The height at which the acceleration due to gravity becomes g/9 (where g = the acceleration due to gravityon the surface of the earth) in terms of R, the radius of the earth, is ?
Jayant Kumar , 11 Years ago
Grade 11
2 Answers
Arun
Last Activity: 4 Years ago
Dear Jayant
GM/(9R2) = GM/(R + h)2
3R = R + h
h = 2R
Hope it helps
In case of any difficulty, please feel free to ask again
Vikas TU
Last Activity: 4 Years ago
Dear student we know that (g')/g) = (R^2)/((R +h)^2 :. (g/9)/(g) = [(R )/(R +h)]^2 :. h =2R`