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        <h1>
            Matter and Gravity
        </h1>

        
        <p class="post-meta"><time itemprop="datePublished">November 5, 2023</time>
        </p>

        <h3 id="the-basics-of-calculating-density-of-matter">The Basics of Calculating Density of Matter</h3>
<p>The average density of the Solar System is about 1 billion times smaller then the Earth&rsquo;s atmosphere.</p>
<p>The above statement puts into perspective the unique nature of space and the universe: It&rsquo;s mostly empty, a vacuum. The Solar System, which includes all the planets, asteroids, and the Sun, which has a mass of roughly 2.0 x $10^{30}$ kilograms, is a very expansive place that stretches far into space until it reaches the Interstellar Medium. Yet, since the Solar System is so large, the density is still quite less than even a gas. This is due to what density really is and how it&rsquo;s calculated. Density is important in the field of astrophysics and can be calculated using the following equation:</p>
<p>$$\frac{M}{V}$$</p>
<p>Where:</p>
<ul>
<li>M = Total mass</li>
<li>V = Volume</li>
</ul>
<p>The average total mass of the Solar System is about 1.9884 x $10^{30}$ kilograms. This is mostly due to the mass of the Sun, since the additional mass of the planets can be ignored for the sake of this example. In addition, the total volume of the Solar System is about 3.9 x $10^{38}$ cubic meters&hellip;that&rsquo;s a lot.</p>
<p>Thus, the average density of the Solar System is $\frac{1.9884\times10^{30}}{3.9\times10^{38}}$ = 5.09 x $10^{-9}$ kilograms per cubic meter.</p>
<p>As you can see, the density is quite low, five billionths of a kilogram in fact.</p>
<blockquote>
<h4 id="note">Note:</h4>
<p>It&rsquo;s important to realize that a material&rsquo;s density changes with temperature. This is intuitive if you think of a liquid&rsquo;s density versus that of a gas.</p>
</blockquote>
<h3 id="calculating-the-gravitational-force">Calculating the Gravitational Force</h3>
<p>The force due to gravity can be calculated in a simple way. The formula is as follows:</p>
<p>$$\mathcal{F} = \frac{G m M}{r^2}$$</p>
<p>Where:</p>
<ul>
<li>G = The Gravitational constant with a value of 6.67 x $10^{-11}$ $\frac{m^3}{Kg \times sec^2}$</li>
<li>m = mass of smaller body in question</li>
<li>M = mass of larger body in question</li>
<li>r = distance between the <em>centers</em> of the bodies in question</li>
</ul>
<p>As you can see from this equation, the force due to gravity is inversely proportional to the square of the distance separating the bodies. As a result, the farther away two bodies are, say a planet and a star, the weaker the gravitational force experienced between them. It&rsquo;s also very important to note that both bodies in a system experience the <em>same</em> force. Gravity is mutual. The reason that the Earth rotates around the Sun instead of the other way around is due to the fact that the Sun has a mass that is about 1 million times larger than the Earth&rsquo;s. Thus, it requires a huge magnitude of force to get it to move in some visible way because of Newton&rsquo;s second law: Force = Mass X Acceleration.</p>
<h3 id="conclusion">Conclusion</h3>
<p>In this brief post, we explored density and gravity, tools that will be used in more advanced calculations later down the road. In upcoming posts, I&rsquo;ll describe to you how to derive gravitational potential energy and how to calculate the mass of the Sun.</p>
<p>For all that and more, I&rsquo;ll see you in the next post. Thanks for reading!</p>


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commit13: added matter and gravity post 2023-11-05 23:03:56 +00:00			`<!DOCTYPE html>`
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			`inlineMath: [['$', '$'], ['\\(', '\\)']],`
			`displayMath: [['$$','$$'], ['\\[', '\\]']],`
			`processEscapes: true,`
			`processEnvironments: true`
			`},`
			`options: {`
			`skipHtmlTags: ['script', 'noscript', 'style', 'textarea', 'pre']`
			`}`
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			`window.addEventListener('load', (event) => {`
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			`<script type="text/javascript" id="MathJax-script" async`
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			`<html>`
			`<!DOCTYPE html>`
			`<html>`

			`<body>`
			`<header class="site-header">`
			`<div class="wrapper">`
			`<a class="muted small" href="https://nathan.freedomland.xyz">Blogarithm</a>`
			`</div>`
			`</header>`
			`</body>`

			`</html>`


			`<body>`
			`<h1>`
			`Matter and Gravity`
			`</h1>`



			`<p class="post-meta"><time itemprop="datePublished">November 5, 2023</time>`
			`</p>`

			`<h3 id="the-basics-of-calculating-density-of-matter">The Basics of Calculating Density of Matter</h3>`
			`<p>The average density of the Solar System is about 1 billion times smaller then the Earth’s atmosphere.</p>`
commit15: fixed typos 2023-11-07 15:17:16 +00:00			<p>The above statement puts into perspective the unique nature of space and the universe: It’s mostly empty, a vacuum. The Solar System, which includes all the planets, asteroids, and the Sun, which has a mass of roughly 2.0 x $10^{30}$ kilograms, is a very expansive place that stretches far into space until it reaches the Interstellar Medium. Yet, since the Solar System is so large, the density is still quite less than even a gas. This is due to what density really is and how it’s calculated. Density is important in the field of astrophysics and can be calculated using the following equation:</p>
			`<p>$$\frac{M}{V}$$</p>`
commit13: added matter and gravity post 2023-11-05 23:03:56 +00:00			`<p>Where:</p>`
			`<ul>`
			`<li>M = Total mass</li>`
			`<li>V = Volume</li>`
			`</ul>`
commit15: fixed typos 2023-11-07 15:17:16 +00:00			`<p>The average total mass of the Solar System is about 1.9884 x $10^{30}$ kilograms. This is mostly due to the mass of the Sun, since the additional mass of the planets can be ignored for the sake of this example. In addition, the total volume of the Solar System is about 3.9 x $10^{38}$ cubic meters…that’s a lot.</p>`
			`<p>Thus, the average density of the Solar System is $\frac{1.9884\times10^{30}}{3.9\times10^{38}}$ = 5.09 x $10^{-9}$ kilograms per cubic meter.</p>`
			`<p>As you can see, the density is quite low, five billionths of a kilogram in fact.</p>`
commit13: added matter and gravity post 2023-11-05 23:03:56 +00:00			`<blockquote>`
			`<h4 id="note">Note:</h4>`
			`<p>It’s important to realize that a material’s density changes with temperature. This is intuitive if you think of a liquid’s density versus that of a gas.</p>`
			`</blockquote>`
			`<h3 id="calculating-the-gravitational-force">Calculating the Gravitational Force</h3>`
			`<p>The force due to gravity can be calculated in a simple way. The formula is as follows:</p>`
commit15: fixed typos 2023-11-07 15:17:16 +00:00			`<p>$$\mathcal{F} = \frac{G m M}{r^2}$$</p>`
commit13: added matter and gravity post 2023-11-05 23:03:56 +00:00			`<p>Where:</p>`
			`<ul>`
commit15: fixed typos 2023-11-07 15:17:16 +00:00			`<li>G = The Gravitational constant with a value of 6.67 x $10^{-11}$ $\frac{m^3}{Kg \times sec^2}$</li>`
commit13: added matter and gravity post 2023-11-05 23:03:56 +00:00			`<li>m = mass of smaller body in question</li>`
			`<li>M = mass of larger body in question</li>`
			`<li>r = distance between the <em>centers</em> of the bodies in question</li>`
			`</ul>`
commit15: fixed typos 2023-11-07 15:17:16 +00:00			<p>As you can see from this equation, the force due to gravity is inversely proportional to the square of the distance separating the bodies. As a result, the farther away two bodies are, say a planet and a star, the weaker the gravitational force experienced between them. It’s also very important to note that both bodies in a system experience the <em>same</em> force. Gravity is mutual. The reason that the Earth rotates around the Sun instead of the other way around is due to the fact that the Sun has a mass that is about 1 million times larger than the Earth’s. Thus, it requires a huge magnitude of force to get it to move in some visible way because of Newton’s second law: Force = Mass X Acceleration.</p>
commit13: added matter and gravity post 2023-11-05 23:03:56 +00:00			`<h3 id="conclusion">Conclusion</h3>`
			`<p>In this brief post, we explored density and gravity, tools that will be used in more advanced calculations later down the road. In upcoming posts, I’ll describe to you how to derive gravitational potential energy and how to calculate the mass of the Sun.</p>`
			`<p>For all that and more, I’ll see you in the next post. Thanks for reading!</p>`




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			`<h3><a href="https://nathan.freedomland.xyzposts">Back to all posts</a></h3>`
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