Having just heard about the asteroid 2010 TK7 in Trojan asteroid seen in Earth's orbit by Wise telescope, I want to know more about its orbit. The BBC article says it moves above and below the ecliptic. I was under the impression that everything orbits in ellipses around the Sun, so what causes this motion?

Also the Wikipedia article about it states that it shuttles between Lagrangian points 3 and 4, but just above that that it orbits 60 degrees in front of Earth which seems contradictory? And that its path oscillates? What does that mean and what exactly is a Lagrangian point?

You can see animations in Earth Finds Dance Partner in a Trojan Asteroid of 2010 TK7.

It is circling the Sun because of the Sun's gravity, but why is it spiraling in its orbit? Does that mean there is an object in the center, around which it spirals? Does it disobey Newton's first law?

A body in motion will continue its motion unless it is opposed by another force.

Also, is there some 3D model or program that can demonstrate its orbit?

  • $\begingroup$ This is the page of one of the discoverers. There are diagrams and videos: astro.uwo.ca/~wiegert/2010TK7 $\endgroup$ – Daggerstab Jul 29 '11 at 17:26

Sorry for the earlier confusion.

The asteroid lies near the fourth Lagrange Point, and orbits around it.

It is still gravity that yanks it around of course, but now we have the combined gravity of the Earth and the Sun (with a modest contribution from the Moon as well).

This orbit may be stable for fairly long periods.

  • $\begingroup$ I understand how the planes are "pivoted" on the sun, but the BBC article seems to suggest a more "visous" movement, and wikipedia uses "oscliating", look at the green line in the BBC article I linked to. $\endgroup$ – Jonathan. Jul 28 '11 at 22:03
  • $\begingroup$ The Bad Astronomer has a good article on this that I think will explain the oscillation to your satisfaction. Check out blogs.discovermagazine.com/badastronomy/2011/07/27/… . I also couldn't find a term "visous" in the BBC article - I was going to suggest you meant "viscous." $\endgroup$ – Stuart Robbins Jul 28 '11 at 23:18

The gravitational interactions of numerous massive objects are too complex to solve exactly. There are simplified solutions where the gravitational pull of one object is ignored. Such solutions are reasonable approximations to the situation where one body is much more massive than the other, such as a star being orbited by a planet or asteroid. It is this model which produces elliptical orbits.

There is a 2nd level of simplified solution where the gravitational pull of two bodies are taken into account and the gravity of a 3rd object is ignored. This is a decent approximation for the case where the 3rd body is far less massive than either of the other two, such as an asteroid relative to a planet and its parent star. This simplification produces slightly more complex results than mere ellipses. The two more massive objects orbit the barycenter between each other, in many cases the barycenter will be within the larger object (especially in cases of stars orbited by planets), in cases of binary stars or binary planets the barycenter may lie between the bodies.

In the case of a star that is much more massive than a planet the planet's orbit approximates an ellipse, but the interactions of a 3rd body are more complicated. There is a volume of space where a 3rd body would orbit only the star, and a much smaller volume of space where a 3rd body would orbit the planet. There are also 5 other volumes where more interesting interactions occur, these are called Lagrangian Points. Along the axis of the planet and star there are 3 such points, one between (L1), one behind the planet (L2), and one opposite the planet and behind the star (L3). Objects at these points can stay at these points indefinitely. However, these points are only quasi-stable, small perturbations of objects at these points will tend to cause them to drift away from them, until eventually they're just in solar orbit. Spacecraft can stay at or near these points because they can perform small trajectory adjustments over time. The other two points lie along the planet's orbit, 60 degrees ahead (L4) or behind (L5) the planet. These points, volumes more precisely, are truly stable, objects near them will tend to stay in the same area over long periods (up to billions of years). Hundreds of thousands of asteroids exist in "lissajous" orbits around these "trojan" points near Jupiter and Saturn, even Mars has a handful.

2010 TK7 appears to exist in an orbit that transitions back and forth between the Earth-Sun L4 region and L3. Given that it may be that this asteroid has become captured around L4/L3 only relatively recently.


The "object" at the center is a gravitational well that the trojan is "spiraling" around. Think of it this way (in fact, you can do this yourself if you really want to): Go to the kitchen and take out a large bowl. Now take a marble (or a pea or a blueberry or something else small and round). If you place it on the side of the bowl, the round object is going to go towards the center, then back up the side, then towards the center, then back up the side, and so on. The only reason that it'll slow down and stop at the center eventually is because of friction.

That's basically what's happening with this Trojan except there isn't actually a bowl there, it's a gravitational sink caused by the interaction of Earth's and the Sun's gravity (called the L4 (Lagrangian) Point).

  • $\begingroup$ you mean blueberry in large bowl with still water, the blueberry will move towards the center and edge, back and forth? $\endgroup$ – physics1 Jul 30 '11 at 17:38
  • $\begingroup$ There is no "still water at the bottom". The L4 Lagrangian point is just like the bottom of a valley in a complex landscape: there is nothing of substance at the "bottom", other than the fact that at L4 (and L5) "the ground always goes up" in every direction. Mathematicians call this "a local minimum of the gravitational field". What matters to the asteroid orbit is not the "bottom", but the shape of the area (gravitational field) surrounding it: that area indeed looks like a bowl and a sufficiently slow object approaching the area would find itself going in circles around the "bottom". $\endgroup$ – Euro Micelli Aug 1 '11 at 21:42
  • $\begingroup$ But there's no mass at the langrangian point so why is there a gravity well? Also in the animation the asteroid does not stay a constant distance in front of earth, it gets further away and then appears to start moving back. Does this langrangian point also move further and closer to earth with time? $\endgroup$ – Jonathan. Aug 2 '11 at 13:08
  • $\begingroup$ The Lagrange points are not minima in the gravitational field they are minima in the in the effective field experienced by a body in a non-inertial reference frame (that is gravity (an actual force) + centrifugal "force" (an illusion attributable to using a frame of reference under acceleration)). And yes, we (that is physicist) harp on about "there is no such thing as centrifugal force", but these quasi-forces are well defined in the formalism of non-inertial frames of reference, and sometimes this is the most natural way to discuss things. $\endgroup$ – dmckee Aug 2 '11 at 23:24

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