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To model the solar system, I took the planets to be point masses, used newtonian mechanics and modelled the orbits as circular (only Mercury's orbit has appreciable eccentricity). The entire system is only considered in 2D. Do all the planets lie on a plane?

How does this limit the model? I don't think it can deal with collisions, since the planets have no dimensions. And Newton's law shouldn't allow the centre of masses to occupy exactly the same point, should it?

I'm not sure when or if planetary motion would require use of relativistic mechanics - maybe the model couldn't simulate the motion of, for example, very fast moving comets because that requires relativity?

Essentially, what are some situations in the solar system that can't be modelled with point masses, circular orbits and newtonian mechanics? What inaccuracies could arise from these limitations?

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  • $\begingroup$ How are you projecting the planets motions exactly? If you are using Newtonian mechanics, the shapes of the orbits should come out as ellipses unless you set things up very carefully. So if you are constraining them to circular orbits, are you actually using Newton's laws to project their motion? $\endgroup$ – tmwilson26 Jan 7 '16 at 13:32
  • $\begingroup$ @tmwilson26 I think the constraint comes from using v = 2*pi*r/T to calculate the initial velocities. Would that force orbits to be circular? It certainly assumes they are. $\endgroup$ – user13948 Jan 7 '16 at 13:36
  • $\begingroup$ Yes, that assumes a constant speed which would force the orbits to be circular. The velocity of the planets changes slightly at different points in their orbits (see Kepler's Laws), and for Mercury, as you've pointed out, it has the most significant deviation because its orbit is the most eccentric (discounting Pluto). But Mars has an eccentricity that is about half of that of Mercury. A table can be found here: hyperphysics.phy-astr.gsu.edu/hbase/kepler.html $\endgroup$ – tmwilson26 Jan 7 '16 at 13:42
  • $\begingroup$ The planets are not quite all confined to a plane but they are close (the plane is called the ecliptic). Besides your circular orbit assumption, you must also be ignoring the interplanetary gravitational attraction. The real n-body gravitational problem is quite complicated. You might want to check this out: physics.stackexchange.com/questions/224800/… $\endgroup$ – Lewis Miller Jan 7 '16 at 14:18
  • $\begingroup$ @LewisMiller Surely that would render the model completely unstable. I am accounting for interplanetary attraction; in fact, the only reason my Sun's orbit works is because I added Jupiter (along with its significant gravity) to the model. At least... I think I am! $\endgroup$ – user13948 Jan 7 '16 at 15:55
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The relativistic corrections are tiny and in the early 20th century, only general relativity was found to matter, and it only matters for the precession of Mercury's perihelion.

But your words indicate that you are neglecting all deviations from circular 2D orbits predicted by Newton's theory.

First, the problem isn't 2D. It's 3D and the planes of the orbits don't coincide. Relatively to the Sun-Earth orbital plane (the ecliptic), the orbital inclinations (angles by which the planes are tilted relatively to each other)

https://en.wikipedia.org/wiki/Orbital_inclination

go from 0.77 degrees for Uranus to 7.01 degrees for Mercury. The latter is a pretty big angle. When the road is inclined by this angle, one may easily feel that it's not flat in the car. The nonzero inclinations are the reason why we don't get the lunar and solar eclipses every month (each of them), not to mention all the other "syzygies" (three celestial bodies are exactly in line).

Then you are neglecting the fact that even for the Newtonian 2D problem, the orbits are not circular but elliptic. Eccentrities – the relative numbers indicating how squeezed the ellipse is, or how far from a circle

http://www.astronomynotes.com/tables/tablesb.htm

are as high as 0.09 for Mars, 0.21 for Mercury, and 0.25 for dwarf planet Pluto. Even Mars' eccentricity 0.09 is so big that you would have trouble to find the planet in the telescope if you assumed the circular shape. And even Earth's 0.017 is enough to substantially contribute to the asymmetry of the seasons. Antarctica has seen cooler record low temperature than the Arctic partly because during the Southern winter, in July, we are further away from the Sun. (Another reason is the Arctic Ocean: the water reduces the overall magnitude of temperature fluctuations.)

Well before people believed that the Sun was at the center of the Solar System, they were highly aware of the non-circular shape of the orbits. They described them by the so-called epicycles (circles on top of circles). It's both simpler and more accurate to describe them as ellipses that obey the three Kepler's laws

https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion

Kepler's laws say that the orbits are elliptic; the Sun is in one of the focal points; the bigger semi-axis and the orbital period scale according to a power law; and the speed of the planet is variable but such that the "speed in terms of the area per unit time" is constant for each planet.

If you described the planetary orbits as inclined i.e. 3D elliptic orbits obeying Kepler's laws, you would be very close to the reality. The biggest omission is the interaction in between the planets themselves, especially Jupiter's action on other planets. The elliptic orbits assume that the Sun is the only object that acts on a given planet. Jupiter adds a significant correction which modifies the shape from exact and periodic ellipses to something slightly aperiodic and even more slightly different from ellipse even locally.

One must also realize that the center-of-mass of the Solar System isn't exactly at the center of the Sun. It's at the "barycenter" which is slightly off the center of the Sun but because the planets are light enough, it's still within the Sun's volume.

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  • $\begingroup$ I'm not neglecting interplanetary interactions. I don't think so, anyway! I'm adding the acceleration caused by each planet together to get a total that I'm using. So if I just changed my starting velocities (and positions) to model ellipses, I should have a pretty good model then? Also , the planets and sun orbit their common centre of mass. $\endgroup$ – user13948 Jan 7 '16 at 16:01
  • $\begingroup$ Hi, if you have at least 3 bodies with nonzero forces in between, none of the trajectories are ellipses let alone circles. The actual trajectories are more complicated - and in fact, impossible to write down using elementary (or not so elementary) functions. They're more general curved. If only the Sun is "big" for a planet, the trajectory will be similar to an ellipse but not quite an ellipse. If you had 2 mutually orbiting stars, the third object - a planet - could very well orbit them along a trajectory whose shape would look like the digit 8, too. $\endgroup$ – Luboš Motl Jan 8 '16 at 12:10
  • $\begingroup$ On one hand, you're saying that you "model" things - which sounds like simulations, a numerical calculation of the differential equations, including the planet-planet forces. But on the other hand, you say that the trajectories are ellipses if not circles. These two statements contradict each other. If you actually simulated the planetary motion, you would know that the trajectories are not ellipses or circles. $\endgroup$ – Luboš Motl Jan 8 '16 at 12:12
  • $\begingroup$ They might not be ellipses in reality. But I thought that was down to general relativity, which forbids perfectly elliptical orbits. Since I know nothing about GR, that is just something I read somewhere. And I am using only newtonian mechanics. And yes, I am doing a numerical calculation, using the velocity verlet algorithm. For comparison purposes I first forced circular orbits and then changed the initial conditions using actual values. This resulted in ellipses. $\endgroup$ – user13948 Jan 8 '16 at 16:24
  • $\begingroup$ Also, having had a quick look around on google, everything I can find says orbits are elliptical. Including Kepler's laws? So is it the relativistic effects that you mean when you say orbits aren't elliptical? $\endgroup$ – user13948 Jan 8 '16 at 16:26

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