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Question

Which what speed mail pass through the center of Earth if falling in a tunnel through the center?

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Solution

We apply the work-energy theorem to the object in question.
It starts from a point at the surface of the Earth with zero initial speed
and arrives at the center of the Earth with final speed vf.
The corresponding increase in its kinetic energy,
1/2mv2f, is equal to the work done on it by Earth's gravity:
Fdr=(Kr)dr.
Thus,

12mv2f=0RFdr=0R(Kr)dr=12KR2

where R is the radius of Earth. Solving for the final speed, we obtain vf=RK/m. We note that the acceleration of gravity ag=g=9.8m/s2 on the surface of Earth is given by

ag=GM/R2=G(4πR3/3)ρ/R2

where ρ is Earth's average density. This permits us to write K/m=4πGρ/3=g/R Consequently,

vf=RKm=RgR=gR=(9.8m/s2)(6.37×106m)=7.9×103m/s


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