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<title>A straight tunnel is dug in the Earth that does not pass through its center. Determine the time of movement of a train with engines turned off through such a tunnel if the influence of the Earth's rotation on the movement of the train and friction are neglected.</title>
$3.2.13.$ A straight tunnel is dug in the Earth that does not pass through its center. Determine the time of movement of a train with engines turned off through such a tunnel if the influence of the Earth's rotation on the movement of the train and friction are neglected.
Similarly to <a href="../3.2.12">3.2.12</a>, we find the external force $F$ as</p><p>
The body, at a distance $x$ from the core, will be affected by the gravitational force of attraction caused by the inner layers of the planet with density $\rho$, forming a sphere of radius $x$. The mass of this part of the earth
$$ M_\oplus = \frac{4}{3}\rho\pi x^3 $$
@@ -86,9 +86,9 @@
Gravitational force acting on a rock at depth $x$
$$ F_G = \frac{GmM_\oplus}{x^2}=mg\frac{x}{R}$$
In this case, only the horizontal component of this force will create a moment. Whence, the resulting external force is equal to
−
$$ F=mg\cos\alpha=mg\frac{x}{R}$$
+
$$ F=mg\cos\alpha=mg\frac{x}{R}$$
From here we find the angular frequency of oscillations
−
$$\omega=\sqrt{\frac{g}{R}}$$
+
$$\omega=\sqrt{\frac{g}{R}}$$
The period of oscillation in a given system
$$ T=2\pi\sqrt{\frac{R}{g}}$$
Since we are interested in the flight time only in one direction, we take half of this period.
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<meta name="description" content="A straight tunnel is dug in the Earth that does not pass through its center. Determine the time of movement of a train with engines turned off through such a tunnel if the influence of the Earth's rotation on the movement of the train and friction are neglected.">
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<title>A straight tunnel is dug in the Earth that does not pass through its center. Determine the time of movement of a train with engines turned off through such a tunnel if the influence of the Earth's rotation on the movement of the train and friction are neglected.</title>
<title>A straight tunnel is dug in the Earth that does not pass through its center. Determine the time of movement of a train with engines turned off through such a tunnel if the influence of the Earth's rotation on the movement of the train and friction are neglected.</title>
$3.2.13.$ A straight tunnel is dug in the Earth that does not pass through its center. Determine the time of movement of a train with engines turned off through such a tunnel if the influence of the Earth's rotation on the movement of the train and friction are neglected.
$3.2.13.$ A straight tunnel is dug in the Earth that does not pass through its center. Determine the time of movement of a train with engines turned off through such a tunnel if the influence of the Earth's rotation on the movement of the train and friction are neglected.
Similarly to <a href="../3.2.12">3.2.12</a>, we find the external force $F$ as</p><p>
Similarly to <a href="../3.2.12">3.2.12</a>, we find the external force $F$ as</p><p>
The body, at a distance $x$ from the core, will be affected by the gravitational force of attraction caused by the inner layers of the planet with 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 affected by the gravitational force of attraction caused by the inner layers of the planet with density $\rho$, forming a sphere of radius $x$. The mass of this part of the earth
$$ M_\oplus = \frac{4}{3}\rho\pi x^3 $$
$$ M_\oplus = \frac{4}{3}\rho\pi x^3 $$
@@ -86,9 +86,9 @@
Gravitational force acting on a rock at depth $x$
Gravitational force acting on a rock at depth $x$
$$ F_G = \frac{GmM_\oplus}{x^2}=mg\frac{x}{R}$$
$$ F_G = \frac{GmM_\oplus}{x^2}=mg\frac{x}{R}$$
In this case, only the horizontal component of this force will create a moment. Whence, the resulting external force is equal to
In this case, only the horizontal component of this force will create a moment. Whence, the resulting external force is equal to
$$ F=mg\cos\alpha=mg\frac{x}{R}$$
$$ F=mg\cos\alpha=mg\frac{x}{R}$$
From here we find the angular frequency of oscillations
From here we find the angular frequency of oscillations
$$\omega=\sqrt{\frac{g}{R}}$$
$$\omega=\sqrt{\frac{g}{R}}$$
The period of oscillation in a given system
The period of oscillation in a given system
$$ T=2\pi\sqrt{\frac{R}{g}}$$
$$ T=2\pi\sqrt{\frac{R}{g}}$$
Since we are interested in the flight time only in one direction, we take half of this period.
Since we are interested in the flight time only in one direction, we take half of this period.
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small>
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small>
