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The height at which the acceleration due...

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

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Given `g/9=g(R/(R+h))^(2)impliesR/(R+h)=1/3`
`3R=R+himplies2R=h`
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