Astro 210 Homework 6 Solutions

# Astro 210 Homework 6 Solutions - Astronomy 210 Spring 2011...

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Unformatted text preview: Astronomy 210 Spring 2011 Homework Set #5: Solutions 1. On the Lookout for the Asteroid of Doom . (a) [5 points] A mass m with speed v has kinetic energy KE = 1 / 2 mv 2 , and if it is a distance d from another mass M , the gravitational potential energy is PE = − GMm/d . The total energy is conserved and is TE = PE + KE = 1 2 mv 2 − GMm d (1) Thus a mass m which falls to earth from a distance r with initial speed v has an initial total energy TE i = PE i + KE i = (2) Then when the object hits the surface of the earth with speed v hit , its total energy is TE f = PE f + KE f = 1 2 mv 2 hit − GMm R ⊕ (3) Since energy is conserved, TE f = TE i , and so 1 2 mv 2 hit − GMm R ⊕ = 1 2 mv 2 − GMm r (4) v 2 hit = v 2 + 2 GM parenleftbigg 1 R ⊕ − 1 r parenrightbigg (5) = v 2 + 2 GM R ⊕ parenleftbigg 1 − R ⊕ r parenrightbigg (6) = v 2 + v 2 esc parenleftbigg 1 − R ⊕ r parenrightbigg (7) (8) where the escape speed is v esc ( r ) 2 = 2 GM/r . But since r ≫ R ⊕ , we have 1 − R ⊕ /r ≈ 1, and thus v 2 hit ≈ v 2 + v 2 esc ≥ v 2 esc (9) and thus v hit ≥ v esc , that is, the impact speed is always at least the escape speed. (b) [5 points] A sphere with radius s and with uniform density ρ has mass m = ρV = 4 π 3 ρs 3 (10) and thus such an impactor with speed v esc has kinetic energy KE = 1 2 mv 2 esc = 1 2 4 π 3 ρs 3 2 GM R = 4 π 3 GMρs 3 R (11) For a medium-sized s = 1 km asteroid with density ρ = ρ rock = 3000 kg / m 3 , this gives KE = 7 . 9 × 10 20 J = 1 . 9 × 10 8 kTon = 1 . 9 × 10 5 MTon (12) This is much larger than a large nuclear weapon and indeed is larger than the entire global nuclear arsenal. Thus a 1 km asteroid impact would be devastating to the region hit, and could threaten global civilization. 2 (c) [5 points] The solar power intercepted by an asteriod is F ( d ) S , where F is the solar flux at distance d , and S is the area of the asteroid’s shadow. But F ( d ) = L ⊙ 4 π (13) and S = πs 2 . Thus the intercepted power is ( s 2 / 4 d 2 ) L ⊙ . If the asteroid has albedo A , then the luminosity L of the sunlight reflected by the asteroid is just A times the intercepted power: L =...
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Astro 210 Homework 6 Solutions - Astronomy 210 Spring 2011...

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