<p>I also asked this question at the History of Science and Mathematics Stack Exchange and got two answers so far.</p>
<p>The answer by Ben Crowell makes a rough calculation that the displacement should be on the order of ten arc seconds.  It also has a plot of the variation in longitude of the observed and predicted positions of Uranus over time, which seem to differ by up to a few tens of arc seconds from predicted positions.</p>
<p>The answer by cktai states that according to his source the deviation between the calculated and the observed orbit was 30 arc seconds in 1835 and 70 arc seconds by 1840.</p>
<p>An arc second is one in 1,296,000, or 0.0000007, of a full circle.</p>
<p>At the average distance between the Sun and Uranus, about 19.22 Astronomical units or AU, a full circle is about 120.76271 AU. Since the orbital period of Uranus is 84.01 years, Uranus travels at an orbital speed of about 1.4374801 AU per year, or about 0.0039356 AU per day, or about 0.0001639 AU per hour, or about 0.0000027 AU per minute, or about 0.000000045 AU per second.</p>
<p>0.000000045 AU is 6.81433 kilometers, so the average orbital speed of Uranus should be about 6.81433 kilometers per second.  The average orbital speed of Uranus is listed as 6.8 kilometers per second.</p>
<p>Earth is at an average distance of 1.0 Astronomical Units from the Sun, so the distance between Earth and Uranus varies between about 17.33 and 21.11 AU, and the average distance is about 19.22 AU.</p>
<p>So since there are 360 degrees in a full circle, at the average distance between Earth and Uranus a degree of arc would be about 0.33545 AU, and an arc minute would be about 0.00559 AU, and an arc second would be about 0.00009318 AU or 13,939.61 kilometers.</p>
<p>Being off position by 30 arc seconds would equal being off position by 418,188.38 kilometers, and being off position by 70 arc seconds would equal being off position by 975,772.88 kilometers.  And that would be approximately the distance traveled by Uranus in about a day.</p>
<p>And I guess that gives some idea of the scale of the deviations which were considered a problem and which led to the discovery of Neptune.</p>

CC BY-SA 4.0
2020-06-25T17:53:51.540Z
2020-06-25T17:53:51.540Z
M. A. Golding
<p>After Uranus was discovered and its orbit calculated, its future orbit was calculated, and its future positions as seen from Earth were calculated.</p>
<p>And observers of Uranus began to notice that Uranus was deviating from its calculated positions.  Eventually the possibility that an undiscovered planet was perturbing the orbit of Uranus was used by Adams and Le Verrier to calculate the orbit of Neptune and led to the discovery of Neptune.</p>
<p>So how large were the deviations in the apparent position of Uranus that were used to calculate the orbit and apparent position of Neptune?</p>

<orbital-mechanics>
How large were the deviations of Uranus's positon from predicted which led to the discovery of Neptune?
M. A. Golding
# How large were the deviations of Uranus's positon from predicted which led to the discovery of Neptune?

(Asked by: username_0 on 2020-06-20)

After Uranus was discovered and its orbit calculated, its future orbit was calculated, and its future positions as seen from Earth were calculated.

And observers of Uranus began to notice that Uranus was deviating from its calculated positions. Eventually the possibility that an undiscovered planet was perturbing the orbit of Uranus was used by Adams and Le Verrier to calculate the orbit of Neptune and led to the discovery of Neptune.

So how large were the deviations in the apparent position of Uranus that were used to calculate the orbit and apparent position of Neptune?

**Tags:** orbital-mechanics


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## 1 answers

### Answer by username_0 on 2020-06-25. Score: 1

I also asked this question at the History of Science and Mathematics Stack Exchange and got two answers so far.

The answer by Ben Crowell makes a rough calculation that the displacement should be on the order of ten arc seconds. It also has a plot of the variation in longitude of the observed and predicted positions of Uranus over time, which seem to differ by up to a few tens of arc seconds from predicted positions.

The answer by cktai states that according to his source the deviation between the calculated and the observed orbit was 30 arc seconds in 1835 and 70 arc seconds by 1840.

An arc second is one in 1,296,000, or 0.0000007, of a full circle.

At the average distance between the Sun and Uranus, about 19.22 Astronomical units or AU, a full circle is about 120.76271 AU. Since the orbital period of Uranus is 84.01 years, Uranus travels at an orbital speed of about 1.4374801 AU per year, or about 0.0039356 AU per day, or about 0.0001639 AU per hour, or about 0.0000027 AU per minute, or about 0.000000045 AU per second.

0.000000045 AU is 6.81433 kilometers, so the average orbital speed of Uranus should be about 6.81433 kilometers per second. The average orbital speed of Uranus is listed as 6.8 kilometers per second.

Earth is at an average distance of 1.0 Astronomical Units from the Sun, so the distance between Earth and Uranus varies between about 17.33 and 21.11 AU, and the average distance is about 19.22 AU.

So since there are 360 degrees in a full circle, at the average distance between Earth and Uranus a degree of arc would be about 0.33545 AU, and an arc minute would be about 0.00559 AU, and an arc second would be about 0.00009318 AU or 13,939.61 kilometers.

Being off position by 30 arc seconds would equal being off position by 418,188.38 kilometers, and being off position by 70 arc seconds would equal being off position by 975,772.88 kilometers. And that would be approximately the distance traveled by Uranus in about a day.

And I guess that gives some idea of the scale of the deviations which were considered a problem and which led to the discovery of Neptune.


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