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Question

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


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 $$ v_{f} $$. 
The corresponding increase in its kinetic energy,
$$ 1 / 2 m v_{f}^{2}, $$ is equal to the work done on it by Earth's gravity: 
$$ \int F d r=\int(-K r) d r . $$ 
Thus,

$$\dfrac{1}{2} m v_{f}^{2}=\int_{R}^{0} F d r=\int_{R}^{0}(-K r) d r=\dfrac{1}{2} K R^{2}$$

where $$ R $$ is the radius of Earth. Solving for the final speed, we obtain $$ v_{f}=R \sqrt{K / m} $$. We note that the acceleration of gravity $$ a_{g}=g=9.8 \mathrm{m} / \mathrm{s}^{2} $$ on the surface of Earth is given by

$$a_{g}=G M / R^{2}=G\left(4 \pi R^{3} / 3\right) \rho / R^{2}$$

where $$ \rho $$ is Earth's average density. This permits us to write $$ K / m=4 \pi G \rho / 3=g / R $$ Consequently,

$$v_{f}=R \sqrt{\dfrac{K}{m}}=R \sqrt{\dfrac{g}{R}}=\sqrt{g R}=\sqrt{\left(9.8 \mathrm{m} / \mathrm{s}^{2}\right)\left(6.37 \times 10^{6} \mathrm{m}\right)}=7.9 \times 10^{3} \mathrm{m} / \mathrm{s}$$


Physics

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