Letting and where h11ErRh=+2E21and h2are the heights of the 2 satellites above the ground. Inserting these into the above gives … 33322122tEEEhhTRRRRgRgππ++=+From Example 6.6.2, RE= 6371 km, h1= 200 mi = 324 km and h2= r2- RE= 42,400 km - 36,029 km. Putting in the numbers … Tt= 4.79 hr (b) Thus, 000180110812tTθ=−=6.33The potential for the inverse-cube force law is …()22kr=VrLetting ur1−=, we have (Equation 6.9.3) 2112dumluV uEd−++=222EVduudml−=−22 22222mlddumE Vmluuml==−−−
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This note was uploaded on 01/06/2012 for the course PHYSICS 4360C taught by Professor Johnanderson during the Fall '11 term at UNF.