Translated 3.2.1-3.2.17

astrosander edited Sep 28, 2024, 1:02 AM

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		+	<meta name="description" content="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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		+	<title>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.</title>
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		+	</head>
		+	<body style="">
		+	<header style="text-align:center;">
		+	<div id = "logo">
		+	<span><img src = "../../img/book.png"><span><span>Savchenko Solutions</span>
		+	</div>
		+	<p class="author">
		+	Solutions of Savchenko Problems in Physics <br>
		+	<i><b>knowledge must be free</b></i>
		+	</p>
		+	</header>
		+
		+	<h3 id="back-link"><a href="../../#3.2">$\leftarrow$Back</a></h3>
		+
		+	<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.
		+	</p>
		+	<center>
		+	<figure>
		+	<img src="https://savchenkosolutions.com/3/3.2.12/statement.png"
		+	loading="lazy" width="150" />
		+	<figcaption>
		+	For problem $3.2.12$
		+	</figcaption>
		+	</figure>
		+	</center>
		+	<p>
		+	</p>
		+
		+	<h3>Solution</h3>
		+	<p>
		+
		+	</p>
		+	<center>
		+	<figure>
		+	<img src="https://savchenkosolutions.com/3/3.2.12/3.2.12_1.png"
		+	loading="lazy" width="200" />
		+	<figcaption>
		+	Body at distance $x$ from the planet's core
		+	</figcaption>
		+	</figure>
		+	</center>
		+	<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.
		+	$$ \boxed{t=\frac{T}{2}=\pi\sqrt{\frac{R}{g}}=42\text{ min}} $$
		+	</p>
		+	<p style="text-align: right; font-style: italic; font-size: 14;">
		+	Dzikan Mikita<br>
		+	Aliaksandr Kanashenka<br>
		+	</p>
		+
		+	<h4>Answer</h4>
		+	<p>
		+	$$t=42\text{ min}$$
		+	</p>
		+
		+
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		+	<br>
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		+	<small> © <strong>Savchenko Solutions</strong>, 2023-2024 <br></small>
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	@@ -0,0 +1,114 @@
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			<html lang="en">

			<head>
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			<meta name="viewport" content="width=device-width, initial-scale=1.0">
			<meta http-equiv="content-language" content="en">
			<meta name="keywords" content="Savchenko Problems in Physics, Savchenko solutions, physics problems, physics olympiad preparation, IPhO, Jaan Kalda">
			<meta name="description" content="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.">
			<meta name="author" content="Aliaksandr Melnichenka">
			<meta name="date" content="2023-10" scheme="YYYY-MM">
			<meta property="og:title" content="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.">
			<meta property="og:image" content="img/logo.png">
			<meta property="og:description" content="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.">
			<meta name="yandex-verification" content="6cfda41f74038368">
			<title>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.</title>
			<link rel="stylesheet" href="https://savchenkosolutions.com/css/css-latex/style.css">
			<link rel="icon" href="https://savchenkosolutions.com/img/logo.png" type="image/png">
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			<script type="text/x-mathjax-config">
			MathJax.Hub.Config({
			extensions: ['tex2jax.js'],
			jax: ['input/TeX', 'output/HTML-CSS'],
			tex2jax: {
			inlineMath: [['$', '$'], ['$', '$']],
			processEscapes: true,
			processClass: 'tex2jax',
			ignoreClass: 'html'
			},
			showProcessingMessages: false,
			messageStyle: 'none'
			});
			</script>
			</head>
			<body style="">
			<header style="text-align:center;">
			<div id = "logo">
			<span><img src = "../../img/book.png"><span><span>Savchenko Solutions</span>
			</div>
			<p class="author">
			Solutions of Savchenko Problems in Physics <br>
			<i><b>knowledge must be free</b></i>
			</p>
			</header>

			<h3 id="back-link"><a href="../../#3.2">$\leftarrow$Back</a></h3>

			<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.
			</p>
			<center>
			<figure>
			<img src="https://savchenkosolutions.com/3/3.2.12/statement.png"
			loading="lazy" width="150" />
			<figcaption>
			For problem $3.2.12$
			</figcaption>
			</figure>
			</center>
			<p>
			</p>

			<h3>Solution</h3>
			<p>

			</p>
			<center>
			<figure>
			<img src="https://savchenkosolutions.com/3/3.2.12/3.2.12_1.png"
			loading="lazy" width="200" />
			<figcaption>
			Body at distance $x$ from the planet's core
			</figcaption>
			</figure>
			</center>
			<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.
			$$ \boxed{t=\frac{T}{2}=\pi\sqrt{\frac{R}{g}}=42\text{ min}} $$
			</p>
			<p style="text-align: right; font-style: italic; font-size: 14;">
			Dzikan Mikita<br>
			Aliaksandr Kanashenka<br>
			</p>

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


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