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Such questions have been asked here and I have read the answers some of which are quite informative. However, I want an answer that's straightforward.

I have gone through the following paper by A B Turner titled The Moon's Orbit Around The Sun.

In this, the author concludes that the curvature of the Moon's orbit around the Sun in the new moon positions, $k_n$ is approximately 1.5 times of the Earth's orbit around the Sun, which is approximately the same as that of the Moon's orbit around the Sun in the full moon positions, $k_f$.

enter image description here

$k_n$$\approx 1.5k_f$, which is intuitive considering the fact that in the new moon positions, the moon experiences gravity from both the Sun and the Earth in almost the same direction.

My question is, then, how does the Moon's orbit manage to not have any concavity, no matter how minute, while transitioning from the full moon to the new moon positions? Isn't that not a mathematical impossibility?

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  • $\begingroup$ This is a good question and a curious paper! I wonder what originally motivated the author to write it, maybe a conversation with a student or a colleague in a hallway. I've posted an answer, let me know if it's satisfactory or if there are lingering issues. Thanks! $\endgroup$ – uhoh Mar 4 at 0:26
  • $\begingroup$ Devil's advocate question: As the Moon's orbit about the Sun is not a plane curve, what exactly do you mean by "concavity" (or its converse, "convexity")? A perhaps overly simple answer is to ignore that the Moon's orbit about the Earth is inclined (by a small amount, ~5°) with respect to the Earth's orbit about the Sun, but that ignores that the Earth's orbit about the Sun also is not a plane curve due to the presence of the other planets. So let's ignore that, too. $\endgroup$ – David Hammen Mar 4 at 8:40
  • $\begingroup$ If you ignore those details, then @MikeG's answer is correct, but he does need to show that this is the case. $\endgroup$ – David Hammen Mar 4 at 8:46
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    $\begingroup$ Highly related, if not a duplicate: Moon's orbit around the Sun. $\endgroup$ – David Hammen Mar 6 at 14:46
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Turner's expression for the radius of curvature ρ is correct, but in case (3), ρ = -0.988 a where the Moon is in the first or last quarter (θ = 7.5° or 22.5°). Where the Moon is full (θ = 15°), ρ should be -0.749 a.

Case (1) models a spacecraft stationed at L1 (b = a / 100, ρ = -0.99 a) or L2 (b = -a / 100, ρ = -1.01 a), roughly 4 times the lunar distance from the Earth.

The case (2) example b value is too large for a stable orbit around the Earth. Instead we can model a geosynchronous satellite with n = 366 and b = a / 3550. Then ρ is +0.0218 a on the day side and -0.0314 a on the night side. This trajectory is alternately concave toward and away from the Sun, resembling Turner's figure 1 but with shorter waves.

Between cases (2) and (3) we can find parameters such that, on the sunny side of the Earth, the ρ expression denominator approaches zero and the trajectory is momentarily straight. This is nearly the case if n = 24 and b = a / 575, corresponding to a 15.2 day orbit at 0.677 lunar distance.

Any curvature in an object's trajectory is due to acceleration in the direction of the net force acting on it. We know from Newton that $$ F = \frac{G m_1 m_2}{r^2} $$

For the Moon, the Sun is ~390 times as far as the Earth but 330,000 times as massive, so the Sun pulls roughly twice as strongly: $F_\oplus$ ≈ 0.45 $F_\odot$. At full moon, both Sun and Earth pull the Moon in the same direction, and the combined force is 1.45 $F_\odot$ toward the Sun. At new moon they pull in opposite directions; the resultant force is only 0.55 $F_\odot$ but still toward the Sun. While the radius of curvature fluctuates, the direction of curvature never reverses.

For a geosynchronous satellite at 0.11 lunar distance, $F_\oplus$ ≈ 38 $F_\odot$. On the night side of the Earth, the resultant force is 39 $F_\odot$ toward the Sun. On the day side, it is 37 $F_\odot$ away from the Sun.

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    $\begingroup$ @uhoh - This is the correct answer. $\endgroup$ – David Hammen Mar 4 at 14:34
  • $\begingroup$ thanks, please ping me when you do, my own down votes make me sad :-) $\endgroup$ – uhoh Mar 5 at 0:28
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    $\begingroup$ it's late but I can at least confirm that when I type in the analytical expression at the bottom of page 118 I can reproduce your values. I really appreciate the extra explanation, thanks for taking the time to write this all up, now I'm happy! :-) $\endgroup$ – uhoh Mar 6 at 14:19
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I think what you may be missing is that $\rho$ is a smoothly changing value. The functions listed are continuous and everywhere differentiable, and as the text says, there is never a sign change, The path the moon follows is not "wavy", even though it does go from behind to in front of the Earth.

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  • $\begingroup$ It's certainly wavy, just not enough in this case to change from concave to convex. $\endgroup$ – uhoh Mar 4 at 0:19
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    $\begingroup$ @uhoh, apologies -- I was using "wavy" as shorthand for a sinusoidal pattern which contains local min/max $\endgroup$ – Carl Witthoft Mar 4 at 14:27
  • $\begingroup$ Interestingly it turns out that there is a sinusoidal pattern which contains local min/max! I wasn't sure until I did a quick calculation. See the edit at the end of my answer. You made me think so +1 :-) $\endgroup$ – uhoh Mar 4 at 14:45
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Prologue

Why is there no concavity in the orbit of the moon around the Sun?

This is a good question and you are of course absolutely right to wonder how it can possibly orbit the Sun without its trajectory being concave to the Sun not only a little bit of the time, but for a substantial fraction of each orbit around the Earth.

Answer

The answer is that the Moon's motion is always concave to the Sun and there's nothing in the article that says anything to the contrary. Curvature doesn't usually have a sign. When we draw problems we can say "curves to the left or right" but $k = 1/R$ where $R$ is the radius of the osculating circle, and we always take the positive value for $R$. If we move to more complex vector notation and include the direction of motion we can talk about the direction of the angular momentum vector but that's beyond the scope of this problem.

In the beginning of the paper the author says explicitly that the Moon's orbit is always concave to the Sun:

Owing however, to the fact that the Moon’s distance from the Earth is very small in comparison with the Earth’s distance from the Sun (about 1/400), and also to the fact that it makes about 13 revolutions around the Earth in a year, it is better to consider the Moon’s path as that of a body describing an orbit around the Sun, and which is constantly being slightly disturbed by the pull of the Earth. Such a path would always be concave to the Sun, but close to the Earth’s orbit, crossing it twice a month.

Figure 1. Incorrect representation of the Moon’s motion.

Figure 1. Incorrect representation of the Moon’s motion.

Epilogue

So how do the Earth-Sun and Moon-Sun distances behave if we forget that they are moving in giant circles?

If we forget that the main concavity is always there and just look at the fluctuation, the Earth gets a little closer and farther during the year, but the Moon is quite busy! It is constantly moving slightly closer and slightly farther. This doesn't mean that the motion is not always concave toward the Sun, but it does mean that the distance between the Moon and the Sun had more than a dozen maxima and minima per year! Who'da thunk it?

solar distances for Earth and Moon for 2020, deviation from mean (km)

Python:

import numpy as np
import matplotlib.pyplot as plt
from skyfield.api import Topos, Loader       
load = Loader('~/Documents/fishing/SkyData')  # avoids multiple copies of large files
data = load('de421.bsp')
ts = load.timescale()
times = ts.utc(2020, 1, np.arange(367))
sun, earth, moon = [data[x].at(times).position.km for x in ('sun', 'earth', 'moon')]
r_earth = np.sqrt(((earth - sun)**2).sum(axis=0))
r_moon = np.sqrt(((moon - sun)**2).sum(axis=0))
plt.figure()
plt.plot(r_earth - r_earth.mean())
plt.plot(r_moon - r_moon.mean())
plt.title('daily solar distances for Earth and Moon for 2020', fontsize=14)
plt.ylabel('Deviation from mean (km)', fontsize=14)
plt.xlabel('Days in 2020', fontsize=14)
plt.show()

Here's a quickie of what the Incorrect View would look like if it were correct per request in comments:

quickie of how the Moon and Earth move around the Sun

Here's a video and a GIF from this answer

Read more about the tools used to make this video in this answer.

GIF below: Screenshots from the YouTube video lagrange points animation.

Screenshots from the YouTube video [lagrange points animation

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    $\begingroup$ A correct version of Figure 1 would be nice. $\endgroup$ – badjohn Mar 5 at 14:39
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    $\begingroup$ @badjohn good point! I've added a bit more. It's great to see a paper written 100 years ago is still generating conversations! This is one of my favorite videos in YouTube. I don't watch it very often, but they did a really nice job of illustrating without talking $\endgroup$ – uhoh Mar 5 at 15:19
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    $\begingroup$ Thanks. I cannot up vote it but only because I already have. $\endgroup$ – badjohn Mar 5 at 15:23
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    $\begingroup$ @badjohn discussing is far more fun than collecting votes, thanks for your suggestions $\endgroup$ – uhoh Mar 5 at 15:24
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    $\begingroup$ @badjohn of course now someone is going to ask why the trace that's trailing the Moon in the video appears to be alternating concave and convex! That's a tough one! I've already asked What do the green lines represent in this Lagrange Point animation? so it's someone else's turn this time. $\endgroup$ – uhoh Mar 5 at 15:28
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My question is, then, how does the Moon's orbit manage to not have any concavity, no matter how minute, while transitioning from the full moon to the new moon positions? Isn't that not a mathematical impossibility?

TL;DR answer: Because the gravitational acceleration of the Moon toward the Sun is about twice the gravitational acceleration of the Moon toward the Earth and because the speed at which the Earth orbits the Sun is about thirty times the speed at which the Moon orbits the Earth.


To get the longer answer, a definition of what "convexity" means is needed. This is easy for a simple closed plane curve: A simple closed plane curve (aka a Jordan curve) is convex if for any two points on the interior of the curve, all of the points on the line segment connecting the two points lie in the interior of the curve.

Unfortunately, neither the Earth's nor the Moon's path about the Sun is closed or planar. To resolve these issues, I'll first do what the referenced paper did, which was to instead investigate coplanar circular orbits of a point mass planet about a star and of a infinitesimal point mass moon about the planet, such that

  • The planet's orbit velocity $v_p$ about the star is inversely proportional to the square root of the distance $r_p$ between the star and the planet,
  • The moon's orbit velocity $v_m$ about the planet is inversely proportional to the square root of the distance $r_m$ between the planet and the moon, and
  • The constant of proportionality for the planet $\left(\frac{{v_p}^2}{r_p}\right)$ is much greater than that for the moon $\left(\frac{{v_m}^2}{r_m}\right)$.

For sufficiently small values of $r_m$, the moon's orbital velocity about the planet will exceed that of the planet about the star, making the moon's path about the star intersect itself:

Non-simple path for sufficiently small moon orbit radius

The loops become small as the moon's orbital distance increases, eventually becoming inward pointing cusps at the point where the moon's orbital velocity about the planet has decreased to where it equals the planet's orbital velocity about the sun. While this curve might or might not be closed, it definitely is not convex due to the inward facing cusps. The cusps broaden into intervals where the primary normal points outward as the moon's orbital radius is increased even further:

Simple but non-convex path

This curve is still non-convex, as exhibited by how the primary normal alternates between pointing inward and outward. the curvature of the moon's path about the star is zero at these transition points: The path, at least instantaneously, is a straight line. This happens because the acceleration vector is parallel to the velocity vector at those transition points.

The intervals where the moon's path about the sun is concave rather than convex shrink as the moon's orbital distance increases even further. At some critical point the intervals of concavity shrink to nothingness. The path is convex everywhere at this orbital distance and beyond:

The Moon's orbit with respect to the Sun is convex

These critical points occur where the acceleration of the moon toward the star is identically zero. This should not be surprising as there is a very close connection between velocity, acceleration, and curvature. In particular, the curvature of a curve at some point is $$\frac{d\hat T}{ds} = \frac{(\vec v \times \vec a)\times \vec v}{v^4} = \kappa \hat N$$ where $\hat T$ is the unit tangent, $\vec v$ are the velocity and acceleration of a point that follows the curve over time, $\kappa$ is the curvature (the inverse of the radius of curvature), and $\hat N$ is the primary normal to the curve. A simpler expression for the curvature is $$\kappa = \frac{||\vec v \times \vec a||}{v^3}$$

Note that the vector $\vec v \times \vec a$ points in the direction of the unit binormal. This suggests a simple metric that extends to non-planar orbits: An orbit about some central point if the cross product between velocity and acceleration with respect to that central point always lies in the same half plane. An even easier metric is to test whether the magnitude of the gravitational acceleration toward the star is greater than that toward the planet.

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  • $\begingroup$ Appreciate the rigor, but any satellite going fast enough for a looping trajectory as in the first image would escape Earth orbit. $\endgroup$ – Mike G Mar 10 at 20:14
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    $\begingroup$ @MikeG - I was referring to stars, planets, and moons in general. For example, Jupiter's two innermost Galilean moons, Io and Europa, orbit just like that, as do four of Saturn's large inner moons (Mimas, Enceladus, Tethys, and Dione). $\endgroup$ – David Hammen Mar 10 at 20:52
  • $\begingroup$ On the other hand, the Earth would need to be a point mass to support such orbits as the object would have to orbit the Earth at a radius (not altitude) of 450 km. $\endgroup$ – David Hammen Mar 10 at 20:55

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