<p>Heliocentric longitude is a direction, not a date.</p>
<p>A Heliocentric longitude of 0 is the direction from the sun towards the vernal equinox (currently in Pices)</p>
<p>You can ask for the date on which the Earth is at a particular Heliocentric longitude.  A careful calculation would require accounting for the eccentricity of the Earth's orbit, and also adjusting for the oddities of the calendar, but since when the Sun in on the vernal equinox (About March 21), the Earth will be at 180 degrees of heliocentric longitude, and longitude will increase by 360/365.24 degrees per day.</p>

CC BY-SA 4.0
2021-01-01T23:27:02.340Z
Sorry that I was not using the exact terms, thanks for clarifying. What do you mean my *oddities of the calendar*?
B--rian
CC BY-SA 4.0
2021-01-01T23:27:25.967Z
Also: Is it really as simple to devide 360 by 365.24, or don't I need to deal with [sidereal time](https://en.wikipedia.org/wiki/Sidereal_time)?
B--rian
CC BY-SA 4.0
2021-01-02T01:09:54.230Z
oddities in the calendar  ie the number of days in the month is varies.  The number of days in a year varies. The average number of days in a calendar year is chosen to be close to (but not exactly equal to) the length of a tropical year.  The orbital period is of the earth is different,  This is why many use julian day numbers, (and convert those to dates only as required)  No, its not enough to divide 360 by 365.25 If you want accuracy you need to deal with Kepler.  However to a first approximation this is probably "good enough" and you can refactor later.
James K
CC BY-SA 4.0
2021-01-01T23:16:29.403Z
2021-01-01T23:16:29.403Z
James K
<p>Being a bit familiar with celestial mechanics, I know that the <a href="https://en.wikipedia.org/wiki/Hohmann_transfer_orbit" rel="nofollow noreferrer">Hohmann transfer orbit</a> is the quickest way to <em>transfer between two circular orbits of different radii around a central body in the same plane</em>. JPL's <a href="https://www.jpl.nasa.gov/edu/teach/activity/lets-go-to-mars-calculating-launch-windows/" rel="nofollow noreferrer">Let's Go to Mars! Calculating Launch Windows</a> explains the geometry and offers a very intuitive guide how to calculate the travel time to Mars, and this can easily be repeated for Venus.</p>
<p>Just for fun, I would like to mark all <a href="https://en.wikipedia.org/wiki/Launch_window" rel="nofollow noreferrer">launch windows</a> for a travel to Venus on a calendar, similar to the list compiled by <a href="http://mentallandscape.com/V_VenusMissions.htm" rel="nofollow noreferrer">Don P. Mitchell</a></p>
<pre><code>1991, Jun 5     Unused Type I window
1993, Jan 5     Unused Type I window
1994, Aug 9     Unused Type I window
1996, Mar 20    Unused Type I window
1997, Oct 15    Cassini flyby   
</code></pre>
<p><strong>Question:</strong> I have a probably minor (astronomical) issue: I am struggling <strong>how to convert <a href="https://en.wikipedia.org/wiki/Ecliptic_coordinate_system#Spherical_coordinates" rel="nofollow noreferrer">heliocentric longitudes</a> into real dates?</strong> I guess I am missing the English terms here. I would appreciate a small reminder how to proceed.</p>

CC BY-SA 4.0
2021-01-01T22:56:06.993Z
I think the "heliocentric longitudes to date" question is an astronomical question, but the other parts would be much better asked on the "Space" stackexchange,  where you can find real rocket scientists.
James K
CC BY-SA 4.0
2021-01-01T23:03:22.183Z
@JamesK I edited out the part which belongs to another site.
B--rian
CC BY-SA 4.0
2021-01-02T01:26:59.530Z
With your edits and the splitting off of the other part [to Space SE](https://space.stackexchange.com/q/49387/12102) this question now seems **on-topic** here.
uhoh
<solar-system><venus><terminology><celestial-mechanics><ecliptic>
Launch windows for Venus transit; how to convert heliocentric longitudes to date?
B--rian
# Launch windows for Venus transit; how to convert heliocentric longitudes to date?

(Asked by: username_0 on 2021-01-01)

Being a bit familiar with celestial mechanics, I know that the Hohmann transfer orbit is the quickest way to transfer between two circular orbits of different radii around a central body in the same plane. JPL's Let's Go to Mars! Calculating Launch Windows explains the geometry and offers a very intuitive guide how to calculate the travel time to Mars, and this can easily be repeated for Venus.
Just for fun, I would like to mark all launch windows for a travel to Venus on a calendar, similar to the list compiled by Don P. Mitchell
1991, Jun 5     Unused Type I window
1993, Jan 5     Unused Type I window
1994, Aug 9     Unused Type I window
1996, Mar 20    Unused Type I window
1997, Oct 15    Cassini flyby   

Question: I have a probably minor (astronomical) issue: I am struggling how to convert heliocentric longitudes into real dates? I guess I am missing the English terms here. I would appreciate a small reminder how to proceed.

**Tags:** solar-system, venus, terminology, celestial-mechanics, ecliptic
> **Comments:**
> **username_1**: I think the "heliocentric longitudes to date" question is an astronomical question, but the other parts would be much better asked on the "Space" stackexchange,  where you can find real rocket scientists.  
> **username_0**: @JamesK I edited out the part which belongs to another site.  
> **username_2**: With your edits and the splitting off of the other part [to Space SE](https://space.stackexchange.com/q/49387/12102) this question now seems **on-topic** here.  

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## 1 answers

### Answer by username_1 on 2021-01-01. Score: 1

Heliocentric longitude is a direction, not a date.

A Heliocentric longitude of 0 is the direction from the sun towards the vernal equinox (currently in Pices)

You can ask for the date on which the Earth is at a particular Heliocentric longitude. A careful calculation would require accounting for the eccentricity of the Earth's orbit, and also adjusting for the oddities of the calendar, but since when the Sun in on the vernal equinox (About March 21), the Earth will be at 180 degrees of heliocentric longitude, and longitude will increase by 360/365.24 degrees per day.

> **Comments:**
> **username_0**: Sorry that I was not using the exact terms, thanks for clarifying. What do you mean my *oddities of the calendar*?  
> **username_0**: Also: Is it really as simple to devide 360 by 365.24, or don't I need to deal with [sidereal time](https://en.wikipedia.org/wiki/Sidereal_time)?  
> **username_1**: oddities in the calendar  ie the number of days in the month is varies.  The number of days in a year varies. The average number of days in a calendar year is chosen to be close to (but not exactly equal to) the length of a tropical year.  The orbital period is of the earth is different,  This is why many use julian day numbers, (and convert those to dates only as required)  No, its not enough to divide 360 by 365.25 If you want accuracy you need to deal with Kepler.  However to a first approximation this is probably "good enough" and you can refactor later.  

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