P15_021 - p 2 = p 1 − Z h ρg dy Assuming ρ = ρ(1 −...

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21. (a) We use the expression for the variation of pressure with height in an incompressible fluid: p 2 = p 1 ρg ( y 2 y 1 ). We take y 1 to be at the surface of Earth, where the pressure is p 1 =1 . 01 × 10 5 Pa, and y 2 to be at the top of the atmosphere, where the pressure is p 2 = 0. For this calculation, we take the density to be uniformly 1 . 3kg / m 3 . Then, y 2 y 1 = p 1 ρg = 1 . 01 × 10 5 Pa ³ 1 . 3kg / m 3 ´³ 9 . 8m / s 2 ´ =7 . 9 × 10 3 m=7 . 9km . (b) Let h be the height of the atmosphere. Now, since the density varies with altitude, we integrate
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Unformatted text preview: p 2 = p 1 − Z h ρg dy . Assuming ρ = ρ (1 − y/h ), where ρ is the density at Earth’s surface and g = 9 . 8 m / s 2 for 0 ≤ y ≤ h , the integral becomes p 2 = p 1 − Z h ρ g ³ 1 − y h ´ dy = p 1 − 1 2 ρ gh . Since p 2 = 0, this implies h = 2 p 1 ρ g = 2(1 . 01 × 10 5 Pa) (1 . 3 kg / m 3 )(9 . 8 m / s 2 ) = 16 × 10 3 m = 16 km ....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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