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Diameter and Depth of Lunar Craters

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Diameter and Depth of Lunar Craters Student Name Alan Brockman HET602 Student ID 1611739 Project Supervisor Barry Adcock [email protected] SAO Project Cover Page Project 14 Project Title Diameter and Depth of Lunar Craters Submitted 15 June 2008 All of the work contained in this project is my own original work, unless otherwise clearly stated and referenced. I have read and understood the SAO Plagiarism Page “What is Plagiarism and How to Avoid It” at http://astronomy.swin.edu.au/sao/students/plagiarism/ 1 Introduction Galileo was the first to attempt measuring mountains of the Moon using a light-tangent method that estimates a peak‟s distance from the lunar terminator at the moment the peak is first (or last) illuminated. He calculated heights of over 7,000 metres, a vast overestimation. Using the same method in 1787, William Herschel used an eyepiece micrometer to determine that lunar hills range in height between 800 and 2,400 metres1. During the same time period, Johann Schroter introduced a new “shadow method” for height determination which was modified by Heinrich Olbers for improved accuracy7. This method has since been used to measure thousands of lunar hills and mountains being replaced only since the advent of remote sensing by lunar probes. AIM The aim of this project was to measure the size and depth of features on the lunar surface using one or more techniques. Two methods are described which utilize digital imagery and trigonometry to calculate the size and depth of selected lunar craters. It is proposed to use these methods and compare their accuracy to published data. METHOD Digital imaging of the Moon This was accomplished using a permanently mounted 14” f/10 Schmidt Cassegrain Telescope and a simple Philips ToUcam web camera (see Figure 1). The large scale image of the moon was accomplished by use of a 4” Takahashi FSQ-106 refractor and a SBIG STL- 11000 CCD camera. Images were taken of selected targets along the Lunar terminator when the Moon was 20 days old. All images were taken in the early hours of May 26 2008 from my observatory in Exmouth, Western Australia. Objects chosen showed good surface structure and were close to the Lunar terminator at time of imaging for maximum relief and shadow length. Statistical analysis comparing paired measurements non-parametrically were performed with the Wilcoxin Signed Ranks Test using SPSS 11.0 for Windows. 2 Figure 1. My equipment set up consisting of a Tak FSQ106 piggy-backed onto a 14” f/10 SCT and mounted on a Paramount ME inside a Sirius Dome. Method 1 for determination of the size (diameter) of Lunar craters The time and date was recorded and used to determine the correct distance and the angle of the Sun to the Lunar surface. Heights of lunar features were calculated by measuring the length of the shadow of the feature and determining the angle of the Sun at the Moon with the help of an Astronomical Almanac. Calculating the height of a lunar feature required solving a series of plane and spherical triangles that related the positions of the Sun and Earth to the position of the feature. One of these triangles involves the point on the Moon where the Earth is directly overhead, which the Astronomical Almanac tabulates as the Earth‟s selenographic latitude and longitude. The Sun‟s selenographic colongitude and latitude were also recorded. Steps 1. A number of potential targets on the lunar surface are selected showing good surface structure and relief. 3 2. Images are taken and the precise time noted to enable distance and angle of the Sun on the lunar surface. 3. The Sun‟s selenographic longitude is taken from published ephemerides to find the length of the target‟s shadow on the lunar surface. 4. The official crater size and latitude and longitude are obtained and compared. Method 2 for determining the heights or depths of Lunar features2 Regions were chosen near the lunar terminator so that the mountains cast long, prominent shadows. Determine the selenographic location of the lunar features The selenographic colongitude is the longitude of the morning terminator on the Moon, as measured in degrees westward from the prime meridian. The morning terminator forms a half-circle across the Moon where the Sun is just starting to rise. As the Moon continues in its orbit, this line advances in longitude and the value of the selenographic colongitude increases from 0° to 359° in the direction of the advancing terminator3. The approximate lunar latitude and longitude (to the nearest degree) of the mountains were found using a lunar atlas3. The selenographic longitude of the morning terminator measured westward from the prime or central meridian is known as the co-longitude. The relationship between selenographic longitude, λ, and co-longitude, c, is as follows: Λ = selenographic longitude of morning terminator = 360º - c = sunrise on the Moon. Measure the apparent length of the shadows Using CCDSoft image processing software, the number of pixels from the rim of a crater to the end of its cast shadow were determined. Several craters were chosen with well-defined rims or walls. The size of these craters were recorded in a line perpendicular to the terminator shadow. Convert the apparent length from pixels to kilometres The image scale per pixel in arc seconds was determined and converted to degrees Correct the apparent length of shadows for tilt of the lunar surface (foreshortening) Tilt correction factor = 1/(cos (latitude) + cos (longitude)). See Figure 2. This correction factor assumes that the observer is directly above the point on the lunar disk which can also be corrected for greater accuracy. Multiply the shadow length by this factor to get the corrected shadow length (km). 4 Figure 2. Correcting for angle tilt4 Formulae used for determination of crater length Image scale, or angular size per pixel, υpixel, υpixel = 206265 x dpixel/F [arcseconds], (Eqn. 1) -3 υpixel = 206265 x 5.55 x 10 /3500 = 0.327 = 0.33 arcsec/pixel Converting apparent length from pixels to kilometres, 1 pixel = 0.33 arc sec = 0.33 arcsec/3600 = 9.16 x 10-5 degrees Earth-Moon distance at 26 May 2008 = D = 395,633.44 km. Linear distance on the Moon covered by one pixel , ldpixel, -5 ldpixels = D [km] x tan 9.16 x 10 [degrees], (Eqn. 2) -5 ldpixels = 395,633.44 x tan (9.16 x 10 degrees) = 0.6325 km. Correction factor for foreshortening due to lunar curvature, cf = 1/[cos (latitude) x cos (longitude)] (Eqn. 3) Distance between two pixels at coordinates (x1, y1) and (x2, y2) can be determined using the 2 2 2 Pythagorean Theorem for 3 points forming a right triangle where (x1 - x2) + (y1 – y2) = z . 2 So that uncorrected length of crater, clupixels = √ z (Eqn. 4) Corrected length of crater, cl = clupixels x cf Length of crater, L (km) = cl x ldpixels (Eqn. 5) 5 Determining a mountains height from the length of its shadow Determine the altitude of the Sun over the mountains at the time of the image Checking an ephermerides using the date and time of the observation will determine the solar colongitude (Co) and the subsolar point Latitude (Bo). The angle of the Sun above the mountain can be calculated by: x = sin(Bo) x sin(latitude) + cos(Bo) x cos(latitude) x sin(Co + longitude) angle θ = arcsin(x) Calculate the height of the mountains using simple trigonometry Now that the shadow length (L) and the angle θ has been determined, the height of the lunar peak can be calculated using simple trigonometry where, cot θ = L/H6 Therefore, height (H) = cot θ/L. See Figure 3. Figure 3. Calculating heights using simple trigonometry4 Optical specifications 14” SCT at f/10 gives a focal length, F, = 3500 mm. With a Phillips ToU webcam, -3 Pixel size, dpixel, = 5.5 um x 5.6 um = 5.55 um (average) = 5.55 x 10 mm Array size 659 x 494 pixels Chip size 2.8 x 3.7 mm Field of view = 2.7 x 3.6 arcmin 6 RESULTS Table 1. Moon Ephemeris for 06:30, 26 May 20084. Figure 4 shows the position of the terminator for the 20 day old Moon in the early hours of May 26 taken with the FSQ-106 at f/5. 7 Figure 4. Day 20 Moon showing position of the terminator and major features Figures 5, 6 and 7 describe some of the more obvious Lunar craters and other features along the terminator. 8 Figure 5. 90°N to 15°N Figure 6. 17°N to 24°S 9 Figure 7. 22°S to 90°S 10 SECTION 1. Determining Crater Size Twelve craters were selected targets for crater size estimation covering varying established size ranges and latitudes. UT date at the start of the imaging session was yyyymmdd.hhmm: 20080525.2241. This enabled determination of the distance to the Moon, D, as 395,633 km. 11 See Figure 8 for the selected craters, W. Bond and Archytas. Time of image: 26 May 2008, 06:53 am. 1. W. Bond [65.3°N, 3.7°E]. Walled plain (158 km)1. cf = 1/(cos 65.3 x cos 3.7) = 1/(0.4179 x 0.9979) = 2.3979 Length of crater is defined by two points with coordinates (xi, yi) of (308, 415) and (300, 292) as determined using the program CCDSoft. 2 2 clupixels = √(308 - 300) + (415 - 292) = √(64+ 15129) = 123.260 L = 123.260 x 2.3979 x 0.6325 = 186.94 km. Diameter of Crater W. Bond = 187 km 2.
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