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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