<p>OK your error is that you did not convert δψ to δψ̇, so the values you get are in radians per revolution, and not in arc seconds per year (or century). You need to apply equation 1024 from the link you provided. This is easily seen by the fact that nowhere in your formulas do you refer to column D (revolution period). Also, I think you will need to convert the values in column D to seconds instead of years.</p>

CC BY-SA 4.0
2023-09-14T18:34:33.977Z
Equation 1024 is in column F. For example in the cell F3, there is a term: =75/D3*J13 They claim that equation 1024 give you results in the arcsec/year (there is even clearly stated in the equation 75/Ti(yr), so years are units used in the calculations).
Bedlasky
CC BY-SA 4.0
2023-09-15T00:22:42.847Z
Ah! Apologies for not noticing that. I have copied your formulas into my own spreadsheet, and I have played around with it. First, I wonder where Pr. Fitzpatrick got his 75 from, as 3π/2 * 180/π to convert to degrees is 270, not 75… When I use 270 instead of 75, and when I use seconds instead of years, and kilometres instead of astronomical units, then I get values that approach the real ones… but divided by 1,000. I’m still investigating where the difference comes from.
Pierre Paquette
CC BY-SA 4.0
2023-09-15T01:02:33.247Z
Still haven’t found the source of the error, and the more I dig, the more confused I get. However, I found this page https://www.mathpages.com/home/kmath280/kmath280.htm and using its formulas properly, I get the right values for the precession, for Mercury at least.
Pierre Paquette
CC BY-SA 4.0
2023-09-15T01:48:46.010Z
OK, so I found a way to make it work with Fitzpatrick’s formulas (your link). You need the orbit radius in **kilometres** and the revolution time in **seconds**, then replace that 75 by 270 as I previously said, and—here’s the kicker— **divide the revolution time by Earth’s revolution time** (so you get 270 / ($\frac {T_i} {T_{Earth}} $) ). Finally, multiply by 3600, because the result is in degrees per year. This gives good results.
Pierre Paquette
CC BY-SA 4.0
2023-09-15T04:07:40.633Z
Oh, I was also wondering were he get 75 from. I've got idea before I was going to bed, so I quickly tried it. I replaced 75/Ti by 3π/(2T) to calculate radians per year. Than I converted them to arcsec/year by multipling with (180/π)*3600 and get results that roughly fits with his. I think the difference is caused by rounding, I'll check that later. Units of semi-major axis and mass shouldn't matter as long as they are the same, you divide them in calculation. So you can use whatever unit of time you need, it gives you result in angle per unit of time.
Bedlasky
CC BY-SA 4.0
2023-09-16T20:26:05.840Z
So the difference isn't caused by rounding. But the equation works good enough for my purposes. Thanks for everyone's help!
Bedlasky
CC BY-SA 4.0
2023-09-14T15:16:52.250Z
2023-09-14T15:16:52.250Z
Pierre Paquette
<p>I found this article about calculating apsidal precession:</p>
<p><a href="https://farside.ph.utexas.edu/teaching/336k/Newton/node115.html" rel="nofollow noreferrer">https://farside.ph.utexas.edu/teaching/336k/Newton/node115.html</a></p>
<p>I tried to calculate rate of apsidal precession from this equation, but I had totally different results from that from paper. My calculated valuese are around 10^-5 arcsec/year. Did I do something wrong? My calculations are as follows:</p>
<p>I did my calculations in Excel to make it easier. The table looks like this:</p>
<p><a href="https://i.sstatic.net/frjQT.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/frjQT.png" alt="" /></a></p>
<p>Bottom part with green and blue cells are calculations of that summations. Green cells are <span class="math-container">$$(Mj/M0)*(ai/aj)^3(1+15/8*(ai/aj)^2+175/64*(ai/aj)^4),$$</span> blue cells are <span class="math-container">$$(Mj/M0)*(aj/ai)^2(1+15/8*(aj/ai)^2+175/64*(aj/ai)^4)$$</span> The rightmost columns is summation of all these terms in the row.</p>
<p>The first row of green cells have following commands (from left to right):</p>
<p>=<span class="math-container">$E$</span>4<em>POWER(C3/<span class="math-container">$C$</span>4;3)</em>(1+15/8<em>POWER(C3/<span class="math-container">$C$</span>4;2)+175/64</em>POWER(C3/<span class="math-container">$C$</span>4;4))</p>
<p>=<span class="math-container">$E$</span>5<em>POWER(C3/<span class="math-container">$C$</span>5;3)</em>(1+15/8<em>POWER(C3/<span class="math-container">$C$</span>5;2)+175/64</em>POWER(C3/<span class="math-container">$C$</span>5;4))</p>
<p>=<span class="math-container">$E$</span>6<em>POWER(C3/<span class="math-container">$C$</span>6;3)</em>(1+15/8<em>POWER(C3/<span class="math-container">$C$</span>6;2)+175/64</em>POWER(C3/<span class="math-container">$C$</span>6;4))</p>
<p>=<span class="math-container">$E$</span>7<em>POWER(C3/<span class="math-container">$C$</span>7;3)</em>(1+15/8<em>POWER(C3/<span class="math-container">$C$</span>7;2)+175/64</em>POWER(C3/<span class="math-container">$C$</span>7;4))</p>
<p>=<span class="math-container">$E$</span>8<em>POWER(C3/<span class="math-container">$C$</span>8;3)</em>(1+15/8<em>POWER(C3/<span class="math-container">$C$</span>8;2)+175/64</em>POWER(C3/<span class="math-container">$C$</span>8;4))</p>
<p>=<span class="math-container">$E$</span>9<em>POWER(C3/<span class="math-container">$C$</span>9;3)</em>(1+15/8<em>POWER(C3/<span class="math-container">$C$</span>9;2)+175/64</em>POWER(C3/<span class="math-container">$C$</span>9;4))</p>
<p>=<span class="math-container">$E$</span>10<em>POWER(C3/<span class="math-container">$C$</span>10;3)</em>(1+15/8<em>POWER(C3/<span class="math-container">$C$</span>10;2)+175/64</em>POWER(C3/<span class="math-container">$C$</span>10;4))</p>
<p>Similarly for other rows.</p>
<p>The bottom row of blue cells have following commands:</p>
<p>=<span class="math-container">$E$</span>3<em>POWER(<span class="math-container">$C$</span>3/C10;2)</em>(1+15/8<em>POWER(<span class="math-container">$C$</span>3/C10;2)+175/64</em>POWER(<span class="math-container">$C$</span>3/C10;4))</p>
<p>=<span class="math-container">$E$</span>4<em>POWER(<span class="math-container">$C$</span>4/C10;2)</em>(1+15/8<em>POWER(<span class="math-container">$C$</span>4/C10;2)+175/64</em>POWER(<span class="math-container">$C$</span>4/C10;4))</p>
<p>=<span class="math-container">$E$</span>5<em>POWER(<span class="math-container">$C$</span>5/C10;2)</em>(1+15/8<em>POWER(<span class="math-container">$C$</span>5/C10;2)+175/64</em>POWER(<span class="math-container">$C$</span>5/C10;4))</p>
<p>=<span class="math-container">$E$</span>6<em>POWER(<span class="math-container">$C$</span>6/C10;2)</em>(1+15/8<em>POWER(<span class="math-container">$C$</span>6/C10;2)+175/64</em>POWER(<span class="math-container">$C$</span>6/C10;4))</p>
<p>=<span class="math-container">$E$</span>7<em>POWER(<span class="math-container">$C$</span>7/C10;2)</em>(1+15/8<em>POWER(<span class="math-container">$C$</span>7/C10;2)+175/64</em>POWER(<span class="math-container">$C$</span>7/C10;4))</p>
<p>=<span class="math-container">$E$</span>8<em>POWER(<span class="math-container">$C$</span>8/C10;2)</em>(1+15/8<em>POWER(<span class="math-container">$C$</span>8/C10;2)+175/64</em>POWER(<span class="math-container">$C$</span>8/C10;4))</p>
<p>=<span class="math-container">$E$</span>9<em>POWER(<span class="math-container">$C$</span>9/C10;2)</em>(1+15/8<em>POWER(<span class="math-container">$C$</span>9/C10;2)+175/64</em>POWER(<span class="math-container">$C$</span>9/C10;4))</p>
<p>Similarly for other rows.</p>
<p>Rate of apsidal precesion is than calculate as: 75/Ti*sum of terms stated above. As you can see, values are in order of 10^-3 to 10^-5, not even numericaly matching results from the paper.</p>

CC BY-SA 4.0
2023-09-13T20:35:19.557Z
Maybe if you show us your calculations (or at least part of them), we might be able to pinpoint what, if anything, you did wrong…
Pierre Paquette
CC BY-SA 4.0
2023-09-13T20:35:40.003Z
Oh! and welcome to AstronomySE!
Pierre Paquette
CC BY-SA 4.0
2023-09-13T22:38:53.667Z
Please elaborate on detail what you tried and where and how that differs from your expectation. Especially also quote all you input values and their corresponding units. It would be lovely if this question would be understandable without the link, should it break.
planetmaker
CC BY-SA 4.0
2023-09-14T00:02:05.350Z
Thanks for welcome! I added some clarification below.
Bedlasky
CC BY-SA 4.0
2023-09-14T07:02:27.963Z
@Bedlasky thanks for the clarification, that will help a lot in understanding the problem. I moved that to this question and I recommend you delete your answer which contains this explanation of your calculations (as it is not an answer, and answers always should be that: answers - that is the difference to a forum here). One can always use the edit button to add something to own postings and suggest this for other ones
planetmaker
<orbital-mechanics><apsidal-precession>
Apsidal precession
Bedlasky
# Apsidal precession

(Asked by: username_0 on 2023-09-13)

I found this article about calculating apsidal precession:

<https://farside.ph.utexas.edu/teaching/336k/Newton/node115.html>

I tried to calculate rate of apsidal precession from this equation, but I had totally different results from that from paper. My calculated valuese are around 10^-5 arcsec/year. Did I do something wrong? My calculations are as follows:

I did my calculations in Excel to make it easier. The table looks like this:

[![](https://i.sstatic.net/frjQT.png)](https://i.sstatic.net/frjQT.png)

Bottom part with green and blue cells are calculations of that summations. Green cells are $$(Mj/M0)\*(ai/aj)^3(1+15/8\*(ai/aj)^2+175/64\*(ai/aj)^4),$$ blue cells are $$(Mj/M0)\*(aj/ai)^2(1+15/8\*(aj/ai)^2+175/64\*(aj/ai)^4)$$ The rightmost columns is summation of all these terms in the row.

The first row of green cells have following commands (from left to right):

=$E$4*POWER(C3/$C$4;3)*(1+15/8*POWER(C3/$C$4;2)+175/64*POWER(C3/$C$4;4))

=$E$5*POWER(C3/$C$5;3)*(1+15/8*POWER(C3/$C$5;2)+175/64*POWER(C3/$C$5;4))

=$E$6*POWER(C3/$C$6;3)*(1+15/8*POWER(C3/$C$6;2)+175/64*POWER(C3/$C$6;4))

=$E$7*POWER(C3/$C$7;3)*(1+15/8*POWER(C3/$C$7;2)+175/64*POWER(C3/$C$7;4))

=$E$8*POWER(C3/$C$8;3)*(1+15/8*POWER(C3/$C$8;2)+175/64*POWER(C3/$C$8;4))

=$E$9*POWER(C3/$C$9;3)*(1+15/8*POWER(C3/$C$9;2)+175/64*POWER(C3/$C$9;4))

=$E$10*POWER(C3/$C$10;3)*(1+15/8*POWER(C3/$C$10;2)+175/64*POWER(C3/$C$10;4))

Similarly for other rows.

The bottom row of blue cells have following commands:

=$E$3*POWER($C$3/C10;2)*(1+15/8*POWER($C$3/C10;2)+175/64*POWER($C$3/C10;4))

=$E$4*POWER($C$4/C10;2)*(1+15/8*POWER($C$4/C10;2)+175/64*POWER($C$4/C10;4))

=$E$5*POWER($C$5/C10;2)*(1+15/8*POWER($C$5/C10;2)+175/64*POWER($C$5/C10;4))

=$E$6*POWER($C$6/C10;2)*(1+15/8*POWER($C$6/C10;2)+175/64*POWER($C$6/C10;4))

=$E$7*POWER($C$7/C10;2)*(1+15/8*POWER($C$7/C10;2)+175/64*POWER($C$7/C10;4))

=$E$8*POWER($C$8/C10;2)*(1+15/8*POWER($C$8/C10;2)+175/64*POWER($C$8/C10;4))

=$E$9*POWER($C$9/C10;2)*(1+15/8*POWER($C$9/C10;2)+175/64*POWER($C$9/C10;4))

Similarly for other rows.

Rate of apsidal precesion is than calculate as: 75/Ti\*sum of terms stated above. As you can see, values are in order of 10^-3 to 10^-5, not even numericaly matching results from the paper.

**Tags:** orbital-mechanics, apsidal-precession
> **Comments:**
> **username_1**: Maybe if you show us your calculations (or at least part of them), we might be able to pinpoint what, if anything, you did wrong…  
> **username_1**: Oh! and welcome to AstronomySE!  
> **username_2**: Please elaborate on detail what you tried and where and how that differs from your expectation. Especially also quote all you input values and their corresponding units. It would be lovely if this question would be understandable without the link, should it break.  
> **username_0**: Thanks for welcome! I added some clarification below.  
> **username_2**: @Bedlasky thanks for the clarification, that will help a lot in understanding the problem. I moved that to this question and I recommend you delete your answer which contains this explanation of your calculations (as it is not an answer, and answers always should be that: answers - that is the difference to a forum here). One can always use the edit button to add something to own postings and suggest this for other ones  

---

## 1 answers

### Answer by username_1 on 2023-09-14. Score: 2

OK your error is that you did not convert δψ to δψ̇, so the values you get are in radians per revolution, and not in arc seconds per year (or century). You need to apply equation 1024 from the link you provided. This is easily seen by the fact that nowhere in your formulas do you refer to column D (revolution period). Also, I think you will need to convert the values in column D to seconds instead of years.

> **Comments:**
> **username_0**: Equation 1024 is in column F. For example in the cell F3, there is a term: =75/D3*J13 They claim that equation 1024 give you results in the arcsec/year (there is even clearly stated in the equation 75/Ti(yr), so years are units used in the calculations).  
> **username_1**: Ah! Apologies for not noticing that. I have copied your formulas into my own spreadsheet, and I have played around with it. First, I wonder where Pr. Fitzpatrick got his 75 from, as 3π/2 * 180/π to convert to degrees is 270, not 75… When I use 270 instead of 75, and when I use seconds instead of years, and kilometres instead of astronomical units, then I get values that approach the real ones… but divided by 1,000. I’m still investigating where the difference comes from.  
> **username_1**: Still haven’t found the source of the error, and the more I dig, the more confused I get. However, I found this page https://www.mathpages.com/home/kmath280/kmath280.htm and using its formulas properly, I get the right values for the precession, for Mercury at least.  
> **username_1**: OK, so I found a way to make it work with Fitzpatrick’s formulas (your link). You need the orbit radius in **kilometres** and the revolution time in **seconds**, then replace that 75 by 270 as I previously said, and—here’s the kicker— **divide the revolution time by Earth’s revolution time** (so you get 270 / ($\frac {T_i} {T_{Earth}} $) ). Finally, multiply by 3600, because the result is in degrees per year. This gives good results.  
> **username_0**: Oh, I was also wondering were he get 75 from. I've got idea before I was going to bed, so I quickly tried it. I replaced 75/Ti by 3π/(2T) to calculate radians per year. Than I converted them to arcsec/year by multipling with (180/π)*3600 and get results that roughly fits with his. I think the difference is caused by rounding, I'll check that later. Units of semi-major axis and mass shouldn't matter as long as they are the same, you divide them in calculation. So you can use whatever unit of time you need, it gives you result in angle per unit of time.  
> **username_0**: So the difference isn't caused by rounding. But the equation works good enough for my purposes. Thanks for everyone's help!  

---

