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Unformatted text preview: Solution for Chapter 10 (compiled by Xinkai Wu) A. 10.3 Order of magnitude estimates (i) Steel wire [C.Y.Mou/90] z=L z=0 Figure 1: Steel wire The weight of the wire creates stress inside, T zz,z + g = 0 T zz = gz The maximum stress is at z = L , T max zz = gL = E , where is the strain and E is the Youngs modulus. Typically the wire would break if the strain > 10 3 . Hence, the maximum length of a wire is : L = T max zz g = E g 10 3 E steel g Plugging in E = 210 GP a , = 7 . 8 g cm 3 for steel we get L 3 10 3 m ,i.e. for a steel wire to break under its own weight it would have to be several kilometers long. (ii) Nonspherical asteroid R R M Figure 2: Nonspherical asteroid 1 Consider first an asteroid of mass M that has a roughly spherical shape with radius R and deviation from sphericity of order R . Then the typical stress due to selfgravity is T F S GM R 2 ( R ) 3 1 ( R ) 2 G 2 R 2 If the stress exceeds the elastic limit the asteroid will deform under it be coming more spherical. The maximum size for a nonspherical asteroid is then given by R max = r T max G 1 For nonspherical asteroids, we can take 1. Further, taking = Earth = 5 g/cm 3 , typical maximum strains 10 3 and maximum stress T max = E 10 8 P a , its then easy to find R max 200 km We conclude that the biggest nonspherical asteroids are several hundred kilometers in size. (iii) Hellium Balloon [N. Niorris/85] First, lets recognize that the best geometry for the tank is spherical. It maximizes the volume to surface area ratio, and therefore means the lightest tank for a given volume of helium. Further suppose that the tank has an inner radius of R and thin walls of thickness d . Then the mass of the tank is M t = 4 t R 2 d = 3 t V t d/R , where V t is the tank volume. Treating the helium in the tank and in the balloon as an ideal gas so P V/T = K Boltzman (number of Helium molecules), we can write V b = P t V t T t T b P b The lifting bouyant force on the balloon is F b = V b ( air He ) g V b air g , with g being the acceleration of gravity (assuming that the gas in the balloon is at the same T and P as the surrounding air). The balloon will lift the tank if the ratio = F b M t g is greater than one. Using the expressions above: = P t P b T b T t R 3 d air t Obviously, the more compressed the gas in the tank is, the bigger the bouyant force. The stress in the tank material is of order = P t R/d and so, the maximum pressure the tank can hold is P max t 10 3 E d R 2 and so = 0 . 3 10 3 E P b T b T t air t Using E = 2 10 6 atm for steel, air t = 1 . 7 10 4 , we get . 1 T b T t Therefore if the helium in the tank was initially at room temperature it wont be possible to lift the tank with a balloon. It may become feasible if the helium in the tank is cooled to a very low temperature, then is warmed when...
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This document was uploaded on 04/17/2010.
 Spring '09
 Physics

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