<p>The formula relating the synodic period of Venus to its sidereal period is</p>
<p><span class="math-container">$$\frac1P = \frac1E + \frac1S$$</span></p>
<p>Where <span class="math-container">$P$</span> is the sidereal period, <span class="math-container">$S$</span> is the synodic period, and <span class="math-container">$E$</span> is the orbital period of Earth (1 year).</p>
<p>Now both E and P can be calculated from Kepler's law, explicitly</p>
<p><span class="math-container">$$P = 2\pi\sqrt{\frac{r_P^3}{\mu}}$$</span></p>
<p>Where <span class="math-container">$r_P$</span> is the semi-major-axis of Venus (distance to sun), and <span class="math-container">$\mu$</span> is the sun's gravitational parameter <span class="math-container">$\mu=1.33\times 10^{20}$</span> in SI units (m,kg,s).  The same formula works for <span class="math-container">$E$</span></p>
<p>So now you can use this law to work out  <span class="math-container">$P$</span> and <span class="math-container">$E$</span> for Venus and Earth in their new positions, and with that find <span class="math-container">$S$</span>. Take care with units you can't use metres and seconds in one part of the formula and km - years in another.</p>

CC BY-SA 4.0
2022-08-18T20:24:35.100Z
Answered as a procedure for find the answer, rather than an answer.
James K
CC BY-SA 4.0
2022-08-19T15:46:17.913Z
Thank you so much for taking the time to post this helpful response and including the math! Much appreciated!
JM Yaden
CC BY-SA 4.0
2022-09-01T19:12:10.463Z
please help! I am working with Kepler's law in your answer above and coming up with two figures that confuse me. When I plug in Venus' actual semi-major axis (108,208,000 km from Wikipedia) I am getting 613, and when I plug in 125,000,000 (my own hypothetical semi-major axis for Venus) I am getting 761. My question is, what do these numbers represent? In your answer you said that P represents Venus' sidereal period, which (according to Wikipedia) is 225 days. Bearing in mind that I am no mathematician and may be making a simple mistake, can you clarify? Thank you!
JM Yaden
CC BY-SA 4.0
2022-09-01T19:15:20.500Z
Sorry I wish I could show you my work here.
JM Yaden
CC BY-SA 4.0
2022-09-01T19:23:47.190Z
Make sure you use the distance in metres  ie 125,000,000,000 then the period will be in seconds.  Convert to days by dividing by 86400.    Here is the calculation for the [real venus](https://www.google.com/search?q=2%2Api%2Asqrt%28108%2C208%2C000%2C000%5E3%2F1.33e20%29%2F86400)
James K
CC BY-SA 4.0
2022-09-01T19:25:30.620Z
Ah ah ah, I knew this had something to do with units, just like you said above. Thank you so much for your help, James! :D
JM Yaden
CC BY-SA 4.0
2022-09-01T19:35:40.353Z
It worked! Actual values render 225 for P, hypothetical value renders 278. Now I will continue to work out the synodic periods for the hypothetical Venus and Earth. Thank you so much, James!
JM Yaden
CC BY-SA 4.0
2022-09-01T20:25:01.030Z
I began by using the proper values for Venus and, rounding up to the nearest day, I end up with a formula that looks like 1/225(P) = 1/366(E) + 1/S. S, as I understand it, is the synodic period of Venus, so I used 584 (days) as a placeholder. Then I noticed that when you add 225 and 366 you get 591, which is very close to 584. So, in my mind, everything had worked out because, of course, I have been rounding up. ANYWAY, this next request clarification will probably seem like kid-stuff to you, but please bear with me...
JM Yaden
CC BY-SA 4.0
2022-09-01T20:27:23.517Z
In order to isolate 1/S, algebraically, don't we have to subtract 1/366(E) from both sides? That creates a formula that looks like 1/225(P) - 1/366(E) = 1/S. But that is confusing to me, because on the left side of the equation we end up with a negative number. Hoping you're not tearing your hair out over this, can you clarify? Thank you!
JM Yaden
CC BY-SA 4.0
2022-09-01T20:29:49.720Z
No, the left hand side is positive!  1/225 is *larger* than 1/366   But you are making another mistake, You also said that you were going to "move earth out 10%" So you will need to calculate the new orbital period of Earth.  It isn't going to be 366.
James K
CC BY-SA 4.0
2022-09-01T20:43:25.913Z
Yes, you're absolutely right, all this is working towards a calculation for the hypothetical Venus and Earth, both of which will have moved 10% out from the Sun. At this time I am just trying to do the math you demonstrated above, using actual values to check my work. Thank you so much for helping. I am going to move on the hypothetical calculations now.
JM Yaden
CC BY-SA 4.0
2022-09-01T21:21:14.757Z
Ok, here are my results, thanks to James K.'s assistance: by adding approximately 16,000,000 km (10,000,000 miles) to both Venus and Earth's semi-major axes, I get a sidereal period of 279 days for a hypothetical Venus (h-Venus) and 426 days for an h-Earth. This means, unless I am mistaken, that the new synodic period for this h-Venus would be approximately 705 days, as opposed to its actual synodic period of approximately 584 days. (For h-Venus I used 125,000,000,000 meters in Kepler's formula, and 166,000,000,000 for h-Earth.) Does this look correct to you, James?
JM Yaden
CC BY-SA 4.0
2022-09-01T21:32:22.463Z
Looks plausible.  Feel free to write up an answer to your own question.
James K
CC BY-SA 4.0
2022-09-01T21:37:00.710Z
Thank you so much, James! Your assistance has been invaluable!
JM Yaden
CC BY-SA 4.0
2022-08-18T20:23:54.157Z
2022-08-18T20:23:54.157Z
James K
<p>Let us say that Earth and Venus are both moved 10 million additional miles from the sun. How would the synodic period of Venus appear to change for an observer on Earth? If 584 days is Venus' current synodic period, how many days would elapse if the planets were adjusted as mentioned?</p>
<p>Thank you so much for your help! Math to help me do it myself is most welcome!</p>

CC BY-SA 4.0
2022-08-18T20:01:51.263Z
Any thought you spend on the problem or researching it yourself?
planetmaker
CC BY-SA 4.0
2022-08-19T16:18:57.597Z
planetmaker is correct.  When you post a question you should say what you have already tried.  That is so we don't repeat what you already know.  Suppose you already knew the 1/p=1/e+1/s law and Kepler's law, that would make my answer completely useless.  So when you post another question be sure to tell us what thoughts you have spent on the problem.  See [ask]
James K
CC BY-SA 4.0
2022-08-19T16:27:24.727Z
James K., thank you for taking the time to play the mediator here. Planetmaker, if I misinterpreted the tone and intent of your comment, I apologize for getting hot.
JM Yaden
CC BY-SA 4.0
2022-08-19T16:37:45.380Z
https://meta.stackexchange.com/help/how-to-ask says what I miss here "Have you thoroughly searched for an answer before asking your question? Sharing your research helps everyone. Tell us what you found and why it didn’t meet your needs. This demonstrates that you’ve taken the time to try to help yourself, it saves us from reiterating obvious answers, and above all, it helps you get a more specific and relevant answer!". Your question makes me believe you know kepler's laws and some math, so something is missing somewhere. But where exactly is your problem and where do you stand?
planetmaker
CC BY-SA 4.0
2022-08-19T17:27:17.843Z
Planetmaker, I appreciate your response here, and both your and James K.'s willingness to give me a little education into how this forum operates. As per your question, James K. has provided me with exactly what I need to continue below, so I think I am good. I am here primarily to get assistance with the mathematics necessary to work out my inquiries. I assure you, I am not here to simply get answers, but to learn more math and astrophysics so that I can help myself in the future. Again, thank you for your patience with my "outburst" above, and I hope to see your comments on my future posts.
JM Yaden
CC BY-SA 4.0
2022-08-19T21:04:54.847Z
No worries. I know it can be frustrating at times to get at terms how a community works. Welcome here. (And you know you can delete your own comments - and it's encouraged to do so when they became obsolete? :D )
planetmaker
CC BY-SA 4.0
2022-08-20T06:18:32.397Z
Thank you for letting me know and thank you again for your understanding. ! I will delete the comment.
JM Yaden
<observational-astronomy><orbital-mechanics><venus><earths-orbit><synodic-period>
How would the synodic period of Venus appear to change if both planets were moved 10 million additional miles away from the Sun?
JM Yaden
# How would the synodic period of Venus appear to change if both planets were moved 10 million additional miles away from the Sun?

(Asked by: username_0 on 2022-08-18)

Let us say that Earth and Venus are both moved 10 million additional miles from the sun. How would the synodic period of Venus appear to change for an observer on Earth? If 584 days is Venus' current synodic period, how many days would elapse if the planets were adjusted as mentioned?

Thank you so much for your help! Math to help me do it myself is most welcome!

**Tags:** observational-astronomy, orbital-mechanics, venus, earths-orbit, synodic-period
> **Comments:**
> **username_1**: Any thought you spend on the problem or researching it yourself?  
> **username_2**: planetmaker is correct.  When you post a question you should say what you have already tried.  That is so we don't repeat what you already know.  Suppose you already knew the 1/p=1/e+1/s law and Kepler's law, that would make my answer completely useless.  So when you post another question be sure to tell us what thoughts you have spent on the problem.  See [ask]  
> **username_0**: James K., thank you for taking the time to play the mediator here. Planetmaker, if I misinterpreted the tone and intent of your comment, I apologize for getting hot.  
> **username_1**: https://meta.stackexchange.com/help/how-to-ask says what I miss here "Have you thoroughly searched for an answer before asking your question? Sharing your research helps everyone. Tell us what you found and why it didn’t meet your needs. This demonstrates that you’ve taken the time to try to help yourself, it saves us from reiterating obvious answers, and above all, it helps you get a more specific and relevant answer!". Your question makes me believe you know kepler's laws and some math, so something is missing somewhere. But where exactly is your problem and where do you stand?  
> **username_0**: Planetmaker, I appreciate your response here, and both your and James K.'s willingness to give me a little education into how this forum operates. As per your question, James K. has provided me with exactly what I need to continue below, so I think I am good. I am here primarily to get assistance with the mathematics necessary to work out my inquiries. I assure you, I am not here to simply get answers, but to learn more math and astrophysics so that I can help myself in the future. Again, thank you for your patience with my "outburst" above, and I hope to see your comments on my future posts.  
> **username_1**: No worries. I know it can be frustrating at times to get at terms how a community works. Welcome here. (And you know you can delete your own comments - and it's encouraged to do so when they became obsolete? :D )  
> **username_0**: Thank you for letting me know and thank you again for your understanding. ! I will delete the comment.  

---

## 1 answers

### Answer by username_2 on 2022-08-18. Score: 5 | Accepted answer

The formula relating the synodic period of Venus to its sidereal period is

$$\frac1P = \frac1E + \frac1S$$

Where $P$ is the sidereal period, $S$ is the synodic period, and $E$ is the orbital period of Earth (1 year).

Now both E and P can be calculated from Kepler's law, explicitly

$$P = 2\pi\sqrt{\frac{r\_P^3}{\mu}}$$

Where $r\_P$ is the semi-major-axis of Venus (distance to sun), and $\mu$ is the sun's gravitational parameter $\mu=1.33\times 10^{20}$ in SI units (m,kg,s). The same formula works for $E$

So now you can use this law to work out $P$ and $E$ for Venus and Earth in their new positions, and with that find $S$. Take care with units you can't use metres and seconds in one part of the formula and km - years in another.

> **Comments:**
> **username_2**: Answered as a procedure for find the answer, rather than an answer.  
> **username_0**: Thank you so much for taking the time to post this helpful response and including the math! Much appreciated!  
> **username_0**: please help! I am working with Kepler's law in your answer above and coming up with two figures that confuse me. When I plug in Venus' actual semi-major axis (108,208,000 km from Wikipedia) I am getting 613, and when I plug in 125,000,000 (my own hypothetical semi-major axis for Venus) I am getting 761. My question is, what do these numbers represent? In your answer you said that P represents Venus' sidereal period, which (according to Wikipedia) is 225 days. Bearing in mind that I am no mathematician and may be making a simple mistake, can you clarify? Thank you!  
> **username_0**: Sorry I wish I could show you my work here.  
> **username_2**: Make sure you use the distance in metres  ie 125,000,000,000 then the period will be in seconds.  Convert to days by dividing by 86400.    Here is the calculation for the [real venus](https://www.google.com/search?q=2%2Api%2Asqrt%28108%2C208%2C000%2C000%5E3%2F1.33e20%29%2F86400)  
> **username_0**: Ah ah ah, I knew this had something to do with units, just like you said above. Thank you so much for your help, James! :D  
> **username_0**: It worked! Actual values render 225 for P, hypothetical value renders 278. Now I will continue to work out the synodic periods for the hypothetical Venus and Earth. Thank you so much, James!  
> **username_0**: I began by using the proper values for Venus and, rounding up to the nearest day, I end up with a formula that looks like 1/225(P) = 1/366(E) + 1/S. S, as I understand it, is the synodic period of Venus, so I used 584 (days) as a placeholder. Then I noticed that when you add 225 and 366 you get 591, which is very close to 584. So, in my mind, everything had worked out because, of course, I have been rounding up. ANYWAY, this next request clarification will probably seem like kid-stuff to you, but please bear with me...  
> **username_0**: In order to isolate 1/S, algebraically, don't we have to subtract 1/366(E) from both sides? That creates a formula that looks like 1/225(P) - 1/366(E) = 1/S. But that is confusing to me, because on the left side of the equation we end up with a negative number. Hoping you're not tearing your hair out over this, can you clarify? Thank you!  
> **username_2**: No, the left hand side is positive!  1/225 is *larger* than 1/366   But you are making another mistake, You also said that you were going to "move earth out 10%" So you will need to calculate the new orbital period of Earth.  It isn't going to be 366.  
> **username_0**: Yes, you're absolutely right, all this is working towards a calculation for the hypothetical Venus and Earth, both of which will have moved 10% out from the Sun. At this time I am just trying to do the math you demonstrated above, using actual values to check my work. Thank you so much for helping. I am going to move on the hypothetical calculations now.  
> **username_0**: Ok, here are my results, thanks to James K.'s assistance: by adding approximately 16,000,000 km (10,000,000 miles) to both Venus and Earth's semi-major axes, I get a sidereal period of 279 days for a hypothetical Venus (h-Venus) and 426 days for an h-Earth. This means, unless I am mistaken, that the new synodic period for this h-Venus would be approximately 705 days, as opposed to its actual synodic period of approximately 584 days. (For h-Venus I used 125,000,000,000 meters in Kepler's formula, and 166,000,000,000 for h-Earth.) Does this look correct to you, James?  
> **username_2**: Looks plausible.  Feel free to write up an answer to your own question.  
> **username_0**: Thank you so much, James! Your assistance has been invaluable!  

---

