Question

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 ?

Answer 1

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)`