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The Loopy Lunatic (Posted on 2021-05-20) Difficulty: 4 of 5
It comes as a surprise to many that the orbit of the moon around the sun has no loops in it. Indeed, it is a convex curve not very different from the orbit of Earth around the sun.

How far away from Earth would our moon have to be for the moon's orbit around the sun to have a loop? How far away for it to be nonconvex?

Assume all orbits are circular and all lie in the same plane (so that "loop" and "convex" have clear planar meanings), the Earth-sun distance is 93 million miles, the Earth's orbit requires 365 days, and the moon's orbit around Earth takes 27 days (and that is constant in this problem). Using such approximations has negligible impact on the problem.

Note that the moon's orbit is "prograde": in the same direction as Earth moves around the sun. Both motions are counterclockwise, viewed from our north pole.

No Solution Yet Submitted by Danish Ahmed Khan    
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soln | Comment 2 of 8 |
I was preparing this problem myself for Perplexus, so I really laughed when I found it posted by DAK. 

The problem is from Stan Wagon's "Loopy Lunatic Problem 1250", with references to: Noah Samuel Brannen in College Mathematics Journal 32 (2001) p 268 and Laurent Hodges in College Mathematics Journal 33 (2002) p 169.

The beauty of this problem is that it is so counterintuitive that the Moon does not make loops. After all, even with Earth and the Moon both orbiting in the same counterclockwise sense, the Moon must pass between the Sun and Earth going in the wrong direction every month. The explanation is that the Earth-Moon system is itself traveling so quickly that the Moon is still moving forward, much like a person walking backwards on a speeding train is still moving forward. The Moon's orbit, it turns out, is almost perfectly circular, in the same sense that Earth's elliptical orbit is almost circular.

The problem is most easily worked in length units where the radius of the Moon's orbit is 1, the radius of the Earth's orbit is d, and we say that the Moon makes p revolutions while the Earth makes one. The solution is below, with the bottom line that: we get loops when d < p, a wiggly (not fully convex) path when p < d < p^2 and a convex path if p^2 < d. The convex path resembles a very smoothed 13 sided polygon.

Here, p = 365/27 ~ 13.4. So, for example, for full convexity, the Earth-Moon distance has to only be less than 93 Million miles/13.4^2 =0.52 M miles, and it is well within this at an actual distance of 0.24 M miles. Loops occur when the Moon orbits further than 93 Million miles/13.4=6.9 M miles from Earth, yet somehow zipping around in 27 days in utter defiance of Kepler's 3rd Law. (This distance, BTW, matches the ~6.9 M miles that tomarken gives!). Examples of the wiggly moon and loopy moon are plotted here

The lunar motion in the frame with the Sun at the origin is given by the parametric equations: 

x(t) = d cos(t) + cos(pt), 
y(t) = d sin(t) + sin(pt), 
with t being the angle theta. 

The definition of convexity for a closed figure is its maintenance of a continually positive signed curvature "k", where k is the inverse of the radius of curvature: 

k = (x'y'' - y'x'')/(x'^2 + y'^2)^(3/2)

Since the denominator is positive, we can ignore it and look for 
curvature as indicated by the sign of the numerator:

x'y''-y'x''

= (-d sin t - p sin pt) (-d sin t - p^2 sin pt) 
 - (d cos t + p cos pt) (-d cos t - p^2 cos pt) 

= d^2 (sin^2 t + cos^2 t) + dp^2(cos t cos pt + sin t sin pt) 
+ dp (cos t cos pt + sin t sin pt)

= p^3 + d^2 + dp(p+1) cos[(p-1)t]. 
where we have used cos[(p-1)t] = cos pt cos t + sin pt sin t

The cos term above oscillates between -1 and 1 as the Moon goes round. So, the curvature of the Moon's orbit stays positive so long as p^3 + d^2 > d p(p+1), and otherwise, it can oscillate with theta.  

This last inequality can be rearranged as: 

(d-p)(d-p^2) > 0. 

If either of these terms in parentheses are negative, the curvature will be at times negative.
 
Deciding which case obtains: convex, wiggly concave and convex, or worse, loops, can be sorted out using velocity vector crossproducts, which I defer to either to a further post or simulation. But the last equation shows it depends on how large d is compared to p or p^2. The orbit is:

fully convex when d > p^2
wiggly (convex & concave) when p < d < p^2 (the cos term matters)
loopy when d < p

While the Moon is well in the first category, not everything is this way within the solar system.  Jupiter's Callisto is not fully convex with d=727 and p=260, and Io does in fact make loops with d=1846 and p=2449.  

I was idly wondering how massive Earth would need to be to support a Moon going round in 27 days at a distance of 6.8 Million miles. 
Moving the Moon (or any mass) from its 0.24 M mile current distance to this distance 28.3 times further would require accompanying 28.3^3 increase in Earth's mass to hold it in orbit via Kepler's 3rd Law, 'a la Newton. At its current radius, this new mass would not make Earth a black hole, just wildly dense.  


Edited on May 22, 2021, 6:42 pm
  Posted by Steven Lord on 2021-05-20 09:42:21

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