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| <span><img src="../../img/book.png"></span><span>Savchenko Solutions</span> | | <span><img src="../../img/book.png"></span><span>Savchenko Solutions</span> |
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| Solutions of Savchenko Problems in Physics <br> | | Solutions of Savchenko Problems in Physics <br> |
| <i><b>knowledge must be free</b></i> | | <i><b>knowledge must be free</b></i> |
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| <h3 id="back-link"><a href="../../#3.2">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../../#3.2">$\leftarrow$Back</a></h3> |
| | | |
| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <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. | | $3.2.12.$ 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. |
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| For problem $3.2.12$ | | For problem $3.2.12$ |
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| <h3>Solution</h3> | | <h3>Solution</h3> |
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| Body at distance $x$ from the planet's core | | Body at distance $x$ from the planet's core |
| </figcaption> | | </figcaption> |
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| 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 | | 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 |
| $$ M_\oplus = \frac{4}{3} \rho\pi x^3 $$ | | $$ M_\oplus = \frac{4}{3} \rho\pi x^3 $$ |
| Gravitational force acting on a rock at depth $x$ | | Gravitational force acting on a rock at depth $x$ |
| $$ F_G = \frac{GmM_\oplus}{x^2}=mg\frac{x}{R} $$ | | $$ F_G = \frac{GmM_\oplus}{x^2}=mg\frac{x}{R} $$ |
| Newton's Second Law | | Newton's Second Law |
| $$ m\ddot{x}(t)=-\frac{mg}{R}x(t) $$ | | $$ m\ddot{x}(t)=-\frac{mg}{R}x(t) $$ |
