# Extract anomalies from two-line element set (TLE) on ISS (International Space Station)

I am faced to a problem of interpretation illustrated on figure below : From wikipedia : Two-line element definition, I can extract the following meaning anomaly : 308.7017 ° (in degrees)

But in the 2 questions, I am asked to find :

1) at which anomaly is the ISS at the two lines epcoh

2) at wchich anomaly is the ISS on December 1st 2018.

I think the answer to question 1) is maybe what I have extracted above (308.7017 °)

But how to compute the anomaly of ISS on December 1st 2018 ?

I don't undertand very well the difference between these two anomalies requested.

If someone could help me, this would be nice.

Regards

UPDATE 1: Sorry, I must precise that I talk about mean anomaly.

• By "anomaly" do you mean "Mean anomaly". (and not the True Anomaly). Do you understand the question to be asking about the mean anomaly at 00:00 on December 1st? (day 335.0) Dec 26, 2018 at 1:23
• @JamesK I don't understand well this notion of meaning anomaly : could you tell me please, if you can, the concept of meaning anomaly, in a simple way ? I saw this is an angle by taking a virtual circular orbit but it causes confusions for me, I don't grasp the subtilities of this notion. Even simple clarifications are welcome, I just want to understand better the 2 questions. thanks Dec 26, 2018 at 1:57

First you should get an idea of what "Anomaly" means. An anomaly is an angle which can describe the position of the satellite in its orbit. The true anomaly is the actual angle made between the position of the satellite and the perigee. The true anomaly changes quickly when the satellite is close to the Earth and slowly when it is far from the Earth.

The mean anomaly is much simpler: it just increases linearly with from 0 to 360. It is zero when the satellite is at perigee. So if a satellite takes 100 minutes to orbit and it is perigee at time t=0, then it will be at 90degrees at t=25, 180degrees at t=50. The mean anomaly is just proportional to time. "Mean" here means "average", not "meaning".

The orbital elements will tell you the mean anomaly at a particular time (given as "days in the year" In your data, at day 320.81254693 of 2018 (the "epoch") the mean anomaly was 308.7017.

You also know that the mean motion is 15.53998777. That is the number of orbits the satellite makes each day. One orbit is 360 degrees, so in 'n' days the mean anomaly will increase by 15.53998777 × 360 × n

You can now find the mean anomaly at any future time. For example 2.5 days after the epoch, at day 323.31254693, the anomaly will be 308.7017 + (15.53998777 × 360 × 2.5). You may need to reduce this angle to the range 0 to 360.

To add accuracy you can also use the fact that the orbital period is slowly changing (It is getting shorter as the ISS falls towards Earth due to drag). But this change is not quick and can be neglected over short periods of time.

Now you can work out how many days pass between the epoch and Dec 1st and you can work out the mean anomaly.

The question is worded strangely. It is an odd use of "which" and is unclear about the time. It should ask "What is the mean anomaly of the ISS at the epoch" and "What is the mean anomaly of the ISS at 00:00 on Dec 18th 2018"

• -@JamesK .Thanks a lot for your very clear explanation. If I want to compute the day for the date 00.00 on Dec 18th 2018, could I do : day(18th 2018 00:00) = (365-31) = 334? Is this a correct calculus ? Dec 26, 2018 at 14:22
• I think you are a day out, because the counting of days starts from 001.0000 at midnight on January 1st, so December 1st is day 335. seehttps://www.epochconverter.com/days/2018 Dec 26, 2018 at 15:56
• -@James K. ok, I am going to compute the mean anomaly for December 1st 00:00 starting from the day 320.81254693 of 2018 (the "epoch"). You will see if my formula is correct. Dec 26, 2018 at 16:49