If 'g' is the acceleration due to gravity on earth then increase in P.E of a body of mass 'm' up to a distance equal to the radius of earth from the earth surface will be ?

1 Answer
Sep 20, 2016

Let

#R->"Radius of the earth"#

#M->"Mass of the earth"#

#m->"Mass of the body"#

#g->"Acceleration due to gravity on earth"#
When the body is on earth surface , considering the force of attraction of earth on it we can write

#mg=(GMm)/R^2" where G = Gravitational constant"#

#=>GM=gR^2.....(1)#

Now the PE of the system when the body is on the surface

#E_s=-(GmM)/R....(2)#

Again the PE of the system when the body is at height h from the the earth surface is given by

#E_h=-(GmM)/(R+h)#

If h =R then

#E_R=-(GmM)/(R+R)=-(GmM)/(2R) ....(3)#

So increase in PE due to shift of the body of mass m from surface to the height equal to the radius (R) of the earth is given by-

#DeltaE_p=E_R-E_s=-(GmM)/(2R)+(GmM)/R=(GmM)/(2R)#
#=(mgR^2)/(2R)=1/2mgR#

# "Inserting "GM=gR^2" from "(1)#