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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

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

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