At the surface of earth, the acceleration due to gravity is g. If an object of mass m is raised from the surface of earth to a height equal to radius of earth (R ). The potential energy gained by the object is

At the surface of earth, the acceleration due to gravity is g. If an o

1.	At the surface of earth, the acceleration due to gravity is g. If an object of mass m is raised from the surface of earth to a height equal to radius of earth (R ). The potential energy gained by the object is
Answer» mgR `(mgR^(2))/(2)` `(+mgR)/(2)` `(-mgR)/(2)` Solution :On surface of EARTH, the GRAVITATIONAL potential energy `U_(2) = -(GMm)/(R )` At a height R from surface of earth, the gravitational potential energy `U_(2) = -(GMm)/((R+R)) = - (GMm)/(2R)` `Delta U = U_(2) - U_(1) = -(GMm)/(2R) - (-(GMm)/(R ))` `= (GMm)/(2R)` `= (gR^(2)m)/(2R) ""(because G = (GM)/(R^(2)),(gR^(2) = GM))` Therefore, `U_(2) - U_(1) = (MgR)/(2)`