A Puzzle Involving the Moon's Orbital Motion

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Summary:

Was Newton's reasoning about the falling apple and the Moon incorrect because he left out the Coriolis acceleration?
This is a puzzle involving the orbit of the Moon around the Earth, based on the well-known story of Newton observing a falling apple and wondering if the same force that made the apple fall could also explain the motion of the Moon.

The story goes that Newton considered the following: the acceleration of the falling apple, near the surface of the Earth; the ratio of the Moon's distance from the Earth's center to the Earth's radius; the inverse square law of gravitation; and the acceleration required to hold the Moon in its orbit. In modern units and taking advantage of modern measurements, he was considering the following numbers (all units are SI, and I am using the Wikipedia page on the Moon's orbit [1] for the Moon's orbital parameters, and the Wikipedia page on the Earth for other parameters [2]):

$$
a_{\text{apple}} = 9.8
$$

$$
\rho = \frac{R_{\text{Moon}}}{R_{\text{Earth}}} = 60.3
$$

$$
\frac{a_{\text{apple}}}{\rho^2} = 0.0027
$$

$$
a_{\text{Moon}} = \frac{v_{\text{Moon}}^2}{R_{\text{Moon}}} = 0.0027
$$

The last two numbers are equal, indicating that indeed the inverse square law of gravitation can explain the Moon's orbit.

But there is a fly in the ointment: what about the Coriolis acceleration? After all, a frame in which the Earth is at rest, which (apparently) is the frame being used in the above calculations, is not an inertial frame, because of the Earth's orbital motion about the Sun. This should give rise to a Coriolis acceleration for the Moon, which (idealizing the Moon's orbit as circular and lying in the plane of the Earth's orbit about the Sun--the corrections due to those things not being quite true don't matter here) will always point outward, away from the Earth. The magnitude of this acceleration is easy to calculate: it is ##2 \Omega v_{\text{Moon}}##, where ##\Omega## is the angular velocity of the Earth's orbital motion about the Sun. So we have

$$
\Omega = \frac{2 \pi}{T_{\text{Earth}}} = 1.99 \times 10^{-7}
$$

$$
a_{\text{Coriolis}} = 2 \Omega v_{\text{Moon}} = 0.000407
$$

Subtracting this from ##a_{\text{Moon}}## gives a net acceleration of ##0.0023##. So taking the Coriolis effect into account, the Moon should not stay in its orbit!

So what gives? Did Newton make a mistake? Or is there an error somewhere in the logic I have just given? If so, what is it?

[1] https://en.wikipedia.org/wiki/Orbit_of_the_Moon

[2] https://en.wikipedia.org/wiki/Earth
 
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  • #2
PeterDonis
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Two notes to clarify items that are not where the solution to the puzzle lies. :wink:

First, the actual non-inertial frame that would be properly used for this problem would be one in which the barycenter of the Earth-Moon system was at rest. Both the Earth and the Moon orbit this barycenter, so the Earth would also be subject to a Coriolis acceleration in a rotating frame centered on the barycenter, but I have left that out because solving the puzzle for the Moon will also solve it for the Earth. Since the distance from the barycenter to the center of the Earth is approximately 1/81 (the Moon/Earth mass ratio) of the Earth-Moon distance, the error involved is only a little more than a percent, an order of magnitude less than the size of the Coriolis acceleration as compared with the gravitational acceleration, so it's not enough to make a difference for this problem.

Also, as noted briefly in the OP, for this problem, treat the Moon's orbit as a perfect circle in the plane of the Earth's orbit about the Sun, i.e., the plane perpendicular to the angular velocity vector of the rotating frame defined by the Earth's orbital motion about the Sun, and treat the Earth's angular velocity about the Sun as constant.
 
  • #3
A.T.
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... what about the Coriolis acceleration? After all, a frame in which the Earth is at rest, which (apparently) is the frame being used in the above calculations, is not an inertial frame, because of the Earth's orbital motion about the Sun. ...
Which rotating frame do you mean here? Rotating around the Sun with to keep the Earth's center at rest? What about the centrifugal force?
 
  • #4
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Which rotating frame do you mean here? Rotating around the Sun with to keep the Earth's center at rest?
Yes.

What about the centrifugal force?
In the rotating frame that is cancelled out by the Sun's gravity. (Technically, this is only true along a tangential curve through the origin, but the error involved in taking it as true over the diameter of the Moon's orbit is small enough not to matter. This is another one of those areas that is not where the solution of the puzzle lies. :wink:)
 
  • #5
A.T.
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Yes.
But in that rotating frame the Moon is moving at a different speed around the Earth, than in a non-rotating frame, so it needs a different centripetal acceleration.
 
  • #6
PeterDonis
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in that rotating frame the Moon is moving at a different speed around the Earth, than in an non-rotating one, so it needs a different centripetal acceleration
Is this your proposed solution to the puzzle?
 
  • #7
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I think that the error is in considering the Earth a rotating frame when it's just an accelerated one (due to sun attraction).
If we choose the reference frame of Earth there is no Coriolis force caused by our rotation around the sun, at most a fictitious force balanced by the attraction of the sun due to the acceleration of Earth toward the sun.
 
  • #8
A.T.
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Is this your proposed solution to the puzzle?
Not, sure. Just wondering what vmoon represents here.

And why assume that Newton's calculation uses a frame rotating around the Sun? You can have the Earth's center at rest, without frame rotation.
 
  • #9
PeterDonis
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If we choose the reference frame of Earth
There is no "the" reference frame of Earth. There are multiple possible reference frames one can define that are centered on the Earth. So you need to specify exactly what reference frame you mean, and why you think using that frame solves the puzzle.
 
  • #10
PeterDonis
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Just wondering what vmoon represents here.
As the puzzle is stated, it represents the orbital velocity of the Moon that can be looked up in the Wikipedia article I linked to.

As is usual with puzzles, however, one should always look for unstated assumptions, and that might lead you to question whether the velocity I just described is the right one to use.

why assume that Newton's calculation uses a frame rotating around the Sun?
Why, indeed? See my comment on unstated assumptions, above.
 
  • #11
A.T.
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As the puzzle is stated, it represents the orbital velocity of the Moon that can be looked up in the Wikipedia article I linked to.
That's presumably the value for an non-rotating frame.

Why, indeed?
Ok, assume a non-rotating frame then, with the Earth's center at rest. No Coriolis there.
 
  • #12
PeterDonis
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That's presumably the value for an non-rotating frame.
You don't have to presume it, you can calculate it.

assume a non-rotating frame then, with the Earth's center at rest. No Coriolis there.
Again, you don't have to assume, you can calculate which numbers are valid for which kind of frame. You are correct that in a non-rotating frame there is no Coriolis force, so if the numbers work out for a non-rotating frame, that would be a solution to the puzzle. (At least, it would if there are no other issues with such a frame.)
 
  • #13
A.T.
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You don't have to presume it, you can calculate it.
Period of:
Moon orbit around Earth (sidereal) 27.322 days
Moon orbit around Earth (synodic) 29.530 days

(27.322 / 29.530)2 ≈ (0.0023 / 0.0027)

So that would explain the different acceleration.
 
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  • #14
PeterDonis
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Period of:
Moon orbit around Earth (sidereal) 27.322 days
Moon orbit around Earth (synodic) 29.530 days

(27.322 / 29.530)2 ≈ (0.0023 / 0.0027)

So that would explain the different acceleration.
Yes, you've got it. Or, to put it another way, the orbital velocity of the Moon in the rotating frame is smaller than the Wikipedia page value; it's about 950 m/s instead of 1000 m/s. So the centripetal acceleration ##v^2 / R## required to keep it in orbit matches the difference of (acceleration due to Earth's gravity minus Coriolis acceleration).
 
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  • #15
PeterDonis
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I think that the error is in considering the Earth a rotating frame when it's just an accelerated one (due to sun attraction).
The "non-rotating frame" that gives the numbers calculated in the OP is indeed not quite an inertial frame, because in an inertial frame (such as one in which the barycenter of the solar system is fixed and not rotating with respect to the stars--at any rate this is much closer to being an inertial frame) the Earth is not moving in a straight line, its path is curved due to the Sun's gravity. That does lead to a follow-up question: why does the non-rotating frame calculation, in which gravity is the only force acting, work so well if that frame is still non-inertial?
 
  • #16
A.T.
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That does lead to a follow-up question: why does the non-rotating frame calculation, in which gravity is the only force acting, work so well if that frame is still non-inertial?
In the Earth (or Earth-Moon barycenter) centered non-rotating frame you have a uniform inertial force -ma (where 'a' is the centripetal acceleration of the origin along it's orbit around the Sun). That force mostly cancels the Suns gravity (which is approximately uniform at this distance). So you can treat the Earth-Moon system as an isolated inertial two body system.
 
  • #17
vanhees71
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I think the solution simply is that Newton's head is not sensitive enough to recognize the very small effects of the inertial forces in the rotating frame ;-)). Sometimes it's good for science to neglect tiny effects and take them into account only later when they are really relevant (e.g., when thinking about the Foucault pendulum, where the daily rotation of the Earth around the axis is important or the tides, where both the rotation of Earth and moon around the Sun as well as the daily rotation of the Earth are important).
 
  • #18
A.T.
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I think the solution simply is that Newton's head is not sensitive enough to recognize the very small effects of the inertial forces in the rotating frame ;-)).
How dare you?!
 
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  • #19
PeterDonis
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In the Earth (or Earth-Moon barycenter) centered non-rotating frame you have a uniform inertial force -ma (where 'a' is the centripetal acceleration of the origin along it's orbit around the Sun). That force mostly cancels the Suns gravity (which is approximately uniform at this distance).
Yes. This is actually the non-rotating frame version of the argument I gave in post #4 for why you can ignore centrifugal force in the rotating frame. The error involved in both cases is the same (the difference in the Sun's gravity over the diameter of the Moon's orbit).
 
  • #20
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This delightful puzzle you posted reminds me of a question I like to ask my work colleagues (non physicists):

"So, the sidereal period of Earth's rotation is not 24 hours. It is 23 hours and 56 mins. Shouldn't that fact mess up our time keeping within a few days?"

I'm always impressed with the person that can explain to me what the difference between the sidereal day and the solar day.

Related to your post, have you ever visited this website?
https://personal.ems.psu.edu/~fraser/Bad/BadCoriolis.html

It's sure to entertain one who is already educated about the Coriolis force.
 

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