Determine the flight time of a stone from one pole of the Earth to the other along a straight tunnel dug through the center. Consider Earth's density constant, its radius equal to 6400 km.

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    <h3> Statement </h3>
    <p>
        $3.2.12.$ Determine the flight time of a stone from one pole of the Earth to the other along a straight tunnel dug through the center. Consider Earth's density constant, its radius equal to $6400$ km.

    <h3>Solution</h3>
    <p>

The body, at a distance $x$ from the core, will be subject to the gravitational force of attraction caused by the inner layers of the planet of density $\rho$, forming a sphere of radius $x$. The mass of this part of the earth $$M_\oplus = \frac{4}{3} \rho\pi x^3$$ Gravitational force acting on a rock at depth $x$ $$F_G = \frac{GmM_\oplus}{x^2}=mg\frac{x}{R}$$ Newton's Second Law $$m\ddot{x}(t)=-\frac{mg}{R}x(t)$$ Let's write the equation of harmonic oscillations $$\ddot{x}(t)+\frac{g}{R}x(t)=0$$ The angular frequency of such oscillations $$\omega=\sqrt{\frac{g}{R}}\Rightarrow T=2\pi\sqrt{\frac{R}{g}}$$ Since we are interested in the flight time only in one direction, we take half of this period. $$\boxed{t=\frac{T}{2}=\pi\sqrt{\frac{R}{g}}=42\text{ min}}$$

Dzikan Mikita
Aliaksandr Kanashenka

    <h4>Answer</h4>
    <p>
        $$t=42\text{ min}$$
    </p>


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