This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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) Nonspherical asteroid R R α 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 αR . Then the typical stress due to selfgravity 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 nonspherical asteroid is then given by R max = r T max αG × 1 ρ For nonspherical 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 nonspherical 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...
View
Full Document
 Spring '09
 Physics, Force, R1 R2, Wire, steel wire, αGρ2 R2

Click to edit the document details