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Jun 17, 2020 at 9:47 history edited CommunityBot
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May 19, 2020 at 16:47 comment added M. A. Golding I don't think that this answer is wrong. Nobody knows anything merely by feeling certain that they know it. Someone has to have proof that something is correct in order to know it. Using the later proved false geocentric theory to deduce that Earth is closet to the Sun in January is not proving and knowing that it is true.
May 17, 2020 at 21:29 comment added userLTK @TonyK I agree with you 100%. The "Mars moved a little bit viewed from 2 different sides of the Earth" is far less accurate than the more geometric measurement of the Venus transit. I wonder if they knew an estimate for Earth's orbital eccentricity when doing the Venus transit, but that's perhaps a different question.
May 17, 2020 at 15:38 history edited Steve Linton CC BY-SA 4.0
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May 17, 2020 at 9:08 comment added TonyK @userLTK: Thanks for that, I wasn't aware of Cassini's contribution. But his estimate was 87 million miles, whereas the estimate obtained from the Venus observations was 93,726,900 miles. The true average distance is 92,955,000 miles, so it depends how you interpret my "reasonably precise".
May 17, 2020 at 5:48 comment added userLTK @TonyK The Venus transit is the better known and more accurate measurement, but the first time a reasonably close estimate was done was in 1672 by Giovanni Cassini. For some reason, Cassini's efforts aren't as well remembered, perhaps due to lower accuracy or perhaps because it preceded Isaac Newton and the desire for more precise measurements. theatlantic.com/technology/archive/2012/09/…
May 16, 2020 at 22:22 comment added TonyK Kepler didn't know the distances to anything like that precision, although he may have known the ratios. The first time we got a reasonably precise knowledge of the absolute size of the solar system was after Captain Cook observed the 1769 transit of Venus from Tahiti. (Notice that your quote doesn't explicitly claim that Kepler knew these distances!)
May 16, 2020 at 20:23 comment added uhoh @robjohn These days I’m totally distracted by slowly figuring out how Loeschian numbers may turn out to be the solution to all my problems.
May 16, 2020 at 20:14 comment added uhoh @robjohn Yep again! I didn't notice that $\cos(\theta)$ is also squared (and therefore equals $\frac{1}{2}(1+ \cos(2 \theta))$) and so doesn't go negative. Thanks!
May 16, 2020 at 16:21 comment added robjohn @uhoh: The major and minor axes differ by 0.014% (1 in 7000), but the ellipse is within 0.007% of a circle (1 in 14000).
May 16, 2020 at 15:24 comment added uhoh Yep that sounds about right! According to this answer the polar equation for an ellipse from its focus is $$r=\frac{a\!\left(1-e^2\right)}{1+e\cos(\theta)}\tag7$$ and from its center is $$r^2=\frac{\overbrace{a^2\!\left(1-e^2\right)}^{b^2}}{1-e^2\cos^2(\theta)}\tag3$$ so it's more like 1 part in 7,000 roughly, but close enough for government work!
May 16, 2020 at 12:46 comment added Steve Linton @uhoh Earth's orbit is very nearly a circle, but with the Sun shifted a bit to one side of the centre. That shift is the 1.7% while the deviation in shape from a circle is the 1 in 10000.
May 16, 2020 at 12:26 comment added uhoh The "...off by about one part in 10,000" line in the quote could use a footnote; I think it's really about 1.7 percent if it's referring to Earth's orbit's deviation from a circle, but maybe I'm misunderstanding?
May 16, 2020 at 10:11 history answered Steve Linton CC BY-SA 4.0