Earth's elliptical orbit

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Why is the earth's orbit elliptical? That is, what force acts on the earth perpendicular to the force of the sun, that causes the tangential accelerations characteristic of an ellipse?
 

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  • #2
Garth
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Hi aranoff and welcome to these Forums!

The real question is: "Why is the Earth's orbit so nearly circular?"

A circle is the ideal limit, Limit e -> 0, of an ellipse. Keplerian motion, explained by Newton's laws, leads us to expect every freely orbiting body to orbit its parent body on an ellipse.

The fact that the Earth's orbit is so nearly circular, e = 0.017, speaks of the averaging of orbital elements of the millions of random collisions of the planetesimals that made up the Earth. The result was very nearly circular.

Proto-planets left with high ellipticity would also be ejected by close encounters with other planets. The eight planets plus dwarf planets left are the end result of such an accretion process.

Garth
 
  • #3
Phobos
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Elliptical orbits are the norm and circular orbits are rare. All the gravitational nudges from the other objects in the solar system can disrupt a perfectly circular orbit. However, the Earth's orbit is fairly circular (http://www.seds.org/billa/tnp/help.html#eccentric [Broken] of only 0.02).
 
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  • #4
D H
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Why is the earth's orbit elliptical? That is, what force acts on the earth perpendicular to the force of the sun, that causes the tangential accelerations characteristic of an ellipse?
Since nobody has answered your question directly, I will. There is no tangential acceleration (at least, not in the simple two body problem of Newtonian mechanics). No tangential acceleration is needed.

Central force motion (systems in which force is a function of distance to the origin only) does not mean the body subject to the central force is constrained to move in a circle. When the force is proportional to 1/r^2, the body is constrained to move along a conic section (circle, ellipse, parabola, or hyperbola).
 
  • #5
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earth's orbit

There is indeed a tangential acceleration. Acceleration is change in velocity. For a circular orbit, the change is perpendicular to the velocity, and parallel to the radius. For an elliptical orbit, the velocity changes in magnitude, and so the acceleration has a component perpendicular to the radius.

Where am I going wrong?

Thanks.
 
  • #6
D H
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There is indeed a tangential acceleration. Acceleration is change in velocity. For a circular orbit, the change is perpendicular to the velocity, and parallel to the radius. For an elliptical orbit, the velocity changes in magnitude, and so the acceleration has a component perpendicular to the radius.

Where am I going wrong?

Thanks.
The acceleration is still directed against the position vector for an elliptical orbit (or a parabolic or hyperbolic orbit, for that matter). What makes you think there has to be a tangential acceleration?

I think you are going wrong by assuming that the velocity vector is normal to the position vector. For a non-circular orbit, the velocity vector almost always has a non-zero radial component. The velocity is normal to the position vector at apofocus and perfocus only.
 
  • #7
D H
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It's fairly easy to show there is no tangential acceleration using cylindrical coordinates. The position of an orbital body with miniscule mass in cylindrical coordinates is

[tex]\vec r = r \hat r[/tex]

Differentiating with respect to time,

[tex]\dot{\vec r} = \dot r \hat r + r \frac d{dt}\hat r
= \dot r \hat r + r \dot{\theta} \hat{\theta}[/tex]

Differentiating once more to get the acceleration,

[tex]\ddot{\vec r} =
(\ddot r -r \dot{\theta}^2) \hat r +
(2\dot r \dot{\theta}+ r \ddot{\theta})\hat{\theta}[/tex]

The term [itex](2\dot r \dot{\theta}+ r \ddot{\theta})\hat{\theta}[/itex] is the tangential acceleration. The next few paragraphs show that this tangential acceleration is zero due to conservation of momentum.

The angular momentum with respect to the origin is

[tex]\vec h = \vec r \times \dot{\vec r} = r^2\dot {\theta} \hat z[/tex]

In any central force motion problem, the force on the body is, by definition, parallel to the position vector. There is no torque on the body since [itex]\vec \tau = \vec r \times \vec F = 0[/itex]. No external torques => constant angular momentum. Thus

[tex]\frac d{dt}(r^2\dot {\theta})
= r(2\dot r \dot {\theta} + r \ddot {\theta}) = 0[/tex]

and the tangential acceleration is zero.
 
  • #8
tony873004
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Earth's eccentricity varies from nearly circular to about 0.06 in cycles of about 400,000 years. Jupiter is the main cause of this. So in addition to the large acceleration vector that is always directed radially towards the sun, perhaps Jupiter supplies the tangental component you're looking for.
 
  • #9
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"Perhaps Jupiter..."

Has anyone done any calculations on the Jupiter effects?
 
  • #10
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earth's orbit

Can one explain why the orbit is an ellipse rather than a circle without invoking other planets?
 
  • #11
russ_watters
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It is true that Jupiter has an effect, but you don't need another object in order for an orbit to be elliptical.

What you are missing, aranoff, is that the speed is changing, but the direction is not always tangential, so the acceleration is always directly toward the center of mass.

An elliptical orbit is the norm - a circular orbit is a special (read: impossibly specific) case of an elliptical orbit.
 
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  • #12
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Let me phrase my comments in another form.

Consider the two-body problem of the earth and sun, each replaced by point masses. Define let the coordinate system so that the orbit is in the x-y plane. At t=0, let the earth be at its closest distance, and moving in the y direction. Solve the equations of motion, using the initial conditions. The solution is a circular path.

The reality is that the orbit of the earth is an ellipse, with an eccentricity ε, which is a function of time. Since the fact of ε(t) is a fact that does not occur in the initial equations, there is no way to find a solution with ε ≠ 0.

I believe that the eccentricity is a result of the many-body problem, taking into account the large planets. There is no analytical solution to the Newtonian gravitational many-body problem. I do not have the tools to solve this problem. I wonder if any of you know about many-body calculations that result in the correct ε(t)?
 
  • #13
russ_watters
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You've set up the problem in such a way as to demand a circular orbit, then asked why the orbit is circular. What if the moon was also moving in the X direction at t=0? What shape would you get then?
 
  • #14
D H
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Let me phrase my comments in another form.

Consider the two-body problem of the earth and sun, each replaced by point masses. Define let the coordinate system so that the orbit is in the x-y plane. At t=0, let the earth be at its closest distance, and moving in the y direction. Solve the equations of motion, using the initial conditions. The solution is a circular path.

The reality is that the orbit of the earth is an ellipse, with an eccentricity ?, which is a function of time. Since the fact of ?(t) is a fact that does not occur in the initial equations, there is no way to find a solution with ? ? 0.

I believe that the eccentricity is a result of the many-body problem, taking into account the large planets. There is no analytical solution to the Newtonian gravitational many-body problem. I do not have the tools to solve this problem. I wonder if any of you know about many-body calculations that result in the correct ?(t)?
This is incorrect.

I don't know what level of math or physics background you have. Solving the two body problem is covered in part in freshman-level physics classes and in full in upper-level undergraduate classes.

All you can say about a planet's orbit is that it is a conic section. A circular orbit is a very special case. Given the masses of the objects and the separation between them, there is only one velocity that results in circular motion. Any other velocity results in some other conic section.
 
  • #15
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This is a reply I received from Sverre Aarseth:

As a researcher in the N-body problem I welcome your query. You are right to pose the problem. Let me try to make some comments.

Suppose we take Earth at present solar distance and add just Jupiter as perturbing body. This is a stable and well behaved system. Assuming we start with circular Earth, the subsequent behaviour would reveal oscillations. I have not done this recently but would guess eccentricity would reach a maximum around 0.05 or so. Adding Saturn would introduce some irregularity but giving essentially similar result.

The actual situation is complicated by the origin of the Solar System. It is generally accepted that the terrestrial planets formed by collisions of smaller bodies. The last few major collisions would be crucial for determining the Earth's starting eccentricity. And it is even more complicated since we believe the orbit may have become more circular due to the presence of gas and accumulation of minor bodies.

I hope the above is helpful. Please don't hesitate to come back to me.

Best regards, Sverre
 
  • #16
Garth
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Which is what I said.

And I concur that the presence of gas, dust and minor bodies adds to the process.

Garth
 
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  • #17
D H
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Aranoff's difficulties are much more basic than describing how addtional bodies perturb the Earth's orbit. He doesn't understand how the two body problem can result in non-circular motion. All: Please keep the perturbations due to multiple bodies or relativity out of this.

Aranoff: That the two-body problem can result in non-circular motion was known even to Kepler, in an ad-hoc manner. Newton developed the mathematics that describes how the two-body problem results in a conic section.
 
  • #18
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No, I do not understand how a two body problem can give non-circular orbits, for then the result is more than the input.
It is not possible to get more out of a set of equations than the stuff that goes into the equations and initial conditions.
 
  • #19
Garth
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No, I do not understand how a two body problem can give non-circular orbits, for then the result is more than the input.
It is not possible to get more out of a set of equations than the stuff that goes into the equations and initial conditions.
It's all in the initial conditions.

Unless the orbital velocity is initially exactly that of the circular velocity and initially exactly normal to the radial direction for both bodies then the two bodies will be on elliptical or hyperbolic orbits.

It's basic orbital dynamics, it is the exactly circular orbit that is impossible to achieve.

Garth
 
  • #20
D H
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Aranoff, could you elaborate on why you think a non-circular orbit is impossible? (It is not impossible; as many have said, it is basic orbital mechanics. In fact, it is the circular orbit that is impossible to achieve.)

It would also help if you could let us know what level of mathematics and physics education you have received.
 
  • #21
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I think I got it. Consider my setup: A mass m is orbiting in the xy plane. At time t=0, the mass is at (x,0). It has a velocity v in the y direction. This is the given. Question: What will the orbit be?

For a certain velocity, the orbit will be a circle. Smaller velocities will not orbit. Larger velocities will be elliptical (or hyperbolic, for very large velocities).

Can one write the eccentricity as a function of velocity - velocity(circle), derived from the equation of motion?

My level: Ph.D. theoretical physics. I have been out of the field for a while, coming back to substitute in high school. My question arose from a student's question.

BTW, how do I write subscripts in these messages?
 
  • #22
D H
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I think I got it. Consider my setup: A mass m is orbiting in the xy plane. At time t=0, the mass is at (x,0). It has a velocity v in the y direction. This is the given. Question: What will the orbit be?
Suppose the mass of the orbital body [itex]m[/itex] is miniscule compared to the mass of the central body [itex]M[/itex] (i.e., [itex]m<<<M[/itex]). The circular orbit velocity is

[tex]\vec v_{\text{circ}} = \sqrt{\frac{GM}r}\hat y[/tex]

What happens for other velocities?
  • A parabolic orbit results if [itex]\vec v = \surd2\,\vec v_{\text{circ}}[/itex].
  • A hyperbolic orbit results if the velocity is greater than the parabolic orbital velocity.
  • An elliptical orbit results for all other velocities, including velocities lower than the circular velocity. If the velocity is larger than the circular velocity (but smaller than the parabolic velocity), the initial point is the perifocus of the orbit. If the velocity is smaller than the circular velocity, the initial point is the apofocus.


Can one write the eccentricity as a function of velocity - velocity(circle), derived from the equation of motion?
This is the specific energy of the orbit,

[tex]\frac{v^2}2 - \frac{GM}r = -1/2\left(\frac{GM}h\right)^2(1-e^2)[/tex]


My level: Ph.D. theoretical physics.
Did you sleep through your junior-level classical mechanics class?

BTW, how do I write subscripts in these messages?
Write your equations in LaTex, as [noparse][tex] LaTeX [/tex][/noparse], or use vBcode, [noparse] subscript [/noparse].
 
  • #23
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I found the answer.

http://www.braeunig.us/space/orbmech.htm

The eccentricity e of an orbit is given by

e = Rp x Vp2 / GM - 1,

where Rp is the periapsis, and Vp is the velocity.

This is the answer to the question I posed: Given the periapsis and the velocity at this point, find the eccentricity.

I cannot find my old mechanics book. I had to search on line. But thanks to you for your comments, which helped me direct my search.

-Sanford
 
  • #24
earth's orbit is elliptical because in December when earth is farthest from sun it is attracted towards galaxy for which Sun & entire solar system itsef has an orbit !
 
  • #25
ideasrule
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First, this thread was started in April 2007. The Earth has followed its elliptical orbit about the Sun almost three times since then. Second, Earth is farthest from the Sun in July, not December. Third, the fact that the solar system is attracted to the galaxy is irrelevant because the acceleration of gravity is independent of mass, so all objects in the solar system would have the same acceleration. Fourth--and this point has been discussed to death already by more than one poster--Newton's second law and the inverse square law of gravitation lead DIRECTLY to the equation of an ellipse.
 

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