<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by tvanflandern</i>
<br /><blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Thomas</i>
<br />But 'a' in Kepler's law is the semimajor axis and not the mean distance. The latter also depends on the eccentricity of the orbit.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">The terms "semi-major axis" and "mean distance" are used interchangeably because they are equivalent when the distance in question is from the focus of the ellipse. The latter does not depend on eccentricity because the average of the minimum and the maximum distances from the focus of the ellipse is the semi-major axis.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote"> But the observations are not being made from the focus of the ellipse (which would be the center of the earth). Although you could consider the observation point a set distance away from the focus, this still introduces ambiguities here in my opinion:
consider for simplicity an observation point which is a distance <i>a</i> (the semimajor axis) away from the focus. Now take first the case of an eccentricity e=1, which would be a line of length 2<i>a</i> with the focus points at both ends. The observation point would then be in the middle of the line and the average distance from this would be <i>a</i>/2. Now consider the eccentricity e=0 i.e. a circle with radius <i>a</i>. The observation point would now be on the circumference of the circle and it is not too difficult to work out that the average distance of all points on the circle is now 4/3*<i>a</i> . For the same semimajor axis, the average travel time of the signal could therefore vary up to a factor 8/3 dependent on the eccentricity of the orbit and I don't see at the moment how you want to distinguish this from an orbit with the same eccentricity but a different semimajor axis (but maybe you can point me to a reference where the method is described in detail).
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Thomas</i><br />Triangulation on the other hand would obviously also depend on the assumptions regarding the presence of any aberration.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by tvanflandern</i><br />How so? The triangle in question would have identical dimensions even if rotated laterally by a small angle. Besides, we don't have to make any "assumptions" about aberration because we can measure changes in the angle of aberration directly for satellites on elliptical orbits, and in that way confirm that aberration is the same as standard theory predicts.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote"> <hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
You've got me confused now: I thought you were saying that there is no aberration for satellites.
On the other hand, if there is an aberration, I can't see a triangulation not being affected by it because the satellite would appear to be in a different place than it actually is (even with a constant speed of light) and in general the effect should be different for the two triangulation points as both distance and the relative velocity vectos are different.
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