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

Sol: $\large g’ = g(\frac{R}{R+h})^2 $

$\large \frac{g}{9} = g(\frac{R}{R+h})^2 $

$\large \frac{1}{3} = \frac{R}{R+h} $

3R = R+h

⇒ 2R = h.