at what altitude above the earth's surface is the free fall acceleration equal to three quarters of its value at the surface ?

1 answer

1. g wherever = some constant k/r^2
   let r = earth radius
   let R = radius with 3/4 g
   then we want R-r
   
   g surface = k /r^2
   g up = k/R^2
   
   g up/g surface = k/R^2/k/r^2 = r^2/R^2 = 3/4
   so
   R/r = sqrt 4 /sqrt 3
   look up r, radius of earth and you can find R, then subtract.

   Answer ID
   524589

   Created
   April 3, 2011 9:29am UTC

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   https://questions.llc/answers/524589