# Relation between launch window and RAAN and Argument Of Perigee

Dear Friends,
Does any dependency exists for achieving particular combination of RAAN ( Right Ascension Of Ascending Node) and argument of perigee with respect to particular launch date and time ? I understand that RAAN would drift because of the regression of nodes. So when and how should it matter to launch to achieve particular RAAN ?

• RAAN is the longitude of the ascending node. This acronym is not common. – SE - stop firing the good guys Jun 4 '17 at 14:01
• @Hohmannfan I meant Right Ascension Of Ascending Node – Soumajit Jun 4 '17 at 14:51
• Those are the same thing. – SE - stop firing the good guys Jun 4 '17 at 16:17

I will try to provide a general relation, assuming a spherical Earth (since my knowledge of the oblate model is not sufficient to produce a good answer).

### Argument of perigee

From this image on wikipedia's article on orbital elements we can find that argument of perigee is essentially independent from the time of launch. No matter what time it is (i.e. how is Earth currently rotated), as long as your launch site location is fixed and the ascent trajectory is fixed1, you will end up with the same AoP every time you launch.

### Longitude of ascending node

For the longitude of ascending node (LAN = RAAN) the time and date does matter - in fact it mostly defines what orbital plane you will end up on. The Earth (or: the body that we're launching from) rotates about the normal to the grey plane (see the above image again). You want to launch a bit2 before the launch site crosses the target plane. Normally you have two such opportunities per day (rotational period) - the northerly and southerly launch windows3. When exactly they happen is quite simple to calculate using spherical trigonometry.

First, let's assume that at the time of launch $T_0$ the Earth's $0°$ meridian is in line with the reference direction (vernal point $\curlyvee$). Our launch site is given by latitude $\delta$ and longitude $\lambda$, and our desired inclination is given by $i$. We can calculate the resulting longitude of the ascending node from Napier's rules for right spherical triangles. Let's look at the figure 4.9 below (source):

We want to find the angle $\Delta\lambda$, because $LAN=\lambda-\Delta\lambda$. We can use the rule "sine of the middle part = the product of the tangents of the adjacent parts" for $b$ and write:

$sin(\Delta\lambda) = tan(90°-i)\cdot tan(\delta)$

This gives us LAN as a function of launch site latitude and longitude and orbital inclination:

$LAN = \lambda - arcsin(tan(90°-i)\cdot tan(\delta))$

To introduce time into this, we simply must provide a term to account for Earth's rotation. Let's say the angular velocity is equal to $\omega [{deg\over s}]$. Then, after time $T_0+t$, our launch site will have rotated by $t\omega$ from the initial position, and our final expression for LAN will look like:

$LAN = \lambda + t\omega - arcsin(tan(90°-i)\cdot tan(\delta))$

1 - The "fixed trajectory" is a rather unrealistic assumption, but the only other choice is to assume it is unpredictable; answer to that would go beyond orbital mechanics and into ascent mechanics and guidance.
2 - Calculating exactly how much is far from trivial and in general requires simulating the entire launch.
3 - Other constraints like launch-site-specific azimuth limitations can make some opportunities infeasible.

• @ Przemek D thanks lot for the insightful answer. – Soumajit Jun 5 '17 at 12:56
• If 2 satellites are launched from the same site into 2 different polar orbits differing only in semi-major axis,will the argument of perigee of the 2 satellites differ ? – Soumajit Jun 5 '17 at 17:58
• @user19925 Theoretically they should not differ - after all, Keplerian orbital elements are independent . However, the ascent itself might take different amount of time depending on the orbit, which might have an effect on AoP. – Przemek D Jun 6 '17 at 6:41
• Posting as answer since I can't add images to comments. From the link in @Przemec_D's answer: $$\cos i = \cos \delta \ \sin \beta \tag{4.43}$$ $$\tan l = \frac{\tan \delta}{\cos \beta} \tag{4.44}$$ $$\tan \Delta \lambda = \sin \delta \tan \beta \tag{4.45}$$ $$\omega = l - \nu \tag{4.46}$$ $$\lambda_1 = \lambda_2 - \Delta lambda \tag{4.47}$$ original image From eqs. 4.37, substituting Delta_lambda from eq. 4.35: $$LAN = \lambda_1 = \lambda_2 - \arctan(\sin \delta \ \tan(90-i))$$ where 90-inc is the beta from eq. 4.35... \@Przemec_D, is this not right? – user39460 Feb 24 at 19:51
• The angle beta is in fact the launch azimuth. And the launch azimuth would be 90deg-inc only if the earth weren't spinning (or if you were launching to a polar orbit, maybe). Generally you would need to correct for the eastward velocity imparted by earth's rotation. For an orbital inclination of 51.6 deg (ISS inclination), the correction would be a few degrees---the kind of error you wouldn't want to risk in an actual launch. Notice that this correction has nothing to do with Przemec's correction at the end of his answer, which seems only to convert the LAN from an inertial frame into a rotati – user39460 Mar 8 at 21:30