Question

 If the acceleration due to gravity at the surface of the earth is g ,the work done in slowly lifting a body of mass m from the earths surface to a height R equal to the radius of the earth is.​

Answer 1

=work done=force*displacement

=gravitational acceleration at height equal to the radius of the earth

=g(r)^2/(r+h)^2

=g(r)^2/(2r)^2

g(r)^2/4r^2

g/4

therfore gravitational acceleration there=

g/4

therefore =gravitational accleration will be equal to g/4 therefore potential energy stored at point=work done

mgr/4=potential energy

which equals work done

therefore work done =mgr/4

therfore option none of these is correct

option mgr cannot be correct because gravitational acceleration at a height equal to the radius of the earth is g/4 and since P.E=Work done=mgr/4

Answer 2

=work done=force*displacement

=gravitational acceleration at height equal to the radius of the earth

=g(r)^2/(r+h)^2

=g(r)^2/(2r)^2

g(r)^2/4r^2

g/4

therfore gravitational acceleration there=

g/4

therefore =gravitational accleration will be equal to g/4 therefore potential energy stored at point=work done

mgr/4=potential energy

which equals work done

therefore work done =mgr/4