n1. Ifg is the acceleration due to gravity on thesurface of the earth,

1.	n1. Ifg is the acceleration due to gravity on thesurface of the earth, the gain in potentialenergy of an object of mass in raised fromthe earth's surface to a height equal to theradius Ro the earth is
Answer» Let R→Radius of the earth M→Mass of the earth m→Mass of the body g→Acceleration due to gravity on earthWhen 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 Es = −GmMR....(2) Again the PE of the system when the body is at height h from the the earth surface is given by Eh = −GmM/R+h If h = R then ER = −GmM/R+R = −GmM/2R....(3) Soincrease in PEdue to shift of the body of massmfrom surface to the height equal to the radius (R) of the earth is given by: ΔEp = ER−Es = −GmM/2R + GmM/R = GmM/2R = mgR^2/2R = 1/2mgR Inserting GM = gR^2 from (1)

n1. Ifg is the acceleration due to gravity on thesurface of the earth, the gain in potentialenergy of an object of mass in raised fromthe earth's surface to a height equal to theradius Ro the earth is

Answer»

Let

R→Radius of the earth

M→Mass of the earth

m→Mass of the body

g→Acceleration due to gravity on earthWhen 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

Es = −GmMR....(2)

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

Eh = −GmM/R+h

If h = R then

ER = −GmM/R+R = −GmM/2R....(3)

Soincrease in PEdue to shift of the body of massmfrom surface to the height equal to the radius (R) of the earth is given by:

ΔEp = ER−Es = −GmM/2R + GmM/R = GmM/2R = mgR^2/2R = 1/2mgR

Inserting GM = gR^2 from (1)

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