Essay About Liquid Mirror Telescope And Liquid Mirror Telescopethe Fact

The Liquid Mirror Telescope
THE LIQUID MIRROR TELESCOPEThe fact that a liquid under uniform (solid body) rotation forms a parabolic free surface profile is known since the times of Sir Isaac Newton. Here we will derive the profile using the differential form of the conservation equations. Consider a liquid in a cylindrical container that rotates with a constant angular speed ω, as illustrated in Fig. 1. The equations of motion for this problem are:Continuity[pic 1]r-momentum[pic 2]θ-momentum[pic 3]        z-momentum[pic 4] [pic 5]where                [pic 6][pic 7]FIGURE 1Originally, the container is at rest and the free surface is flat, see Fig. 1b. Once the container is forced to rotate (at t = 0+) with a constant ω, the fluid will also begin to rotate. Due to viscosity, the fluid layers in contact with the solid walls will start to move first and then the rest of the fluid elements will begin to rotate. The liquid surface under the influence of the centrifugal acceleration vθ2/r and gravity g will gradually deform. When a sufficient time has elapsed, the system will reach a steady state where there will be no more changes with time of the flow properties (∂/∂ t  = 0). In this state, the fluid will have only rotary motion (vr = vz = 0). Because the system is symmetric, all flow properties will enjoy a symmetry about the z-axis (∂/∂θ = 0). There are no agents present to influence the rotary fluid motion in any other direction except in the radial. Consequently, vθ is expected to be a function of r alone. Under the previous hypothesis continuity is automatically satisfied and the rest of the equations reduce into:r-momentum[pic 8]                                                                                (1)θ-momentum[pic 9]                                                                (2)z momentum[pic 10]                                                                        (3)The θ-momentum is the well-known Caucy equation. The general solution is found assuming vθ = const r n[pic 11][pic 12][pic 13]As r → 0, vθ → ∞ , therefore B must be equal to zero or vθ = A r. Application of the boundary condition; r = Ro, vθ = vθο = A Ro →  A = vθο / Ro = ω . Then  vθ = ω r. From Eq. (3) we have,