A body of mass is taken from earth surface to the height h equal to twice the radius of earth (Re). the inerease in potential energy will be: ( g = acceleration due to gravity on the surface of Earth)

Question

A body of mass is taken from earth surface to the height h equal to twice the radius of earth (Re). the inerease in potential energy will be: ( g = acceleration due to gravity on the surface of Earth) (A) (1/2) m g Re (B) 3 m g Re (C) (2/3) m g Re (D) (1/3) m g Re

Tardigrade Admin · Accepted Answer

U =(- GM e m / r ) U i =(- GM e m / R e ) U f =(- GM e m /( R e + h ))=(- GM e m / R e +2 R e ) (- GM e m /3 R e ) Increase in internal energy Δ U = U f - U i =(2/3) ( GM e m / R e ) (2/3) ( GM e / R e 2) mR e =(2/3) mgR e

Q. A body of mass is taken from earth surface to the height $h$ equal to twice the radius of earth $(R_{e})$ . the inerease in potential energy will be: ( $g$ = acceleration due to gravity on the surface of Earth)

$\frac{1}{2} m g R_{e}$

$3 m g R_{e}$

$\frac{2}{3} m g R_{e}$

$\frac{1}{3} m g R_{e}$

Q. A body of mass is taken from earth surface to the height h equal to twice the radius of earth (Re​). the inerease in potential energy will be: ( g = acceleration due to gravity on the surface of Earth)

21​mgRe​

3mgRe​

32​mgRe​

31​mgRe​

U=r−GMe​m​ Ui​=Re​−GMe​m​ Uf​=(Re​+h)−GMe​m​=Re​+2Re​−GMe​m​ 3Re​−GMe​m​ Increase in internal energy ΔU=Uf​−Ui​ =32​Re​GMe​m​ 32​Re2​GMe​​mRe​ =32​mgRe​

Q. A body of mass is taken from earth surface to the height $h$ equal to twice the radius of earth $(R_{e})$ . the inerease in potential energy will be: ( $g$ = acceleration due to gravity on the surface of Earth)

$\frac{1}{2} m g R_{e}$

$3 m g R_{e}$

$\frac{2}{3} m g R_{e}$

$\frac{1}{3} m g R_{e}$