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4 Pages

### P7_sol

Course: AST 111, Fall 2009
School: Rochester
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Word Count: 1444

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Set Problem #7, AST111 Solutions 1) The Voyager 1 spacecraft detected nine active volcanoes on Io. Assume that, on average, there are nine volcanoes active on Io, and that the average eruptive rate per volcano is 50km3/yr. The radius of Io is 1820 km. a) Estimate the average resurfacing rate on Io in cm/yr. b) How long does it take to completely renew the upper kilometer of Ios surface? The surface area of Io is A...

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Set Problem #7, AST111 Solutions 1) The Voyager 1 spacecraft detected nine active volcanoes on Io. Assume that, on average, there are nine volcanoes active on Io, and that the average eruptive rate per volcano is 50km3/yr. The radius of Io is 1820 km. a) Estimate the average resurfacing rate on Io in cm/yr. b) How long does it take to completely renew the upper kilometer of Ios surface? The surface area of Io is A = 4 R 2 = 4.2 107 km 2 . The resurfacing rate is (50km3 / yr 9)/(4.2 107 km3 ) = 1.1cm/yr . The time to resurface this upper kilometer is 1km/(1.1105 km/yr)=9.2 104 years. 2) Which object is better screened against meteorites, Titan (1bar surface pressure) or the Earth (also 1bar)? Assume that a meteor is stopped by an atmosphere if it encounters its mass in gas as it passes through the atmosphere. Consider a meteorite with density 4g cm-3. What diameter meteors are stopped by Titans atmosphere, and by Earths atmosphere? The effective temperature for the Earth is 288K and that for Saturn is 95K. Titans mass is 1345.51023g and its radius is 2575km. Titan and Earths atmosphere are both dominated by N2. As we found from our previous problem set the column of mass per unit area can be written as P0 / g where P0 is the surface pressure and g is the gravitational acceleration at GM 6.7 108 1.34 1026 g = = 135 cm s -2 . The acceleration (2575 105 cm) 2 R2 is about 1/7th of that on the Earth. Since the two have similar surface pressures, this implies that Titan has about 7 times the column of mass per unit area in its atmosphere compared to that of the Earth. 1bar = 106 dynes cm -2 . The mass per unit area on the Earth is P0 / g = 106 dynes cm -2 /1000cm s -2 = 103 g cm -2 . The mass of a meteor with the surface. For Titan g = and the surface area is A = d 2 / 4 . So the mass per unit area 6 M / A = 2 d / 3 . Setting this equal to the atmospheric column we find the the 3P atmosphere will stop a meteor with diameter less than d = 0 . Now we solve for the 2 g diameter for the Earth and for Titan. The Earths atmosphere would stop a meteor with 3 106 dynes cm-2 = 375cm . Titan has the same surface diameter less than d = 2 4g cm -3 103 cm s -2 pressure but an surface gravity that is 7 times smaller. So Titan will stop a meteor that is 7 times bigger or with a diameter of 2600cm. diameter d is M = d 3 3) Consider the impact between an iron meteoroid (density 7 g cm-3) with a diameter of 300m and the Moon. a) Calculate the kinetic energy involved if a meteroid hits the surface of the Moon at a velocity of 12 km/s. mv 2 v 2d 3 7g cm -3 (12 105 cm) 2 (300 102 cm)3 E= = = = 7 1025 erg=7 1018J 2 12 12 An empirical scaling relation exists for crater diameters D = 1.8 m 0.11 p 1/ 3 g p 0.22 d m 0.13 Ek 0.22 (sin )1/ 3 where m and p are the densities of the meteor and planet g p is the gravitational acceleration on the surface of the planet is the angle of impact from horizontal d m is the diameter of the meteor, Ek is the kinetic energy of the meteor. All units are mks for this relation, so D is in meters. b) Estimate the size of the crater formed by a head-on collision, and one where the angle of impact with respect to the horizon is 30o. To use this I need to put everything in mks. GM 6.7 108 734 1023 g gp = 2 = = 163 cm s -2 = 1.6m s -2 5 2 R (1737 10 cm) 3 m = 7 10 kg m -3, p ~ 3 103 kg m -3 Plugging into to the empirical formula: D = 1.8 (7 103 )0.11(3 103 ) 1/ 3 (1.6) 0.22 (300)0.13 (7 1018 )0.22 (sin 90)1/ 3 = 8800m If the angle of impact is 30o then the crater is smaller or about 7000m large. 4) PS 5.13. The secondary craters related to a crater of a given size on Mercury typically lie closer to the primary crater than do the secondary craters of a similarly sized primary on the Moon. Presumably, this is the result of Mercurys greater gravity reducing the distance which ejecta travel. a) Verify this difference quantitatively by calculated the throw distance of ejecta launched at a 45o angle with a velocity 1 km/s from the surface of Mercury and the Moon. The equatorial value of Mercurys surface gravity is 3.7 m s-2 as compared to 1.6 on the Moon. Mercurys surface gravity is 2.3 times as large as that on the Moon. The horizontal upward and components of the initial velocity are cos(45o) and sin(45o) times the total velocity or 0.7km/s=7000m/s. The total time of flight would be t~2vz/g or t ~ 2 7000 / 3.7 = 3780 sec on Mercury. Multiply by the horizontal component of the velocity to get the ejecta travel distance or 2.6 107 m . Since the Moon has a smaller surface gravity the travel time is longer and the distance is larger (2.3 times as large) or 6 107 m . b) Typical projectile impact velocities are greater on Mercury than they are on the Moon. Explain why. (Hint: consider both orbital motions and escape velocities). Does this difference counteract the surface gravity effect mentioned above? (Hint: consider the empirical scaling law for craters of the same diameter on the two bodies, and consider assumptions for predicting the ejecta velocity). Surprisingly ejecta curtains on Mercury are observed to be smaller than those on the Moon (for similar diameter craters) despite the fact that higher kinetic energy impacts are required to create the same size crater on Mercury as on the Moon. 1) Objects in the inner Solar system typical move faster than objects in the outer GM solar system v 2 ~ . Because Mercury is at 0.4 AU and the Moon is at 1AU d we expect approach velocities to differ by 0.4 or a factor of about 0.6. The velocity of an object in a circular orbit at Earths radius is 30km/s and at Mercurys radius is 47km/s. 2) Objects incoming with small velocities relative to Mercury or the Moon will have impact velocities approximately equal to the escape velocity. For the Moon the escape velocity is 2GM 2 6.7 108 735 1023 g = = 2.4km/s . R 1737 105 cm For Mercury the escape velocity is ve = 2GM 2 6.7 108 3.3 1026 g = = 4.2km/s R 2440 105 cm The escape velocity from Mercury is about twice as high as that on the Moon. How do the escape velocities compare to the approach velocities? Note the orbital motions far exceed the escape velocities. So objects in moderately eccentric orbits will approach the Moon and Mercury with velocities that are well above the escape velocity. Consequently the velocity of impact is primarily set by the approach velocity and not the gain in velocity as the object descends down onto Mercury or the Moon. ve = The gravity effect mentioned above refers to two craters of the same size diameter. Empirical scaling laws predict that crater size depends on surface gravity as well impact velocity and density of planet or satellite or D 1/ 3 ( Ek / g ) 0.2 . Assume we have a crater of the sa...

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