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	<title>Newtons Laws EX 41 - Revision history</title>
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	<updated>2026-04-20T21:11:54Z</updated>
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		<title>imported&gt;Patrick: Created page with &#039;The Moon moves around the Earth in 27.32 days. Therefore the period of the uniform circular motion of the Moon is &lt;math&gt;T = 27.32 days = 2.36 \times 10^6 s&lt;/math&gt;. The diagram be…&#039;</title>
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		<updated>2011-05-27T05:52:12Z</updated>

		<summary type="html">&lt;p&gt;Created page with &amp;#039;The Moon moves around the Earth in 27.32 days. Therefore the period of the uniform circular motion of the Moon is &amp;lt;math&amp;gt;T = 27.32 days = 2.36 \times 10^6 s&amp;lt;/math&amp;gt;. The diagram be…&amp;#039;&lt;/p&gt;
&lt;p&gt;&lt;b&gt;New page&lt;/b&gt;&lt;/p&gt;&lt;div&gt;The Moon moves around the Earth in 27.32 days. Therefore the period of the uniform circular motion of the Moon is &amp;lt;math&amp;gt;T = 27.32 days = 2.36 \times 10^6 s&amp;lt;/math&amp;gt;. The diagram below illustrates the Moon&amp;#039;s orbit around the Earth.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
[[image:N_APP_EX_41_SOLN.png|TOP]]&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The Moon is in uniform circular motion and therefore it has radial acceleration. To determine its acceleration we first need to determine its speed&lt;br /&gt;
&lt;br /&gt;
 &amp;lt;math&amp;gt;v = {2 \pi r \over T}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
and substitute this value into the expression for radial acceleration:&lt;br /&gt;
&lt;br /&gt;
 &amp;lt;math&amp;gt;a_r = {v^2 \over r} = {({2 \pi r \over T})^2 \over r} = {4 \pi^2 r \over T^2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;br&amp;gt;&lt;br /&gt;
The gravitational attraction by Earth is the force that causes this acceleration and thus, we can write Newton&amp;#039;s 2nd Law&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{G M_E m_M \over r^2} = m_M a_r = m_M {4 \pi^2 r \over T^2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
By solving for &amp;lt;math&amp;gt;M_E&amp;lt;/math&amp;gt; and then by substituting the values for the gravitational constant, the radius of the Moon&amp;#039;s orbit and the Moon&amp;#039;s period into this equation, we find the mass of Earth to be&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;M_E = {4 \pi^2 r^3 \over G T^2} = {4 \pi^2 (3.8 \times 10^8)^3 \over (6.67 \times 10^{-11})(2.36 \times 10^6)^2} = 6.01 \times 10^{24} kg&amp;lt;/math&amp;gt;&lt;/div&gt;</summary>
		<author><name>imported&gt;Patrick</name></author>
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