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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

Answers (1)
S safeer

@Santhosh

The accelerator due to gravity at a height h from the earth given as \frac{g}{9}

\frac{GM}{r^{2}}=\frac{GM}{R^{2}} \frac{1}{9}

r=3R

The height above the ground is 2R.

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