# The height at which the acceleration due to gravity becomes g/9 in terms of the radius of the earth (R) is

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.