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soln10

# soln10 - Solution for Chapter 10(compiled by Xinkai Wu A...

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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 + ρg = 0 ⇒ T zz = ρgz The maximum stress is at z = L , T max zz = ρgL = E , where is the strain and E is the Young’s modulus. Typically the wire would break if the strain > 10- 3 . Hence, the maximum length of a wire is : L = T max zz ρg = E ρg ≈ 10- 3 E steel ρg Plugging in E = 210 GP a , ρ = 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) Non-spherical asteroid R R α M Figure 2: Non-spherical asteroid 1 Consider first an asteroid of mass M that has a roughly spherical shape with radius R and deviation from sphericity of order αR . Then the typical stress due to self-gravity is T ≈ F S ≈ GM R 2 × ρ ( αR ) 3 × 1 ( αR ) 2 ≈ αGρ 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 non-spherical asteroid is then given by R max = r T max αG × 1 ρ For non-spherical asteroids, we can take α ≈ 1. Further, taking ρ = ρ Earth = 5 g/cm 3 , typical maximum strains ≈ 10- 3 and maximum stress T max = E ≈ 10 8 P a , it’s then easy to find R max ≈ 200 km We conclude that the biggest non-spherical asteroids are several hundred kilometers in size. (iii) Hellium Balloon [N. Niorris/85] First, let’s 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 πρ t R 2 d = 3 ρ 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 ( ρ air- ρ He ) g ≈ V b ρ 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 κ = F b M t g is greater than one. Using the expressions above: κ = P t P b T b T t R 3 d ρ air ρ 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 σ = 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 κ = 0 . 3 × 10- 3 E P b T b T t ρ air ρ t Using E = 2 × 10 6 atm for steel, ρ air ρ t = 1 . 7 × 10- 4 , we get κ ≈ . 1 × T b T t Therefore if the helium in the tank was initially at room temperature it won’t 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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soln10 - Solution for Chapter 10(compiled by Xinkai Wu A...

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