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Unformatted text preview: Homework 1 University Astronomy University Astronomy 1017301 Homework Assignment 1 Instructor: Dr Andrew Robinson Due Date: 4pm Tuesday, 21 September 2006. To tackle some of the following problems you will need to do some research, using either the textbook or other resources, to find various pieces astronomical data and the values of certain physical constants. For questions requiring calculations, show all important steps. 1. Suppose that a highly advanced (and highly evil) interstellar civilization hijacks our Sun by opening a spacetime wormhole into which the Sun instantaneously vanishes. What period of time would elapse before we on Earth noticed the Sun disappear from the sky (it is daytime). t = c/1 AU 8m 20s
Apart from the sudden disappearance of our source of light and heat, what other serious consequence would there be for Earth? How long would it be before Earth feels this effect? Sun's gravitational force would vanish. Since according to Einstein, information cannot travel faster than the speed of light, Earth would leave its orbit after 8m 20s.
2. The largest of Mars' two moons is Phobos. Jupiter's largest moon is Ganymede. On paper, estimate the angular diameters of these satellites as seen from the surfaces of their respective planets. (OK, Jupiter doesn't have a solid surface, but we'll take a point in the outer cloud layer for this purpose). Considering your results, determine if a total solar eclipse is possible on either planet. Note: relevant data can be found in the appendices of Universe, or from this web site: http://www.nineplanets.org/. 10/6/06 1 Homework 1 University Astronomy Look up the diameter of each moon (D) and the moonplanet distance (a) in each case. The latter will usually be measured from the center of the planet and so we need to subtract the planet's radius (r), since we are observing from the surface. Now apply the small angle formula to determine the angular diameters: = D/(ar) radians. Converting to degrees, we find Phobos from surface of Mars: = 0.11 Ganymede from surface of Jupiter: = 0.15 A total eclipse is possible if the angular diameter of the moon is at least equal to the angular diameter of the Sun as seen from the planet. This is again calculated by applying the small angle formula using the diameter of the Sun and the planetsun distance (the radius of the planet is insignificant in this case). You should find the following: Sun seen from Mars = 0.35 Sun seen from Jupiter = 0.10 Therefore Phobos cannot produce a total eclipse on Mars, but Ganymede can on Jupiter. 3. Solar heating. The average amount of radiant energy from the Sun arriving at the top of Earth's atmosphere (the Solar Constant) is 1373 W m2. Estimate the radiant energy flux falling on a flat 1 m2 patch of ground located at 43 N latitude at the following times: (a) Noon on the Summer Solstice (b) Noon on the Winter Solstice Calculate the fractional difference in flux between the Summer and Winter Solstices, relative to the solar constant. Compare this with the fractional difference in radiant flux arising from the slight ellipticity of Earth's orbit (recall that the SunEarth distance varies by 3%). Notes: assume that the Earth is a perfect sphere and that no absorption of solar radiation occurs in the atmosphere. In general the radiant flux passing through a surface of area A is FsolAcos, where Fsol is the solar flux arriving at Earth and is the angle between the normal to the surface and the incoming rays. (Lecture 3, slide 10). Since Fsol is given and A=1 m2, we simply need to determine in each case. At noon on the Winter solstice, incoming rays make an angle of 66.5 to the normal to the Earth's surface at latitude 43 (see diagram below)
10/6/06 2 Homework 1 University Astronomy illuminated surface North Pole 43+23.5=66.5 Incoming sunlight 43 Ecliptic plane Equator 23.5 Similarly, on the Summer solstice incoming sunrays make an angle 4323.5=19.5 with the normal to the ground. The fractional difference in flux is just (FsummerFwinter)/Fsol=cossummer coswinter) = 0.54. The fractional difference in flux due to the Earth's elliptical orbit is most easily found by differentiating the equation relating flux and luminosity: F/r =2L/(4r3). Then F/Fsol = 2r/r. Since r/r =0.03, the corresponding fractional difference is 0.06 i.e., 9x less than that due to the variation in the apparent height of the Sun.
4. An observer located on the Greenwich Meridian (longitude 0) observes that a particular star passes his/her celestial meridian at midnight UT (Universal Time). Exactly 5h 10m 24s later, the same star passes the celestial meridian of another observer. What is the longitude of this observer? If the observer's latitude is N 43d 10m, where is he/she located? Earth rotates 15/h hence in 5.1733 h, it rotates 15x5.1733 = 77.6 or 77 36'. Since Earth rotates Eastward, 2nd observer must be 77 36' West of Greenwich. Observer location is Latitude N 43d 10m, W 77 36' Rochester, NY (as if you hadn't guessed...) 5. Set up a planetarium program so that you are observing from Rochester at 23:00 (local time) on September 15 2005. Use the Find function to locate the following stars (a) Polaris 3 10/6/06 Homework 1 University Astronomy (b) Capella Determine the coordinates of each star in both the horizon (Azm & Alt) system and in the equatorial system (RA & dec). Write them down. Now skip forward in time 5 hours (to 4 am on Sept 16) and remeasure the coordinates, Write them down and compare the new coordinates with the previous set. The time skip caused the coordinates to change in some cases but not in others. Why? If you don't have Starry Night, Sky View Caf (http://www.skyviewcafe.com/) will also do the job. Polaris 23:00 9/16: RA 02 31 Dec +89 15; Azm 00 57 Alt +43 20 04:00 9/17: RA 02 31 Dec +89 15; Azm 00 00 Alt 43 52 Capella 23:00 9/16: RA 05 16 Dec +45 59; Azm 40 42 Alt 15 55 04:00 9/17: RA 05 16 Dec +45 59; Azm 70 08 Alt 61 29 RA & Dec do not change because equatorial coords fixed relative to sky Azm & Alt change because horizon system is measured relative to local horizon & zenith Azm & Alt do not change much for Polaris because it is near NCP
6. At approximately what time of day (dawn, noon, dusk, midnight) does the Moon rise and set when it is in the following phases: (a) Full (b) 1st Quarter Explain your answers with the aid of diagram. (a) Moon rises at dusk and sets at dawn. (b) Moon rises at noon and sets at midnight. See Universe Ch. 2 Fig. 7. Electromagnetic radiation can be regarded either as a wave or as a stream of photons. Calculate the energy (in Joules) of individual photons at the following wavelengths: (a) 1 nm (Xray band) (b) 500 nm (visible) (c) 10 m (Infrared) (d) 1 m (radio) A 100 W domestic light bulb radiates at a wavelength of about 1 m. Approximately many photons does it emit per second? 4 10/6/06 Homework 1 Note: 1 m = 106 m; 1 nm = 109 m. University Astronomy Each photon has energy E = h, where h = Planck's constant (=6.626x1034 Js) and = frequency. Since c = for an EM wave, we can write this in terms of wavelength: E = hc/. Remembering to convert all wavelengths to meters, we find the following photon energies: (a) (b) (c) (d) 2x1016 J 4x1019 J 2x1020 J 2x1025 J Using the same method a light bulb radiating at a wavelength of 1 m emits photons of energy 2x1019 J. A 100 W bulb therefore radiates (100 J/s)/( 2x1019 J) = 5x1020 photons/s. 8. Estimate the following. (a) The number of photons emitted per second by a 100 W domestic light bulb radiating at a wavelength of 1 m. (b) The number of photons emitted per second by the Sun assuming that it radiates at a wavelength of 0.5 m. (c) The number of photons from the Sun that fall on Earth per second. (Hint: what is the projected area of the Earth as seen from a distant point in space?) As in Qu. 6, the energy of the emitted photon is E = hc/. If the luminosity of the source is L (units are J/s), then the number of photons emitted is N = L/E, where E has units of Joules. (a) Wavelength = 1 m, so photon energy E=2x1019 J. A 100 W bulb therefore radiates N=(100 J/s)/(2x1019 J) = 5x1020 photons/s. (b) Wavelength = 0.5 m, so photon energy E=4x1019 J. The Sun therefore radiates N=(4x1026 J/s)/(4x1019 J) = 1045 photons/s. (c) Photons are subject to the inverse square law, so the number of photons per m2 reaching Earth is F = N/(4r2), where r = 1 AU = 1.5x1011 m. The number actually "collected" by Earth is n = F x Earth's crosssectional area = F x R2, where R = 6.4x106 m is Earth's radius. Hence n = (N/4)x(R/r)2 = 4.5x1035 photons/s. 10/6/06 5 ...
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