<?xml version="1.0" encoding="UTF-8"?><!-- generator="wordpress/2.0.5" -->
<rss version="2.0" 
	xmlns:content="http://purl.org/rss/1.0/modules/content/">
<channel>
	<title>Comments on: Physics derivation of Kepler&#8217;s rule from Newton&#8217;s laws</title>
	<link>http://globalpioneering.com/wp02/physics-derivation-of-keplers-rule-from-newtons-laws/</link>
	<description>Transfer scientific authority to people</description>
	<pubDate>Fri, 25 Jul 2008 19:16:46 +0000</pubDate>
	<generator>http://wordpress.org/?v=2.0.5</generator>

	<item>
		<title>by: Pioneer1</title>
		<link>http://globalpioneering.com/wp02/physics-derivation-of-keplers-rule-from-newtons-laws/#comment-20213</link>
		<pubDate>Fri, 02 May 2008 13:07:39 +0000</pubDate>
		<guid>http://globalpioneering.com/wp02/physics-derivation-of-keplers-rule-from-newtons-laws/#comment-20213</guid>
					<description>Hi Doug,

Thanks for your comment. It will probably take me months to understand Terry Tao's derivation. I remember that I marked it as interesting when I first saw it but I could never find the time to study it carefully. Also, I consider E=mc2 to be a definition and I'm not sure if one should or could derive definitions. But I may be wrong. Since you mention a geometric interpretation for kinetic and potential energies I thought I comment on that. My understanding is that physicists call Kepler's rule "conservation of energy" and they write it as (for unit mass)

&lt;blockquote&gt;&lt;a href="http://www.codecogs.com" rel="nofollow"&gt;&lt;img src="http://www.codecogs.com/eq.latex?\textrm{Kinetic&space;energy}=\frac{R^2}{T^2}=\frac{1}{R}=\textrm{Potential&space;energy}" alt="\textrm{Kinetic energy}=\frac{R^2}{T^2}=\frac{1}{R}=\textrm{Potential energy}" border="0"/&gt;&lt;/a&gt;&lt;/blockquote&gt;

So they labeled R2/T2 = kinetic energy and 1/R= potential energy (for unit mass). Since this is Kepler's rule, it has a simple geometric interpretation. Given two orbits R_0 and R I have this geometry:

&lt;blockquote&gt;&lt;img src="http://www.alphysics.com/Images/conservation-of-energy2.JPG" alt="Conservation of energy" /&gt;&lt;/blockquote&gt;

AE = R_0
BE = w_0
AD = R
DC = w

Then by Kepler's rule

&lt;blockquote&gt;&lt;a href="http://www.codecogs.com" rel="nofollow"&gt;&lt;img src="http://www.codecogs.com/eq.latex?\frac{R_0^2\,&space;\omega_0^2}{R^2\,&space;\omega^2}=\frac{R}{R_0}" alt="\frac{R_0^2\, \omega_0^2}{R^2\, \omega^2}=\frac{R}{R_0}" border="0"/&gt;&lt;/a&gt;&lt;/blockquote&gt;

R^2 and R_0^2 are the areas swept by the radii. Please let me know what you think.</description>
		<content:encoded><![CDATA[<p>Hi Doug,</p>
<p>Thanks for your comment. It will probably take me months to understand Terry Tao&#8217;s derivation. I remember that I marked it as interesting when I first saw it but I could never find the time to study it carefully. Also, I consider E=mc2 to be a definition and I&#8217;m not sure if one should or could derive definitions. But I may be wrong. Since you mention a geometric interpretation for kinetic and potential energies I thought I comment on that. My understanding is that physicists call Kepler&#8217;s rule &#8220;conservation of energy&#8221; and they write it as (for unit mass)</p>
<blockquote><p><a href="http://www.codecogs.com" rel="nofollow"><img src="http://www.codecogs.com/eq.latex?\textrm{Kinetic&space;energy}=\frac{R^2}{T^2}=\frac{1}{R}=\textrm{Potential&space;energy}" alt="\textrm{Kinetic energy}=\frac{R^2}{T^2}=\frac{1}{R}=\textrm{Potential energy}" border="0"/></a></p></blockquote>
<p>So they labeled R2/T2 = kinetic energy and 1/R= potential energy (for unit mass). Since this is Kepler&#8217;s rule, it has a simple geometric interpretation. Given two orbits R_0 and R I have this geometry:</p>
<blockquote><p><img src="http://www.alphysics.com/Images/conservation-of-energy2.JPG" alt="Conservation of energy" /></p></blockquote>
<p>AE = R_0<br />
BE = w_0<br />
AD = R<br />
DC = w</p>
<p>Then by Kepler&#8217;s rule</p>
<blockquote><p><a href="http://www.codecogs.com" rel="nofollow"><img src="http://www.codecogs.com/eq.latex?\frac{R_0^2\,&space;\omega_0^2}{R^2\,&space;\omega^2}=\frac{R}{R_0}" alt="\frac{R_0^2\, \omega_0^2}{R^2\, \omega^2}=\frac{R}{R_0}" border="0"/></a></p></blockquote>
<p>R^2 and R_0^2 are the areas swept by the radii. Please let me know what you think.
</p>
]]></content:encoded>
				</item>
	<item>
		<title>by: Doug</title>
		<link>http://globalpioneering.com/wp02/physics-derivation-of-keplers-rule-from-newtons-laws/#comment-20150</link>
		<pubDate>Thu, 01 May 2008 23:13:18 +0000</pubDate>
		<guid>http://globalpioneering.com/wp02/physics-derivation-of-keplers-rule-from-newtons-laws/#comment-20150</guid>
					<description>Hi Pioneer1,

I would like to comment on a modification of first figure [circle] in item #2 Acceleration.

Perhaps you could draw the figure from this description.
Let c be the radius of the crcle.
From orthogonal X and Y axes, with the origin at the center of the circle, draw the chord.
Do this only for the usual Quadrant I [ , ], although this could be done for the other three quadrants.
Reflect the right triangle so that a square is formed.

Compare this to the &lt;a href="http://en.wikipedia.org/wiki/Gini_coefficient" rel="nofollow"&gt; Gini coefficient&lt;/a&gt; diagram.
GOTO the Terence Tao &lt;a href="http://terrytao.wordpress.com/2007/12/28/einsteins-derivation-of-emc2/" rel="nofollow"&gt; Einstein’s derivation of E=mc^2&lt;/a&gt; discussion.

From my perspective, this may bring economics into relativity:
From the wiki page:
a - the shaded segment labeled gini index may be O(v^3) and
b - the remaining portion of the lower right triangle may be E’ or perhaps bonding energy [?];
From the constructed diagram:
c - may correspond to the upper triangle in Quadrant I,
d - the lower triangle in Quadrant I may correspond to the kinetic energy,
e - the entire square may correspond to potential energy.

Insight - maybe?
Rigor - no.
Geometry in GR - perhaps? - when letting m=1, v=c, a=c^2.

The chord may be the relative velocity of orthogonal particle movement or c*sqrt(2).
The diameter is 2c which may be the relative velocity of two particles traveling 180 degrees from each other, although the vector sum of energy is zero.</description>
		<content:encoded><![CDATA[<p>Hi Pioneer1,</p>
<p>I would like to comment on a modification of first figure [circle] in item #2 Acceleration.</p>
<p>Perhaps you could draw the figure from this description.<br />
Let c be the radius of the crcle.<br />
From orthogonal X and Y axes, with the origin at the center of the circle, draw the chord.<br />
Do this only for the usual Quadrant I [ , ], although this could be done for the other three quadrants.<br />
Reflect the right triangle so that a square is formed.</p>
<p>Compare this to the <a href="http://en.wikipedia.org/wiki/Gini_coefficient" rel="nofollow"> Gini coefficient</a> diagram.<br />
GOTO the Terence Tao <a href="http://terrytao.wordpress.com/2007/12/28/einsteins-derivation-of-emc2/" rel="nofollow"> Einstein’s derivation of E=mc^2</a> discussion.</p>
<p>From my perspective, this may bring economics into relativity:<br />
From the wiki page:<br />
a - the shaded segment labeled gini index may be O(v^3) and<br />
b - the remaining portion of the lower right triangle may be E’ or perhaps bonding energy [?];<br />
From the constructed diagram:<br />
c - may correspond to the upper triangle in Quadrant I,<br />
d - the lower triangle in Quadrant I may correspond to the kinetic energy,<br />
e - the entire square may correspond to potential energy.</p>
<p>Insight - maybe?<br />
Rigor - no.<br />
Geometry in GR - perhaps? - when letting m=1, v=c, a=c^2.</p>
<p>The chord may be the relative velocity of orthogonal particle movement or c*sqrt(2).<br />
The diameter is 2c which may be the relative velocity of two particles traveling 180 degrees from each other, although the vector sum of energy is zero.
</p>
]]></content:encoded>
				</item>
</channel>
</rss>
