# Mars' orbital period

The orbital period of Mars, is, as anyone can find at Wikipedia, $$T=686.98$$ d, and the semi-major axis of its orbit is $$a=2.2794\cdot10^{11}$$ m.

This gives $$T=2\pi\sqrt{\dfrac{a^3}{GM}}=686.84\text{d}$$. Of course, $$M$$ is the mass of the Sun.

I'm no physicist, but a mathematician. But I think that this discrepance (about 3h) should have been detected much before 20th century.

Just guessing: I think that the main cause of this discrepance is the high eccentricity of the orbit. Of course, Jupiter must have also much to say about that. I'm tried to find a more precise formula for the orbital period for elliptic orbits (in the context of classical physics). Is there any?

• I am sure it was detected by the 1800s. Orbital mechanics was very precise. Einstein famously explained a $43$ seconds of arc per century precession in the orbit of Mercury. That is really just the unexplained part of a 5000 arc sec/century precession. The rest is caused by things like perturbations from the gravitational attraction of Jupiter and other planets. There is no simple precise formula for the orbits. Jan 26 at 15:29
• @user2425 - Being a ‘computer’ used to be a human profession. Jan 26 at 15:40
• @user2425 - Gauss invented least squares fitting of noisy data for this case. Jan 26 at 15:54
• One problem right off the bat is that G is only known/accurate to 3 decimal places, this alone could account for a difference of 3h. Further, all of the masses in our solar system were back-calculated based on some value of G that has been adjusted over time. So if the solar mass that you are using was estimated using even a slightly different value of G than the one that you are currently using, then you can easily get discrepancies like this. I ran into this problem when I was writing my own orbital simulation and realized my parameter references had slight differences in them. Jan 27 at 18:59
• @RBarryYoung It's not quite that bad. The most recent (2017) CODATA value for G has a relative uncertainty of $2.2\times10^{-5}$. However, the OP is also using wikipedia values, which are oftentimes inconsistent with one another. Never (almost never) use $GM$ for solar system bodies; see my answer. Jan 28 at 23:09

For the "pure" two-body problem in classical mechanics, the period of the orbit is completely determined by the semi-major axis $$a$$, and is completely independent of the orbit's eccentricity. In this context, the "more precise formula for the period of elliptical orbits" that you are looking for simply doesn't exist; Kepler's Third Law is it.

The actual properties of planetary orbits differ from the "pure" Keplerian orbits largely due to the gravitational influence of other planets. I would expect that for Mars, most of the perturbations are due to Earth and Jupiter (Jupiter is obviously more massive but Earth is closer on average.) The actual calculation of these effects is a broad topic that I'm not as familiar with as I would like to be. I'll merely note that properties such as the eccentricity of the planet's orbit will generally influence the size and amount of perturbations it experiences; but they also depend on the size of the perturbing planet (i.e., Jupiter/Earth) and its orbital properties as well.

As a bonus fun fact, the orbit of Mars is now so precisely determined that we can infer the masses of some of the larger asteroids from their influences on Mars's orbit.

• Jupiter wins this battle, but you're right that they are more or less "comparable." Even at the minimum distance between earth and mars (0.5 AU) and the maximum distance between mars and Jupiter (6.7 AU), Jupiter exerts more force. 318 times the mass divided by (13.4 times the distance) squared is 1.8 times the force. If I use a more typical earth/mars distance of 1AU I get 7 times the force from jupiter. Jan 26 at 22:36
• @AXensen: I am not so sure. To change an orbit it is necessary to move a planet relative to the Sun. Earth when it is between Mars and Jupiter, pulls on both in opposite directions. But Jupiter pulls on both in the same direction and only the difference matters. Jan 29 at 1:18

At first, using (presumably) more accurate NASA fact sheet data about Mars and the Sun, namely semi-major mean Mars axis of $$1.52366231~\text{AU}$$ and Sun mass, this discrepancy reduces to about an hour :

$$\tag 1 \left|2\pi × \sqrt{\frac{(1.52366231~AU)^3}{G × 1988500 × 10^{24} kg}} − 686.980~\text{days}\right| \approx 51~\text{min}$$

Second, this Kepler period formula is for modelling orbital period for two point masses circling around COM. Our and presumably most part of planetary systems, are composed from many bodies, $$N \gt 2$$, so for modelling periods of such planets you need to solve N-body problem which in very special cases have some specific solutions, but in general is unsolvable yet (get a Nobel prize if you will) and so can be attacked only with numerical methods.

Third. Planetary orbits are not carved in a stone and they may change in the long run, like the paper Existence of collisional trajectories of Mercury, Mars and Venus with the Earth describes. Basic idea is that they have made numerical simulations of our Solar system $$5~Gyr$$ into the future and have found that in a set of $$\approx 2500$$ initial orbits $$1\%$$ of solutions have produced instabilities which changed Solar system in a mandatory ways. Hence, stability is just a nice dream. Planetary systems seems now more like "quantum wave-functions" if you let me to say so. Sure, probability of some orbit "eigenvalue" is high enough to call it "relatively stable", but never forget that it's not $$100\%$$ and never will be.

EDIT

To illustrate chaotical part of our solar system, I will include a nice chart from this article which shows how solar system barycenter moves over the course of $$100~\text{years}$$ relative to the Sun (I just re-colored different trajectory patches with different colors for better following with eyes trajectory changes, otherwise hardly seen):

HTH !

• Sun looses about 0.0000000000000000001% total mass per second due to radiation, so it's a very tiny effect, not much into play of planet orbits evolution. Most part of planet orbits changing over time is just because of many planets gravity inter-play. Just Google for N-body problem and you will get many nice charts. So after gazillion of years some orbits may shrink, some - to expand, some planets may escape and be free, some will change orbital inclination angle and etc. Because it's not static. Jan 26 at 18:12
• So, Jupiter could weaponize Mercury against the inner solar system? Jan 29 at 1:25
• Except for very small solar system bodies, it's best not to use $G*M$ for such bodies. Instead use the observable product, typically denoted by $\mu$. (The mass is calculated via $M=\mu/G$). For the Sun, the solar gravitational mass parameter is known to twelve or so places of precision. Using $GM$ throws that precision under the bus. Jan 29 at 12:58
• One source of your discrepancy is the value of 1.52366231 au that you used for Mars' semi-major axis length. NASA GSFC's planetary fact sheets aren't as chock full of errors as is Wikipedia, but that's not saying much. They have a good number of factual errors; this is one of them. I used 1.52371034 au, which comes from the place (JPL) that calculates planetary ephemerides for all of NASA. Jan 29 at 15:30
• Another error is using 686.980 days as the length of a Martian year. Never trust numbers from Wikipedia. That duration is the length of a Mars sidereal year, per NASA planetary fact sheets. It would be far better to use the Mars anomalistic year, which is 686.9957 days (or maybe 686.9956 days) per multiple sites. That knocks the discrepancy down to 4.5 minutes, which essentially means there is no discrepancy. See my answer for details. Jan 29 at 15:32

Just guessing: I think that the main cause of this discrepance is the high eccentricity of the orbit.

That's a bad guess. The source of your error is using $$G$$ and $$M_\text{Sun}$$. Do not do that for solar system bodies. For solar system bodies, especially the Sun, the Earth, and any other bodies to which humanity has sent vehicles pr has observed objects orbiting the body in question, it is far, far better to use the gravitational mass parameter $$\mu_{\text{body}}\equiv G M_{\text{body}}$$. The gravitational mass parameter is observable. The mass itself is not. For the Sun, the uncertainty in $$\mu$$ is in the twelfth decimal place. You have ruined that very high precision by using low precision and inconsistent values of $$G$$ and $$M_{\text{Sun}}$$. A much better estimate is $$T_{\text{Mars}} = 2\pi\sqrt{\frac{{a_{\text{Mars}}}^3}{\mu_{\text{Sun}}}} = 2\pi\sqrt{\frac{(2.2794382\text e11\,\text{m})^3}{1.32712440041279419\text e20\,\text m^3/\text s^2}} = 686.9926 \operatorname{days}$$ Sources of the numbers:

This result differs by about 18 minutes from the wikipedia value of 686.980 days for a Martian year. This however raises a key question: What is a year?

So what is a "year"?
Digging into the source of the quoted wikipedia value, that is the length of the Martian sidereal year. This is how long it takes Mars to orbit with respect to the "fixed" stars. There are other metrics that qualify as a "year". A tropical year is how long it takes for a planet to repeat its seasons. A Mars tropical year is 686.9725 days long, per Mars24 Sunclock — Time on Mars. Yet another "year" is the anomalistic year, which is the length of time from one perihelion passage to the next. For Mars, this is 686.9957 days from the same site as the tropical year. (Other sources provide a value of 686.9956 days.) The sidereal, tropical, and anomalistic years differ from one another due to axial and apsidal precession.

The OP is using Kepler's laws, which pertain to how the mean anomaly progresses over time. Mean anomaly repeats after one anomalistic year. The correct "year" to use with regard to using Kepler's laws is the anomalistic year. There's a 0.0031 days (4.5 minutes) difference between the value calculated above and the posted values for the anomalistic year.

There is no discrepancy.

• Sure you greatly clarified things, down to minor difference. BUT... Does $4.5~minutes$ falls strictly into the orbital period, sun mass, other data,- measurement uncertainty ? Can it be proved ? (I somehow believe that $\text{few minutes}$ discrepancy range is the zone out of reach for Kepler laws and kicked-in relativity and/or N-body numerical solution could explain this tiny dissimilarity. No ? Jan 29 at 15:53
• @AgniusVasiliauskas As far as I'm concerned, that 4.5 minute discrepancy is in the noise for an approximate calculation, which is what Kepler's laws are. For precise values, one should use an ephemeris such as NASA JPL's DE440 or DE441. Those Development Ephemerides do not use Kepler's laws. Jan 29 at 16:22

Kepler's third law states that the period is only dependent on the semi-major axis, not the eccentricity (not obvious at all!) I am pretty certain the general relativity corrections are small out there.

As for accurate orbits, the standard Keplerian elliptical orbit is known to be accurate to a few hours, giving five significant figures. The Nasa JPL DE405 ephemerides do a bit of extra fitting that get kilometer resolution and hence presumably an error of a few minutes.

• DE405 was released in 1998. We've advanced a bit since then. ;) See The JPL Planetary and Lunar Ephemerides DE440 and DE441, by Park et al (2021) doi.org/10.3847/1538-3881/abd414 Jan 27 at 0:43
• @PM2Ring Regarding We've advanced a bit since then -- just a little bit. NASA JPL regularly updates their Development Ephemerides to better support planned missions. There have been 16 missions to Mars since the release of DE405 (I'm excluding some failed missions) -- and several concomitant updates to the Development Ephemerides. Jan 29 at 16:01
• @David Indeed! And it's no coincidence that Ryan Park is not only involved in developing the DEs, he was also a principle investigator on the Dawn mission to Ceres, and the Juno mission, which is currently gathering excellent data on the gravity in the Jovian system. Jan 29 at 16:15
• The other planets (mostly Jupiter and the Earth) plus general relativity cause Mars to undergo an apsidal precession of about 1628 arc seconds per century. The GR contribution to this number is 1.4 arc seconds per century -- small (very small), but it still exists. Jan 29 at 16:36

I have read Kepler's orbit is essentially the shape and orbital period of an two-body orbit with a precession of the perihelion caused by perturbative effects of bodies outside the two-body system. You could use numerical methods. I've never used it in a gravity problem, but I'm thinking notions of perturbation theory might work.

$$\frac{1}{r}=A\cos \theta + B \sin \theta+C$$ satisfies Newton's gravity equation for the two body problem.

$$\frac{1}{r}=A\cos \theta + B \sin \theta+C+f(\theta)$$

Plug that into Newton's Force equation for gravity and you end up with an differential equation in terms of $$f(\theta)$$ that probably isn't pretty. Still, the resulting problem maybe be more tractable than a straight numerical approach.