# How much work (in Joules) is done on a 1kg object to lift it from the center of the Earth to its surface? Note: The gravity force in Newtons on a 1 kg object at distance `r` from the center of the Earth is given by `F(r)=0.0015r` also note that radius of Earth is `R=6,371km`

## Expert Answers Because the gravitational force

`F(r) =0.0015*r`

acts toward the center of the Earth, it is parallel to the lifting distance of the object. Thus the infinitesimal work done to lift the object from `r` to `r+dr` is just

`dW = F(r)*dr`

To obtain the total work necessary to lift the object from the center of the Earth to its surface we need to integrate the above expression

`int_0^W dW =int_0^RF(r)dr`

where `R=6371 km` is the Earth radius (given in km).

`W =int_0^6371 0.0015*r*dr =0.0015*r^2/2 (0->6371) =30442.23 J`

The work necessary to lift a mass of 1 kg from the center of the Earth to its surface is 30442 J

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