Abody of mass 'm' is taken from the earth's surface to the height equal to twice the redius (R) of the earth . The change in potential energy of the body will be:- (a) 1/3mgR

(b) mg2R (C) 2/3 mgR (D) 3mgR PLEASE EXPLAIN ME .........SIR....

Option C is the correct answer. answer is $u=\frac{2}{3}mgR$
mass = m, height = 2R, potential energy (u) = ?
At height the value of acceleration due to gravity decreases. so the value of  
acceleration due to gravity g' = $\frac{g}{1+h/R}=\frac{gR}{h+R}$
The value of potential energy (u)  = mgh = $\frac{mgRh}{R+h}=\frac{mgR2R}{3R}=\frac{2}{3}mgR$ here h =2R