# How much work (in Joules) is done on a 1 kg object to lift it from the center of the Earth to its surface? The radius of the Earth is 6371 km.

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### 1 Answer

The gravitational force of attraction acting on an object of mass m is given by `F = (m*g*r)/R_e` where r is the distance of the object from the center of the Earth and `R_e` is the radius of the Earth.

The work required to bring an object with mass 1 kg from the center of the Earth to the surface is given by :

`int_0^(R_e) m*g*r/R_e dr`

= `(m*g/R_e)*(r^2/2)_0^(R_e)`

= `(m*g/R_e)*(1/2)*(R_e^2 - 0)`

= `(m*g*R_e)/2`

The value of g is 9.8 m/s^2 and `R_e` = 6371 km. Using these in the formula derived for the work done gives (1*9.8*6371*10^3)/2 = 31.217*10^6 J

**The work done to bring an object with mass 1 kg from the center of the Earth to the surface is approximately 31.217*10^6 J**

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