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The height at which the acceleration due to gravity becomes g/9 (where g = the acceleration due to gravity on the surface of the earth) in terms of R, the radius of the earth, is ?

The height at which the acceleration due to gravity becomes g/9 (where g = the acceleration due to gravity
on the surface of the earth) in terms of R, the radius of the earth, is ?

Grade:11

2 Answers

Arun
25763 Points
one year 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
14149 Points
11 months ago
Dear student  
we know that (g')/g) = (R^2)/((R +h)^2
:. (g/9)/(g) = [(R )/(R +h)]^2 
:. h =2R`
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