# 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.

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...

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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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