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Assume the earth were a solid body and a tunnel was dug to the center of the Earth. If...

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grade82 | Student | eNoter

Posted June 22, 2013 at 3:49 AM via web

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Assume the earth were a solid body and a tunnel was dug to the center of the Earth. If a person were to go to the center how much work would be required to come back to the surface. (Radius of Earth is 6000 km and mass of person is 100 kg)

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted June 22, 2013 at 5:14 AM (Answer #1)

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The force with which two bodies with mass M1 and M2 are attracted towards each other due to the gravitational force of attraction between them is given by `F = (G*M1*M2)/r^2` where r is the distance between their center of gravity.

The acceleration of an object towards the center of the Earth at the surface is 9.8 m/s^2. This is equal to `(G*M_e)/(R_e)^2` . Assume the Earth to be a body of uniform density. As the object moves towards the center, the effective mass of the Earth decreases according to the relation `M = M_e/((4/3)*pi*(R_e)^3)*(4/3)*pi*r^3`= `(M_e*r^3)/(R_e)^3` where r is the distance from the center of the Earth.

The force of attraction at a distance r from the center is `F= (G*(M_e*r^3)/(R_e)^3*m)/r^2` `= (G*M_e*m*r)/(R_e)^3`

The work done to move an object from the center of the Earth to the surface is: `W = int_0^(R_e) (G*M_e*m*r)/(R_e)^3 dr`

`= [(G*M_e*m*r^2)/(2*(R_e)^3)]_0^(R_e)`

= `(G*M_e*m*(R_e)^2)/(2*(R_e)^3)`

= `(G*M_e*m)/(2*R_e)`

But g = 9.8 = `(G*M_e)/(R_e)^2`

=> `W = (9.8*100*6000)/2`

= 2940000 J

The work required to bring a person with mass 100 kg from the center of the Earth is 2940000 J

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