Calculation of Mean anomaly

[Nikogori]

I'm trying to calculate Mean anomaly which appears in my textbook but I can't get correct answer.

semi major axis a = 26552.305 km
eccentricity e = 0.747411
mean anomaly at epoch M_0 = 24.70°
inclination i = 63.96°
longitude of ascending node Ω = 33.79°
argument of periapsis ω = 243.99°
t – t_0 = 4707393.045 sec

Mean anomaly I get from above equation is M = 711.6°

eccentric anomaly E is shown in the textbook.
E = 157.69°
therefore, required mean anomaly should be M = 141.4°

Can someone tell me where I went wrong? I'm not a student, I have no one to ask math questions...
Thanks in advance.

 

[Urwumpe]

Your formula is correct, if that relieves you a bit. Its mean motion ([math]\sqrt{\frac{\mu}{a^3}}[/math]) multiplied by time passed since reference ([math]t-t_0[/math]) shifted by the reference mean anomaly [math]M_0[/math]
Well, you know how the mean anomaly is defined geometrically? Maybe it helps remember that the mean anomaly maps your real elliptic orbit on a perfect circular orbit that has the same orbit period.

Eccentric anomaly on the other hand is the angle to a virtual point that is created by taking your elliptic orbit and stretching it into a circle.

I also get completely different numbers than you do with the same input numbers, which leads me to the conclusion that [math]t-t_0[/math] is not the time of E. I even get different numbers if I "forget" converting the angles into radians.

My results so far:

M from time|351.6°
M from E|157.41°

 

[martins]

I get M = 141.4 deg with the OP's equation (assuming mu is for Earth, which wasn't actually specified).
All is well ...

Edit: well, actually M = 3.9381e+04 deg which might matter if you need to count orbits ...

Edit2: As an attempt at a postmortem, I *do* get 711.6 if if forget a set of brackets and only convert the second term to degrees:

sqrt(mu/a^3)*(t-t0) + M0/RAD = 711.6 (deg? rad?)

Is that what happened?

Edit3: Sorry, the above was assuming that M0 had been converted to radians to start with. So maybe all that happened was that you forgot that the first term gives a result in radians, not degrees.

 

[Nikogori]

Your formula is correct, if that relieves you a bit. Its mean motion ([math]\sqrt{\frac{\mu}{a^3}}[/math]) multiplied by time passed since reference ([math]t-t_0[/math]) shifted by the reference mean anomaly [math]M_0[/math]

Thank you. Actually my textbook says [math]\sqrt{\frac{a^3}{\mu}}[/math]. It took me a day to realize it was a typo.

So maybe all that happened was that you forgot that the first term gives a result in radians, not degrees.

Thank you! That was what actually happened to me. Now I got the correct answer.

 

[martins]

Using degrees as angular units sometimes makes for easier human reading, but is useless for any trigonometric calculations. To avoid confusion, I'd recommend to

- convert all input angular data to radians (or angular velocities to radians/s, etc.)
- perform your calculations
- as an optional final flourish, convert results back to degrees for presentation

 

[Urwumpe]

Edit3: Sorry, the above was assuming that M0 had been converted to radians to start with. So maybe all that happened was that you forgot that the first term gives a result in radians, not degrees.

One reason more for getting a Matlab license at work instead of using Excel for this. :facepalm: