The height in terms of radius of the earth $(R)$, at which the acceleration due to gravity becomes $

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The height in terms of radius of the earth $(R)$, at which the acceleration due to gravity becomes $\frac{g}{9}$, where $g$ is acceleration due to gravity on earth's surface, is $\_\_\_\_$ .

Accepted Answer

Sol. [Key Concept: $g_{h} = \frac{g}{(1 + h/R)^{2}}$; set it equal to $g/9$, so $1 + h/R = 3$.]$${\ g_{h} = \frac{g}{\left( 1 + \frac{h}{R} \right)^{2}} }{\frac{g}{9} = \frac{g}{\left( 1 + \frac{h}{R} \right)^{2}} }{1 + \frac{h}{R} = 3 }{h = 2R }$$

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