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Two planets A and B move around the Sun in elliptic orbits with time periods $T_A$ and $T_B$ respectively. If the eccentricity of the orbit of B is ε and its distance of closest approach to the Sun is R, then the maximum possible distance between the planets is?


Attempt at a solution

I think i am almost there but this is what i have so far. Using the eccentricity and R, I found the expression for the semi major axis of B as $R_{SB}$=$\frac{R}{1-\epsilon}$ and using the relation for time periods, i have the expression for $R_{SA}$=$[\frac{T_A}{T_B}]^{2/3}$$\frac{R}{1-\epsilon}$ Now my doubt is whether to assume that two planets have sun at the same focus(say, on the right side ...picture concentric ellipses) in which case the maximum distance is when one of them is closest to the sun and the other is farthest OR if the two planets have sun at different focii (one to the left and other to the right, in this the orbits overlap partly; in either of the case, i cannot find a perfect expression for the maximum distance!


enter image description here

These are the two cases i am picturing, of course there might be many more oriented differently in the 2D. In the diagram on the left, max distance is when both are at their apogee. In the diagram on the left, its when one is at the apogee and the other is at the perigee. Is it possible to find an expression for distance which i can differentiate to find the max value?


Answer is $\frac{1+\epsilon}{1-\epsilon}$(1+($\frac{T_A}{T_B})^\frac{2}{3})$R

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  • $\begingroup$ Looking at your sketch, which option do you think gives the greatest possible separation? Isn't it obvious that the LHS gives the greater separation? Of course you can do it mathematically by defining the relative orientation $\theta$ of the orbits and the positions $\phi_1, \phi_2$ of each planet on each orbit, writing an expression for the square of the distance between the planets, and minimize this wrt $\theta, \phi_1, \phi_2$. But that is tedious, and not as convincing as a diagram with a geometrical argument. $\endgroup$ – sammy gerbil Oct 5 '16 at 13:19
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Make a sketch of the possible relative orientations of the 2 orbits. The planets will have maximum separation if the orbits are oriented with major axes aligned but set at $180^{\circ}$ - ie apogees on opposite sides of the focus.

enter image description here

In the diagram above, F is the common focus (the Sun), and the 2 planets can be anywhere on elliptical orbits A and B. While holding the larger orbit A fixed, vary the orientation of orbit B by rotating it about fixed point F. Planet B can then lie anywhere within circle C (shaded in pink), while A can lie anywhere on ellipse A. Without any need for a calculation you can see by eye ('by inspection') that the maximum separation is when the planets are at positions $a$ and $b$.

The distances $aF$ and $Fb$ are the apogees of orbits A and B. The apogee of B is $(1+\epsilon)R_{SB}$ where $R_{SB}=\frac{R}{1-\epsilon}$ is the semi-major axis of B. Let the eccentricity of orbit A be $\delta$. Then the apogee of A is $(1+\delta)R_{SA}$ where $R_{SA}$ is the semi-major axis of A. $R_{SA}$ is related to $R_{SB}$ via Kepler's Third Law : $(\frac{T_A}{T_B})^2=(\frac{R_{SA}}{R_{SB}})^3$. The maximum possible distance ab is therefore
$(1+\epsilon)R_{SB}+(1+\delta)R_{SA}=(1+\frac{1+\delta}{1+\epsilon}(\frac{T_A}{T_B})^{\frac23})\frac{1+\epsilon}{1-\epsilon}R$.

If we assume that $\delta=\epsilon$ then you get the expression in your answer key, viz. $(1+(\frac{T_A}{T_B})^{\frac23})\frac{1+\epsilon}{1-\epsilon}R$. However there is no requirement in the question to have $\delta=\epsilon$. The maximum possible value $\delta$ could take is 1, giving aF its maximum value of $2R_{SA}$. Then the maximum possible distance ab is $(1+\frac{2}{1+\epsilon}(\frac{T_A}{T_B})^{\frac23})\frac{1+\epsilon}{1-\epsilon}R$.

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  • $\begingroup$ Yes, thats true and i did mention trying the two cases pertaining to that orientation before the edit. But it presumes the orientation of the orbits themselves. What i want to find is the expression for maximum distance b/w the planets in which there is no indication of the orbit orientation. $\endgroup$ – Prasad Mani Oct 5 '16 at 6:08
  • $\begingroup$ Of course. I did know that. I just could not find the expression for that max distance using the above variables viz. semi major axis of A and perigee, apogee (and hence major axis of B). And in the question, the orientation of the two orbits are not given wrt each other. So assuming(correctly) for the case you mentioned, i would get one answer. But for the other case(my RHS), i would get another one $\endgroup$ – Prasad Mani Oct 5 '16 at 13:59
  • $\begingroup$ The question is asking for the maximum possible separation of the planets - ie for any orientation of the orbits. $\endgroup$ – sammy gerbil Oct 5 '16 at 14:07
  • $\begingroup$ Okay could you please tell me using the variables that i calculated the expressions for, what would be the max distance? Because i dont see it without the knowledge of eccentricity of A and its perigee and apogee $\endgroup$ – Prasad Mani Oct 5 '16 at 14:09
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    $\begingroup$ The answer given in the key assumes that orbit A has the same eccentricity as orbit B. This assumption is not required by the question as given by you. The eccentricity of A can have any value from 0 to 1. The maximum distance aF occurs when the eccentricity of A is 1. So I think my answer is correct. $\endgroup$ – sammy gerbil Oct 5 '16 at 17:17
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The center of mass is at the focus of each orbit. Assuming that the sun is far more massive than either of the planets, then the sun is at the focus for both orbits. But note that, for a given elliptical orbit, the shape of the orbit doesn't change if you switch the center of mass from one focus to the other (it's symmetric). Keep in mind that just because the Sun is at the focus for both, the orientation of the orbits is not necessarily aligned.

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  • $\begingroup$ Please check out the edit that i made, i have a feeling i wasnt clear in my description before $\endgroup$ – Prasad Mani Oct 5 '16 at 6:13
  • $\begingroup$ @PrasadMani I think you're making the problem more complicated than you need to. You don't need any differentiated optimization --- based on the diagram you added, you can already figure-out the "worst-case scenario" (maximum distances). Based just on the geometry of the problem, how can you express the maximum distance? (hint: first find the locations of each orbit at the maximum separation relative to the common focus). $\endgroup$ – DilithiumMatrix Oct 5 '16 at 14:40

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