The equation for mean anomaly is

$$M = \sqrt{\frac{\mu}{a^3}} (t-T_0)$$

How could I extract information from a TLE in order to calculate a mean anomaly? Would I use this equation, or something different?

  • $\begingroup$ This should be on topic and clear? Is this homework? $\endgroup$
    – Muze
    Mar 30, 2019 at 17:01
  • $\begingroup$ @Muze see answers to Does this site have anything like a homework policy? We haven't needed a special tag so far. $\endgroup$
    – uhoh
    Mar 31, 2019 at 1:27
  • 2
    $\begingroup$ I could understand this question perfectly as asked and I suspect it was clear to most of the close-voters as well. I would have liked to have written an answer. I'm voting to **re-open and I've made an edit to make the question even clearer. $\endgroup$
    – uhoh
    Mar 31, 2019 at 1:29

1 Answer 1


The following is an approximate answer only, for the following reasons and probably other reasons as well:

  1. Values found in TLE's are not real Keplerian elements. They have the "look an feel" and the names may sounds similar, but they are not. Instead, they are generated specifically to be interpreted by SGP4, an algorithm or program that interprets TLEs and generates coordinates and times. However, they are close enough that they can be used to approximate Keplerian orbits.

  2. Real orbits around the Earth are not Keplerian. The biggest effects are the Earth's oblateness as expressed in the $J_2$ coefficient (found inside SGP4 along with many others), and atmospheric drag.

$$M = \sqrt{\frac{\mu}{a^3}} (t-T_0)$$

is also commonly written as

$$M = n(t-T_0)$$


$$n = \sqrt{\frac{\mu}{a^3}} = \frac{T}{2\pi}$$

and $T$ is the orbital period.

To get mean motion n:

Luckily $n$ can be found quickly from the TLE. On line 2, in field 8 (columns 53–63 numbered starting from 1) is the mean motion in units of revolutions per day. See Celestrak documentation and NORAD Two-Line Element Set Format and NASA page Definition of Two-line Element Set Coordinate System for more on this.

Searching https://celestrak.org/satcat/search.php for the catalog number 25544 finds that the current TLE for the ISS is:

ISS (ZARYA)             
1 25544U 98067A   19090.17372685  .00002490  00000-0  47463-4 0  9991
2 25544  51.6431  29.0343 0002449 133.6498  81.3192 15.52464104163136
                                            xxxxxxx xxxxxxxxxxx

and the mean motion is therefore 15.52464104 revolutions per $24$ hour day. Multiply that by $2\pi$ and divide by $24 \times 3600$ and you get $1.1289837556 \times 10^{-3}$ radians per second or $6.4686004335 \times 10^{-2}$ degrees per second.

edit: Addressing @McMissile's question directly and explicitly, if you want the period in every day units, it's just $T = 1/n$ in (24 hour) days: 1/15.52464104 = 0.06441372766 days or 1.54592946389 hours or 92.755767833 minutes, etc.

To get time of periapsis T₀:

Start with the epoch, which is contained in an old fashioned format in line 1 columns 19–20 and 21–32. The first is the last two digits year. First two digits will be 19 for values of 57 and larger (1957 being Sputnik 1) and will be 20 for lower values. So for this ISS TLE the year is 2019 and the day number is 090.17372685 (which starts from 1.0, not 0.0). To convert to seconds, use something like the answers to Convert TLE times (decimal days) to seconds after epoch. Call the result $T_{epoch}$.

Next, obtain from the TLE the mean anomaly at epoch from line 2, field 7, columns 44–51 which is 81.3192 degrees.

From your epoch, subtract the time it took to advance from 0.0 to 81.3192 degrees to get $T_0$:

$$T_0 = T_{epoch} - \frac{\text{mean anomaly}}{\text{mean motion}}$$

You now have $T_0$ and $n$. For any time measured in the same units as the units you converted your TLE's epoch, you get the mean anomaly at time $t$ from:

$$M = n(t-T_0)$$

From NASA's Definition of Two-line Element Set Coordinate System enter image description here


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