# Longitude of the Ascending nodes (calculation, wrong data perhaps?)

by Implicant
Tags: ascending, calculation, data, longitude, nodes
 P: 6 Question: Calculate the Longitude of the Ascending node for a satellite of Tundra Orbit. Satellite's longitude of perigee is 21.8 degrees and its argument of perigee is 270 degrees. Satellite must cover an Earth station with 21.8 degrees longitude and 63.5 degrees latitude. My approach So my approach so far is like this: Suppose that Tundra orbit inclination is 63.4: azimuth of the earth base (b) = asin(cos(63.4)/cos(63.5)) ~ 90 degrees. dl = atan(sin(63.5)*tan(90)) = indeterminable but for Ω: l1 = l2 - dl. (l2=21.8 deg) I am not physicist and i don't get this, at all. If anyone could shed light please.
 P: 6 come on people, 111 views and no responses ? At least tell me something, like question is wrong , or my methodology is totally wrong , or something...
 Mentor P: 13,646 What's the relation between longitude of ascending node, longitude of perigee, and argument of ascending node?
P: 6

## Longitude of the Ascending nodes (calculation, wrong data perhaps?)

longitude of perigee = Longitude of the ascending node + argument of perigee .

but how to calculate longitude of perigee , with the given data ?
 Mentor P: 13,646 What's to calculate? The longitude of perigee is a given: "Satellite's longitude of perigee is 21.8 degrees and its argument of perigee is 270 degrees."
 P: 6 oh, god, i dont think that its sooo easy. Should i say that longitude of ascending node is 21.8 deg-270 deg ? i that all ? and the rest data ?
 P: 6 also, is it normal for Longitude of ascending node (what a WEIRD NAME JESUS) to be negative ? edit: so i read that if its negative you can add +360 deg. Well , again, is it that simple? I have written like two papers full of crap, obviously when the answer was in front of me.
 Mentor P: 13,646 It's that simple. Why that other data? Perhaps it's needed for other questions on the same problem, but perhaps they just wanted to see if you could spot the relevant information. That's how the real world works, after all. Lots of data, but only a small part is relevant.
P: 6
 Quote by D H It's that simple. Why that other data? Perhaps it's needed for other questions on the same problem, but perhaps they just wanted to see if you could spot the relevant information. That's how the real world works, after all. Lots of data, but only a small part is relevant.
Dude, thanks a lot, a tone i meant. You just cant imagine how complicated i thought it was. And worst of all , this ludicrous thing ate my time from doing some odes that i like, hell.

cheers
HW Helper
P: 2,270
 Quote by Implicant longitude of perigee = Longitude of the ascending node + argument of perigee . but how to calculate longitude of perigee , with the given data ?
Be careful, here.

longitude of perigee = right ascension of the ascending node + argument of perigee (in equinoctial elements). This is important, since in other coordinate systems, you could be adding a broken angle (the two angles won't be in the same plane, especially with an inclination of 63.4 degrees).

Sometimes, "longitude of ascending node" is used interchangeably with "right ascension of ascending node" and sometimes "longitude of ascending node" is referenced to the prime meridian (earth fixed). Make sure you know what they mean by "longitude of perigee" and "longitude of ascending node".

The fact that the satellite has to cover an Earth station with a longitude of 21.8 degrees suggests that this is Earth fixed and you want the center longitude of the orbit to be 21.8 degrees. (In practice, the center longitude could vary quite a bit and still be visible from 21.8 degrees longitude, but I assume they gave you the longitude of the Earth station for a reason.)

The longitude will vary because you have an inclined orbit (63.4 deg) and an elliptical orbit (eccentricity around .256?). The longitude of the satellite will be 21.8 degrees when at perigee, at apogee, and one other point in between (i.e. - a warped figure 8). The longitude of the satellite when it crosses the ascending node will be East of 21.8 degrees (greater than 21.8 degrees). This is different than for a true geosynchronous orbit (eccentricity approximately 0) where the ascending node also matches the center longitude.

Since the argument of perigee is 270 degrees, true anomaly must be 90 degrees at the ascending node. If you convert true anomaly to mean anomaly, you can figure out your position in time. This is a two step process, converting from true anomaly to Eccentric anomaly, and from Eccentric Anomaly to Mean Anomaly.

The Earth rotates at a constant velocity and it completes one revolution in the same amount of time the satellite completes one orbit.

So the difference between true anomaly and mean anomaly tells you how far East the satellite drifted relative to the Earth's surface by time it reached the ascending node. Add that angle to the longitude you were at a perigee and you'll know what longitude the satellite must be over when you reach the ascending node (at least a rough approximation).

I'm sure there's some set of equations that does this easier, but I wouldn't know what they are off the top of my head. Additionally, this method can't be particularly accurate, since I know 21.8 degrees longitude will pass under the satellite somewhere between perigee and apogee in addition to the perigee and apogee points. To do this right, you need to do a true geocentric to Earth Fixed transformation (longitude, latitude, height, local velocity, azimuth, and flight path angle).

Note: This happens to get a very accurate estimate for longitude of ascending node - about a tenth of a degree off. But that's kind of lucky. Maximum East longitude is 60.9 degrees -a drift of 39.1 degrees from the center longitude. It's just already drifting back to the West by time it reaches ascending node. And while it is fairly easy to figure out the variation in longitude due to eccentricity (2e, in radians), I'm not sure there's an easy way to figure out the variation of both combined (I don't have my good book with me, but there's no easily found way to do both).

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