dvVkdtmk t=+−DThus ()1lndtkVm kt Cmkt==−−−∫DDvVwhere Cis a constant. +1Now, vso 0@0t1lnCVm=DHence lntm−=−DD, 0mtkε≤≤DLet ybe the displacement at the time tso 2lntyVdtCm−+∫DDIntegrating the above expression by parts 2lnttdttVkCmmkt−−+−∫DDD21tmVtVdtCkt−+−+−∫DDln2lntVmVtVtmk tC−++−+DDDDlnsince 0y=at t, 0=2lnVmmkC−=and we have ( )lntVttkm−−DDDAt burn-out mktTDso (a) 1ln1mVyDkεε+−−D(b) cannot exceed 1.0 although it can approach 1.0 for small payloads Thus maxlim1yyk→D7.27From eqn. 7.7.5,
This is the end of the preview. Sign up
access the rest of the document.
This note was uploaded on 01/06/2012 for the course PHYSICS 4360C taught by Professor Johnanderson during the Fall '11 term at UNF.