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Question about two-body problem

  1. Jul 20, 2010 #1
    Good morning,

    First of all, a brief description of the two-body problem:
    The general solution of the two-body problem is:
    Consider a system of two bodies of mass m and M. There are no internal forces other than that of gravity, and there are no external forces acting on the system.
    The position vectors of each body are, respectively, [tex]\vec{r}_m[/tex] and [tex]\vec{r}_M[/tex], measured from an arbitrary inertial frame of reference.
    The vector [tex]\vec{r}=\vec{r}_m-\vec{r}_M[/tex] is the position of m with respect to M (a vector from M to m, pointing to m).
    The gravitational accelerations of m and M are, respectively:
    [tex]\ddot{\vec{r}}_m=-G\frac{M}{r^2}\hat{r}[/tex], where [tex]r=|\vec{r}|[/tex] and [tex]\hat{r}=\frac{\vec{r}}{r}[/tex] is a unit vector with the direction of [tex]\vec{r}[/tex], and the two dots above represent the second derivative with respect to time.
    [tex]\ddot{\vec{r}}_M=G\frac{m}{r^2}\hat{r}[/tex]
    Subtracting the latter from the former:
    [tex]\ddot{\vec{r}}_m-\ddot{\vec{r}}_M=-G\frac{M}{r^2}\hat{r}-G\frac{m}{r^2}\hat{r}[/tex]
    Therefore:
    [tex]\ddot{\vec{r}}=-G\frac{(M+m)}{r^2}\hat{r}[/tex],
    which is the equation of relative motion of m with respect to M.
    This leads to:
    [tex]r=\frac{p}{1+e\cos{\nu}}[/tex], which describes a conic section (for example, an ellipse), where r is the distance from body m to body M - which is at one focus -, p is the semi-latus rectum, e is the eccentricity and [tex]\nu[/tex] is the angle measured from the periapsis to the radius vector of body m.

    Now, my question:
    The equations above describe the trajectory of body m, but relative to body M (that is, a focus of the conic section). Here, M is the origin of this coordinate system.
    If I were to describe the trajectory of body m with respect to the center of mass of the system, it would make no difference if I used the equations above, since, generally, the central body has a mass M much greater than m, therefore the motion of M with respect to the center of mass is negligible.
    But what if m is comparable to M (and the center of mass is located outside of M)? In this case, the motion of M with respect to the center of mass wouldn't be negligible. How could I describe, then, the motion of m and M with respect to the center of mass? Can I still use the equation above, somehow?

    Thank you in advance.
     
  2. jcsd
  3. Jul 20, 2010 #2
    Both original equations of motion reference a relative position vector that is a function of the absolute position vector of BOTH bodies. In other words, you have a coupled system of differential equations, which must be integrated simultaneously to generate a solution for each body.
     
  4. Jul 20, 2010 #3

    Filip Larsen

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    Using the combined mass, G(M+m), is valid for a barycentric solution, even when the masses are comparable.
     
  5. Jul 20, 2010 #4
    Sure it is... and when you decompose the relative position vector into its constituents and substitute into the equations of motion given above, you'll find that the (M+m) factor remains. But you still have to integrate the coupled system to evolve the positions of both bodies in an inertial frame.
     
  6. Jul 20, 2010 #5

    Filip Larsen

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    I'm not really sure why you would say this. Having comparable masses is included in the normal solution of the two-body problem (as long as you use the sum of their masses as the mass).

    Given that you have the polar equation for a conic section, like the last equation mentioned in the original post, complete with the relationship between true and eccentric anomaly if you need position as function of time, you can easily go from the planet-to-planet radius vector, r, to the barycentric radii, r1 and r2, from the definition of the center of mass (r = r1+r2 and r1m1-r2m2 = 0) to get

    [tex]
    r_{1,2} = \frac{m_{2,1}}{m_1+m_2}r
    [/tex]

    At least, that is how I understand the original question.
     
  7. Jul 20, 2010 #6
    Thank you for the responses.

    That's what I thought.
    If, for example, I was running a computer simulation and wanted to keep the center of mass fixed, could I use the procedure below, for example?
    - Calculate the true anomaly as a function of time;
    - Calculate the radius of m with respect to M using the polar equation;
    - Calculate the coordinates of the center of mass with respect to M (which would be moving if M was fixed, since the frame of reference in M is non-inertial);
    - Move the whole system in order to keep the center of mass always in the same position;
    - Draw the two bodies.
    If I used this procedure, I would see the correct movement of the two bodies along the center of mass, wouldn't I?
     
    Last edited: Jul 20, 2010
  8. Jul 20, 2010 #7
    I made the explicit point because it appeared in the original post that the author had painted himself into a corner: by taking the (perfectly valid) step of combining the two equations of motion, he arrived at a solution for RELATIVE positions (one body with respect to another) when what he said he wanted were ABSOLUTE positions (both with respect to the inertial frame).

    Your suggestion is perfectly valid, and actually answers his question more precisely, as he had specified the ORIGIN of his coordinate frame to be the barycenter of the system. So long as you want an analytical solution to a two-body problem, you're done. But I thought it might be useful for him to take a step back and see his problem within the larger context, where a useful solution depends on the ability to recognize a coupled system as such and the elliptical two-body solution as merely a special case.

    Perhaps I muddied the water by attempting to clear up a confusion that never existed.
     
  9. Jul 20, 2010 #8
    Well, what's a simulation?

    When I run a simulation, I evolve a state vector from an initial condition through a progression of states according to a transformation expressed by a system of coupled differential equations. Simulations--particularly non-linear ones--are subject to an accumulation of integration error.

    When you have an analytic solution like the one above, you can generate data for a million years with NO accumulation of error.

    In a numerical simulation, I would calculate metrics like the distance between bodies as a function of my state values (position and velocity for each body). In an analytic solution, I can calculate all values directly as a function of time.

    The bottom line is that it really doesn't matter HOW you represent your system, so long as whatever scheme you use is isomorphic with all the others. The scheme you outlined above will work just fine... aside from mathematical validity, the only criteria that matter are computational efficiency and representational convenience.
     
  10. Jul 20, 2010 #9
    Thank you for the clarification.
    Very good, then, to know that this scheme is valid.
    My knowledge of calculus is very limited, and I'm not very familiar with differential equations. But it would be very interesting to know how I would proceed in a solution with differential equations like you said.
     
  11. Jul 20, 2010 #10

    Filip Larsen

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    Yes you would.

    Alternatively, since the barycentric radii are fixed ratios of the planet-centric radius, you can calculate the barycentric positions directly using two corrected semi-latus rectum values p1 and p2 and the two "barycentric polar angle" [itex]\nu[/itex] and [itex]\nu + \pi[/itex] (as the two masses will always be on opposite side of the CM), for instance,

    [tex]
    (x_1,y_1) = r_1 (\cos \nu, \sin \nu)
    [/tex]

    and

    [tex]
    (x_2,y_2) = r_2 (\cos(\nu+\pi), \sin(\nu+\pi)) = -r_2(\cos \nu, \sin \nu)
    [/tex]
     
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