Aberration of Starlight

<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 /><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 />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.<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 &lt;i&gt;a&lt;/i&gt; (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&lt;i&gt;a&lt;/i&gt; 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 &lt;i&gt;a&lt;/i&gt;/2. Now consider the eccentricity e=0 i.e. a circle with radius &lt;i&gt;a&lt;/i&gt;. 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 1.3333...*&lt;i&gt;a&lt;/i&gt; . For the same semimajor axis, the average travel time of the signal could therefore vary up to a factor 2.6666... 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).<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">All the GPS satellite orbits are good approximations of a circle (low eccentricity). Orbit analysis can take the eccentricity into account too. But let’s stick with the concepts instead of the complications and assume a zero-eccentricity orbit, which some GPS satellites very nearly are. We are then free to choose satellites passing overhead for our measurements. In those cases, the signal time from the satellite to the ground plus a constant representing the presumed travel time from the ground to the Earth’s center represents the semi-major axis to good approximation. If there were any aether wind, it would have to show up in such measurements made at different times of the day. But it doesn’t.<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">[tvf]: 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">[Thomas]: You’ve got me confused now: I thought you were saying that there is no aberration for satellites.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">There is no <i>stellar</i> aberration for satellites (20”). There is still the small aberration that exists between any two bodies in the universe with a relative transverse motion. For our Moon, this is 0.7”. For Earth satellites, this can be up to 5”. For GPS satellites, it is about 2”.<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">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 vectors are different.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">The aberration is parallel to the Earth’s surface, so it does not affect measures of the satellite’s height above the Earth’s surface. It can only affect measures of the satellite’s angular coordinates, but not its distance. -|Tom|-

[Note that I will be on travel most of June. But if follow-ups are needed, I'll get to them eventually.]