is the orbital period? How fast is this object moving when it is closest to the Sun? b2 e = 1 e is 0.4 and b is 4. Solving yields b/a = 0.917, so a=4.36 AU. a 2 Aphelion distance is a*(1+e) = 6.11 AU Orbital period just comes from P 2 = a 3. So P= 9.1 years. Perihelion velocity is given by GM a 1+e 1 e = 22 km/s 11. a) The event horizon, or Schwarzschild radius, is the distance from a (non-rotating!) black hole at which even light can not escape. Calculate the Schwarzschild radius for a black hole of 10 solar masses? Hint: The escape velocity is the speed of the light at that radius. v escape = 2 GM so setting vescape to the speed of light, c (3x10 8 m/s), and putting in the R appropriate values in the appropriate unites for G (6.67x10-11 mks) and M (2x10 31 kg), we can solve for the radius outside a 10 solar mass point mass where the velocity equals the speed of light. R s = 2 GM. With the values here we get Rs = 30 km. c 2 b) A 2-meter tall astronaut is falling towards this Black Hole and is currently at a distance of 1000 Km. What is the tidal acceleration (F/m) experienced by the astronaut from head to toe? Compare that acceleration with the Earth's gravitational acceleration at the surface of g=9.8 m/s 2. The tidal acceleration across DR is 2 GM Δ R where R is 106 meters and DR is 2 meters so a is R m/s 2, more than 400x gravitational acceleration, so the astronaut experiences a pull on the feet 400 times greater than their weight on Earth. Not a good situation.

5 12) Proxima Centauri is the closest star to the Sun and is a part of a triple star system. It has the epoch J coordinate 14h30m, -62d41m. The brightest member of the system, Alpha Centauri has J coordinates of 14h40m, -60d50m. a) What is the angular separation of Proxima Centauri and Alpha Centauri assuming the sky is flat over this angular scale? The difference in RA (14h40m vs. 14h30m) is 10 minutes of time. The difference in Dec (-60d50m vs. -62d41m is 1d51m or 111 arcminutes. Since the RA difference and Dec difference are two sizes of a right triangle and we are interested in the length of the hypotenuse we just need to get the units consistent and then take the square root of the squares of the two sides. Converting 10 minutes of RA time to a true angle in arcminutes requires multiplying by 15 and by cos(dec). So the actual arcminutes of RA is 10 * 15 * cos(60) = 10 * 15 * ½ = 75. So the total distances is ( )=134 arcminutes. b) Proxima Centauri has a parallax of 0.77 arcseconds. How far from Alpha Centauri is Proxima in units of AU assuming their separation is entirely in the plane of the sky? With a parallax of 0.77, Alpha Centauri is 1/0.77 = 1.3 parsecs away. Using the small angle equation, the separation between Alpha and Proxima in AU is then 1.3 parsecs x AU/pc x 134 /(60*57.3) = 10,000 AU