This might be a completely wrong question, but this is bothering me since many days ago. Given the mass (Sun) curves the space around it, gravitation is the result of such curved space (Correct me if I am wrong, source: A Documentary film). Given any point on a circle with center same as center of the mass, curvature in the space should be equal (Intuition).

Planets rotate around the Sun because of the curve in the space they should follow a circular path and the distance between planet and Sun should be at a distance.

Given the fact that earth has a elliptical orbit around the sun, and the distance between Earth and Sun varies according to position of the earth. Why do we have a elliptical orbit not a circular orbit.

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    $\begingroup$ Related: physics.stackexchange.com/q/25110 $\endgroup$ – HDE 226868 Mar 8 '15 at 15:01
  • $\begingroup$ It is almost perfectly circular. The deviation is about 3%. When it starts getting closer to the Sun it also accelerates that in turn gives it speed to eventualy start distancing it self from the Sun. But the gravity pull from the Sun slows it down enough as it tries to escape so it falls back and does not flay off. $\endgroup$ – Ruskes Jun 29 '15 at 8:19

Because orbits are general conic sections. Why this is true is another fascinating question in and of itself, but for now I'll just assume it. The point is that circular orbits are special examples of general orbits. It's perfectly possible to get a circular orbit, but the relationship between the bodies' velocities and separation needs to be exactly right. In practice it rarely is, unless we plan it that way (e.g, for satellites).

If you threw a planet around the sun really hard its path would be bent by the sun's gravity, but it would still eventually fly off at a tangent. Throwing it really hard would make it almost go straight, since it moves by the sun so quickly. As you reduce the speed, the sun gets to bend it more and more, and so the tangent is flies off on gets angled more and more towards moving backwards. So general hyperbolas are possible orbits. If you move it at the right speed, then it'll be just slow enough that other tangent points 'exactly backwards', and here the motion will be a parabola. Less than this and the planet will be captured. It doesn't have enough energy at this point to escape at all.

A key realization here is that the path should change continuously with the initial speed. Imagine the whole path traced out by a planet with a high velocity. An almost-straight hyperbola, say. Now as you continuously lower the velocity, the hyperbola bends more and more (continuously) until it bends "all the way around" and becomes a parabola. After this point, you'll have captured orbits. But they have to be steady changes from the parabola. All captured orbits magically being circles (of what size anyway, since they have to start looking like parabolas at some point?) wouldn't make any sense. Instead you get ellipses that get shorter and shorter as you get slower. Keep doing this, and those ellipses will come to a circle at some critical speed.

So circular orbits are possible, they're just not general. In fact, I'd say the real question is why the orbits are often so close to circular, since there are so many other options!

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    $\begingroup$ As for that last remark: in fact, only the objects in the inner Solar system have near-circular orbits, due to frictional losses in the primordial disk, and more importantly, transfer of angular momentum from the inner to the outer parts of the disk. A large number of objects in the Solar system (if not most) actually have highly elliptical orbits. $\endgroup$ – Rody Oldenhuis Jul 4 '13 at 15:29
  • $\begingroup$ But weren't the planets created from the sun itself? Where's the question of throwing the planets into the sun's gravitational field, then? $\endgroup$ – mikhailcazi Jul 4 '13 at 16:19
  • $\begingroup$ @mikhailcazi Throwing them "away", or "at an angle" works just as well; they just have to be moving through its gravitational field somehow. The only reason I used "throwing in" instead of the historically accurate option is because I don't like looking at half of a parabola. $\endgroup$ – Robert Mastragostino Jul 4 '13 at 18:43
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    $\begingroup$ @mikhailcazi - "But weren't the planets created from the sun itself?" Nope. The current best theory is that the planets were formed from the same protocloud as the sun did, but they were formed by (relatively speaking) the outer portions. So they formed at the same time as the sun, not from the sun. See here hubblesite.org/hubble_discoveries/discovering_planets_beyond/… for a quicky read. $\endgroup$ – WhatRoughBeast Jun 2 '15 at 22:37
  • $\begingroup$ The orbits cannot be more elliptical, because in the formation of the planets from nebula, there is rotation in the nebula clouds, and bodies having random elliptical paths will collide, so thats why, they are so close to circular. $\endgroup$ – Adesh Tamrakar Jan 23 '17 at 18:49

This is because nature is not often as perfect as we tend to imagine it to be :)

In reality, when close to the Sun, the Earth has a little "too much speed" for it to stay that deep in the Sun's gravity well. In other words, the local spacetime curvature induced by the Sun is not strong enough to keep the Earth as close to it as it is, given that it also moves sideways; the Earth will start to "climb out" of the gravity well.

Building on your intuitions from everyday life, you know that if you move fast at the bottom of a hill, you'll be able to go up the hill, at the expense of your speed. You'll go higher and higher, but also slower and slower. The same holds in the context of the Earth. The further out of the gravity well it climbs, the slower it will move. At a certain point it will have climbed so far out that it actually moves too slow to stay as far out of the well as it is; it will start falling back in.

If you consider also the Earth's sideways motion and assume the Earth's motion is exactly perpendicular to the line of sight between the Earth and the Sun at the point where it moves slowest, it is not hard to imagine that the point where the Earth has its lowest speed is on the exact opposite side of the Sun as where it will have its highest speed. From that, it is not hard to imagine that the net result of all this will be an elliptic orbit, rather than a perfectly circular one.

Now, this does not mean it is not possible to have circular orbits. Of course, if the Earth would have had the exact amount of energy required to stay on the exact same distance from the Sun at all times, the orbit would have been circular. But this is nice in theory, but practically impossible to realize given that even the slightest perturbation from Jupiter or asymmetry in the Sun or whatever would cause the Earth's orbit to start deviating from a circle. Practically speaking, a circular orbit is a limit case that you can get arbitrarily close to, but never quite reach.

I think this video will help. Look at the motion of the balls in the cone; virtually the same principles apply for this case.

For completeness: there are many details I have omitted here that influence the exact shape of the orbit -- the "hill" analogy (or bowling-ball-on-a-rubber-sheet analogy, as it is often depicted in movies or documentaries) is an inaccurate and incomplete representation of the true nature of spacetime. Also, the other planets, remote galaxies, in-homogeneity of the Sun's gravity field, etc. distort the Earth's local spacetime. But all these effects are relatively small, and can safely be ignored to understand the ellipticity of the orbit.

  • $\begingroup$ Your answer is partial. Please explain what happens when the Earth is moving in to the Elliptical part, and why it turns back. Does it looses speed ? If so why ? $\endgroup$ – Ruskes Jun 29 '15 at 7:56

Note: This answer was written for the original version of the question, in which the OP asked about spherical orbits (in which the plane of the orbit may change due to angular-momentum exchange), rather than circular/eccentric ones (in which the plane of the orbit stays constant in time). This answer is not relevant for the question as it currently stands, but I'll leave it here anyway, in case it's ever useful.

The technical reason is conservation of angular momentum; if the Earth were to move out of its orbital plane, that would change the direction (though not necessarily magnitude) of the angular momentum vector.

A less technical way of thinking about it is just with Newton's second law: the acceleration of an object is proportional to the force on that object. In particular, the acceleration and the force are in the same direction.

Let's say that the direction the Earth is traveling is "forward" and the Sun is on the "left". Then, to move in a spherical orbit rather than a circular or elliptic orbit, the Earth would have to move "up" or "down". But since we just defined "forward" as the direction it is currently going, that means Earth would have to accelerate "up" or "down". And Newton's second law tells us that would require a force in the up or down direction. The Sun's gravity doesn't give us a force in this direction; it only pulls to the "left". It's possible for the Sun to be pulling too hard or too soft to keep the Earth going in a circle, so Earth can go a little to the left or right (and in fact it does go left and right in an ellipse), but never up or down. So the Earth will continue gliding along in its orbital plane, rather than a more general spherical orbit.

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    $\begingroup$ I think OP meant to ask why the orbit isn't circular instead of elliptical, but they used the wrong terminology in their last sentence. So while your answer is correct, I don't think it gets at what the OP is asking. Most likely their confusion stems from their idea of what "curvature of spacetime" means (probably inspired by popular analogies). I would mention that this curvature is in correspondence with the Newtonian laws of gravity, from which elliptical orbits naturally follow (in combination with conservation of energy & angular momentum). $\endgroup$ – Wouter Jul 4 '13 at 13:26
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    $\begingroup$ @Mike I've made a mistake in my question, I originally meant to be circular orbit and not spherical orbit. $\endgroup$ – aravind ramesh Jul 4 '13 at 13:37
  • $\begingroup$ Ah, I see. Sorry. I work with orbits of black holes, where we do actually talk about (quasi-)spherical orbits, because there is exchange of angular momentum. I'll edit the answer to note that it is not relevant after your correction. $\endgroup$ – Mike Jul 5 '13 at 0:26
  • $\begingroup$ Makes sense when on the closer part to the Sun, the continuous acceleration of falling in to the Sun would keep it going straight. But that does not explain when the Earth is at the maximum distance (on the ellipse) from the Sun). Does the Earth then slows down so it gets pulled back. $\endgroup$ – Ruskes Jun 29 '15 at 7:53

Maybe it's best to forget space curvature as described in general relativity; while that theory implies major changes to the foundations of celestial mechanics, it is set up in a way that reproduces Newtonian mechanics when gravitational fields are of moderate strength, and leads to only very tiny adjustments of the actual orbit calculations.

Movement of an object in a fixed gravitational field (which is a realistic assumption for the movement of a satellite around a much more massive body, whose own perturbation can then be ignored) is described by a second order differential equation: at each point in its orbit the acceleration of the satellite is given through Newton's law by the gravitational field at that point. Calling them Sun and planet for ease of description, the orbit of the planet will then be determined by its position and velocity at some (arbitrary) starting time. It is easy to see that movement will remain in the plane determined by the Sun and the initial planet position, and containing its initial velocity vector. Thus each state is essentially given by two position coordinates and two velocity coordinates. Rotational symmetry allows us to forget about one degree of freedom, and describe the initial conditions by the radial position (initial distance of the planet from the Sun), and the tangential and radial components of the velocity. Instead of the tangential velocity component one can also give the instantaneous angular velocity, obtained from it by dividing by the radius.

Since a circular orbit is characterised by the fact that the radial component of the velocity is always zero, having no radial component of the initial velocity is a necessary condition for getting a circular orbit. This already answers the question why orbits are not always circular. However, the condition is not sufficient: in a circular orbit the angular velocity must be in a precise relation to the radius in order to ensure that the the gravitational acceleration, which depends only on the radius, matches the centripetal acceleration for a circular orbit, which also depends on the angular velocity. This explains why satellites just above the Earth's atmosphere (which must be given circular orbits for practical reasons) all have the same rotation period (about 1.5 hours). This also explains why non-circular orbits, which do have some point where the radial component of the velocity becomes zero, do not become circular from that point on: their angular velocity is either too low or too high at that point.

When the angular velocity is too high, the planet will "spin out" and its radius starts increasing (possibly going off to infinity if the velocity exceeds the escape velocity at that point, though one would not speak of a planet in this case). If it is too low, it will "drop out" of a circular trajectory and start falling. In the latter case its angular velocity will increase (since its product with the radius squared must be constant, as the constant angular momentum is proportional to it), giving in the rotating frame an increasing centrifugal force that at some point overtakes the gravitational force (even though the latter also increases) at which point the negative radial velocity is maximal and starts decreasing. When the radial velocity become zero again, the planet has reached it closest point to the Sun. But the angular velocity is now too large for a circular orbit, even at the smaller radius attained; the planet is now in a "spin out" situation described above. Events then repeat in reverse order, until a point of maximal radius is again attained. What is remarkable and can only be explained by a detailed computation (which I will not give; I'm not sure I even could), is that this point is the same point in space as the starting point: the curve closes and the planet orbit turns out to be an ellipse.


For systems involving central forces, the orbit can be any of the conic sections. And which conic it will be is determined by the total energy of objects revolving around the centre of force. If the total energy is negative ( as in the case of planets), the eccentricity of the orbit will be either zero or one. If eccentricity is equal to zero, it will be a circle and if its less than one, its an ellipse. And for our planetary system, the calculated eccentricity is found to be less than one.


Comets are coming from almost infinite distance towards Sun. Due to Sun's Gravitational force they should be pulled into Sun rather than forming Elliptical or Hyperbolic trajectory.

If object is starting travel from closer distance at considerable speed, we can visualize formation of Elliptical orbit. For object coming from infinite distance getting pulled by Sun's Gravity, should end up falling into Sun

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    $\begingroup$ Over long enough timelines, maybe they'll get pulled into the sun. But as it stands, your answer doesn't really answer the question (because you're talking about comets and not planets). $\endgroup$ – Kyle Kanos Mar 8 '15 at 14:57
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    $\begingroup$ Almost infinite distance. Not even close! Due to Sun's Gravitational force they should be pulled into Sun rather than forming Elliptical or Hyperbolic trajectory. No, for the same reason that the planet's don't crash into the Sun. $\endgroup$ – HDE 226868 Mar 8 '15 at 15:00

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