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 ?

Question

Leo Abeyta · Accepted Answer

Allow me to #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, when the body is on the surface, the PE of the system #E_s=-(GmM)/R....(2)# Once more, the system's PE at height h from the earth's surface is provided by #E_h=-(GmM)/(R+h)# When h equals R, then #E_R=-(GmM)/(R+R)=-(GmM)/(2R)
....(3)# Therefore, the increase in PE caused by the body of mass m shifting from the surface to a height equal to the earth's radius (R) is as follows: #DeltaE_p=E_R-E_s=-(GmM)/(2R)+(GmM)/R=(GmM)/(2R)#
#=(mgR^2)/(2R)=1/2mgR# # "Inserting "GM=gR^2" from "(1)#

Amelia Adams · Answer

undefined The increase in potential energy (P.E) of a body of mass 'm' when raised to a distance equal to the radius of the Earth from the Earth's surface is equal to mgR, where 'g' is the acceleration due to gravity on Earth and 'R' is the radius of the Earth.