Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7The Solar System MissionsExposure timesLight from a source is collected by a 6 inch telescope by a high school science class. Rather than using an eyepiece, the image is recorded by a camera on film. Since the source is fairly dim, a 30 minute time exposure is required to adequately record the image. If the same source is observed with the 200 inch Mt. Palomar telescope, what exposure time will be requried to produce the same photo, assuming all else is equal?Exposure time is determined by the number of photons collected. Best guess – exposure time α 1/ d2Number of photons α d2Therefore, (exposure time 1) (d12 ) = (exposure time 2) (d22 )Exposure time 2 = (d1/d2)2 (exposure time 1) = 0.027 min = 1.62 sec.planets not shown to scale >> Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune PlutoMean Distance from the Sun (AU) 0.3871 0.7233 1 1.524 5.203 9.539 19.19 30.06 39.48Sidereal period of orbit (years) 0.24 0.62 1 1.88 11.86 29.46 84.01 164.79 248.54Mean Orbital Velocity (km/sec) 47.89 35.04 29.79 24.14 13.06 9.64 6.81 5.43 4.74Orbital Eccentricity 0.206 0.007 0.017 0.093 0.048 0.056 0.046 0.010 0.248Inclination to ecliptic (degrees) 7.00 3.40 0 1.85 1.30 2.49 0.77 1.77 17.15Equatorial Radius (km) 2439 6052 6378 3397 71490 60268 25559 25269 1160Polar Radius (km) same same 6357 3380 66854 54360 24973 24340 sameMass of planet (Earth=1) 0.06 0.82 1 0.11 317.89 95.18 14.53 17.14 0.002Mean density (grams/centimeter³ ) 5.43 5.25 5.52 3.95 1.33 0.69 1.29 1.64 2.03Body rotation period (hours) 1408 5832 23.93 24.62 9.92 10.66 17.24 16.11 153.3Tilt of equator to orbit (degrees) 2 177.3 23.45 25.19 3.12 26.73 97.86 29.6 122.46Number of observed satellites 0 0 1 2 >28 30 24 8 1Io orbits Jupiter with a period of 1.77 days. At a time when Jupiter is 8 x 108 km, Io is at its furthest position from Jupiter as observed from the earth. Further, as viewed from the earth, the angle between Jupiter and Io is 108”. Based on this information, what is the mass of Jupiter?EarthJupiterIo108”108” = 0.03o Tan (0.03) = 0.000524 = R / 8 x 108 km → R = 418,879 km = 4.2 x 108 mR8 x 108 kmIo orbits Jupiter with a period of 1.77 days. At a time when Jupiter is 8 x 108 km, Io is at its furthest position from Jupiter as observed from the earth. Further, as viewed from the earth, the angle between Jupiter and Io is 0.109”. Based on this information, what is the mass of Jupiter?108” = 0.03o Tan (0.03) = 0.000524 = R / 8 x 108 km → R = 418,879 km = 4.2 x 108 m1.77 days = 1.53 x 105 secM = v2 R / G = (2πR/T)2 R/G = 4π2R3 / T2 G = 4 π2 (4.2 x 108 )3 / (1.53 x 105 ) (6.6 x 10-11) = 1.8 x 1028 kg = 314 times the mass of the earthAs seen from the earth, Jupiter has an angular diameter of .01 degrees. What is the density of Jupiter?EarthJupiter0.01 degreesTan (0.01) = 0.000174 = R / 8 x 108 km → D = 139,200 km =Radius approximately equal to 70,000 km. 8 x 108 kmAs seen from the earth, Jupiter has an angular diameter of .01 degrees. What is the density of Jupiter?Tan (0.01) = 0.000174 = R / 8 x 108 km → D = 139,200 km =Radius approximately equal to 70,000 km.Density = m / V = m / ( 4/3 π R3 ) = (1.8 x 1028 kg ) / (4/3 π (7 x 107 m)3 ) = 1.25 x 103 kg/m3 = 1.25
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