1.

If g is the acceleration due to gravity on the surface of the earth, the gain in potential energy of an object of mass m raised from the earth's surface to a height equal to the radius R of the earth is

Answer»

`(MGR)/(4)`
`(mgR)/(2)`
mgR
2mgR

Solution :Gravitational potential ENERGY of an object of mass ms at any point at a distance r from the centre of the earth is
`U=-(GMm)/(r )`
Where M is the mass of the earth.
At the surface of the earth, r=R
`:. U_(1)=-(GMm)/(R )=-mgR(.: g=(GM)/(R^(2)))""`..(i)

At a height R from the surface of the earth, r=2R
`:. U_(2)=-(GMm)/(2R )=-(mgR)/(2)""` ....(ii)
Gain in potential energy, `DeltaU=U_(2)-U_(1)`
`DeltaU=-(mgR)/(2)-(-mgR)`(Using (i)and (ii))
`=(mgR )/(2)`.


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