Q: If g is acceleration due to gravity on the surface of the earth, having radius R, the height at which the acceleration due to gravity reduces to g/2 is

(a) R/2

(b) √2 R

(c) R/√2

(d) (√2-1)R

Ans:(d)

Sol:

$ \displaystyle g= \frac{GM}{R^2}$

$ \displaystyle g’ = \frac{GM}{(R+h)^2} $

Since g’ = g/2 ;