<p>The value obtained is in degrees per day (which I didn’t know); this is simply multiplied by 3600 to convert to seconds per day, then by 36525 to convert to seconds per century.</p>

CC BY-SA 4.0
2021-04-30T16:36:53.860Z
The reason your answer came out in degrees per day is that you plugged in $n$ with those units.  Assuming you use the same units for both parts of the ratio Rsun/a, the only thing left that has units is $n$, so that will set the units of your answer.
Eric Jensen
CC BY-SA 4.0
2021-04-30T02:50:45.053Z
2021-04-30T02:50:45.053Z
Pierre Paquette
<p>In <em>Precession of Mercury’s Perihelion from Ranging to the MESSENGER Spacecraft</em> (<a href="https://ui.adsabs.harvard.edu/abs/2017AJ....153..121P/abstract" rel="nofollow noreferrer">https://ui.adsabs.harvard.edu/abs/2017AJ....153..121P/abstract</a>), one finds the precession of Mercury’s perihelion <strong>due to the oblateness (quadrupole) of the Sun</strong> to be:</p>
<p><span class="math-container">$ \displaystyle \dot \varpi_{J_2} = \frac {3}{2} \frac {nJ_2}{(1-e^2)^2} \left( \frac {R_\odot}{a} \right)^2 \left(1 - \frac {3}{2} sin^2 i \right) $</span></p>
<p>(equation 3, p. 2). A few lines down, we read that it amounts to about 0.03″ per century—later refined to 0.0286″ per century.</p>
<p>However, when I do the calculation, with <span class="math-container">$ a $</span> = 57.90905 Gm, <span class="math-container">$ e $</span> = 0.20563, <span class="math-container">$ i $</span> = 3.38°, <span class="math-container">$ R_\odot $</span> = 696,342 km (all four from Wikipedia), <span class="math-container">$ n $</span> = 4.09°/d (calculated from formula 33.6 in Meeus 1998), and <span class="math-container">$ J_2 $</span> = 2.25 × 10⁻⁷ (paper’s abstract), I get 0.01855″ per century.</p>
<p>What am I doing wrong?</p>

CC BY-SA 4.0
2021-04-29T23:38:33.833Z
Plugging in those numbers, I got 0.02844" per century, which is roughly 1.53 times your value and fairly close to theirs, so I think the parameter values you chose are definitely right.
HDE 226868
CC BY-SA 4.0
2021-04-30T02:34:25.390Z
Can you please detail your calculation? Once you calculate $ \dot \varpi $, you need to convert to arcseconds per century; that might be where I screw up…
Pierre Paquette
CC BY-SA 4.0
2021-04-30T02:43:24.543Z
[Here's a Wolfram Alpha link](https://www.wolframalpha.com/input/?i=%283%2F2%29*%284.09+degrees+per+day%29*%282.25*10%5E%28-7%29%29+*+%281-%28.20563%29%5E2%29%5E%28-2%29+*+%28%28696%2C342+km%29%2F%2857.90905*10%5E9+meters%29%29%5E2+*+%281+-+%283%2F2%29*%28sin%283.38+degrees%29%29%5E2%29&assumption=%22UnitClash%22+-%3E+%7B%22degrees+per+day%22%2C+%7B%22AngularDegreesPerDay%22%7D%7D&assumption=%22ListOrNumber%22+-%3E+%7B%22Number%22%2C+%22696%2C342%22%7D) confirming the answer, although it's given in milliarcseconds per year.
HDE 226868
CC BY-SA 4.0
2021-04-30T02:50:23.400Z
Ha! That’s what I had wrong. The value obtained is in degrees per day (which I didn’t know); this is simply multiplied by 3600 to convert to seconds per day, then by 36525 to convert to seconds per century. Excellent. Thanks! :)
Pierre Paquette
<apsidal-precession>
Apsidal Precession: What Am I Getting Wrong?
Pierre Paquette
# Apsidal Precession: What Am I Getting Wrong?

(Asked by: username_0 on 2021-04-29)

In *Precession of Mercury’s Perihelion from Ranging to the MESSENGER Spacecraft* (<https://ui.adsabs.harvard.edu/abs/2017AJ....153..121P/abstract>), one finds the precession of Mercury’s perihelion **due to the oblateness (quadrupole) of the Sun** to be:

$ \displaystyle \dot \varpi\_{J\_2} = \frac {3}{2} \frac {nJ\_2}{(1-e^2)^2} \left( \frac {R\_\odot}{a} \right)^2 \left(1 - \frac {3}{2} sin^2 i \right) $

(equation 3, p. 2). A few lines down, we read that it amounts to about 0.03″ per century—later refined to 0.0286″ per century.

However, when I do the calculation, with $ a $ = 57.90905 Gm, $ e $ = 0.20563, $ i $ = 3.38°, $ R\_\odot $ = 696,342 km (all four from Wikipedia), $ n $ = 4.09°/d (calculated from formula 33.6 in Meeus 1998), and $ J\_2 $ = 2.25 × 10⁻⁷ (paper’s abstract), I get 0.01855″ per century.

What am I doing wrong?

**Tags:** apsidal-precession
> **Comments:**
> **username_1**: Plugging in those numbers, I got 0.02844" per century, which is roughly 1.53 times your value and fairly close to theirs, so I think the parameter values you chose are definitely right.  
> **username_0**: Can you please detail your calculation? Once you calculate $ \dot \varpi $, you need to convert to arcseconds per century; that might be where I screw up…  
> **username_1**: [Here's a Wolfram Alpha link](https://www.wolframalpha.com/input/?i=%283%2F2%29*%284.09+degrees+per+day%29*%282.25*10%5E%28-7%29%29+*+%281-%28.20563%29%5E2%29%5E%28-2%29+*+%28%28696%2C342+km%29%2F%2857.90905*10%5E9+meters%29%29%5E2+*+%281+-+%283%2F2%29*%28sin%283.38+degrees%29%29%5E2%29&assumption=%22UnitClash%22+-%3E+%7B%22degrees+per+day%22%2C+%7B%22AngularDegreesPerDay%22%7D%7D&assumption=%22ListOrNumber%22+-%3E+%7B%22Number%22%2C+%22696%2C342%22%7D) confirming the answer, although it's given in milliarcseconds per year.  
> **username_0**: Ha! That’s what I had wrong. The value obtained is in degrees per day (which I didn’t know); this is simply multiplied by 3600 to convert to seconds per day, then by 36525 to convert to seconds per century. Excellent. Thanks! :)  

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## 1 answers

### Answer by username_0 on 2021-04-30. Score: 1

The value obtained is in degrees per day (which I didn’t know); this is simply multiplied by 3600 to convert to seconds per day, then by 36525 to convert to seconds per century.

> **Comments:**
> **username_2**: The reason your answer came out in degrees per day is that you plugged in $n$ with those units.  Assuming you use the same units for both parts of the ratio Rsun/a, the only thing left that has units is $n$, so that will set the units of your answer.  

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