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We know that Kepler's equation is $$M = E - e\sin E$$ Here, $e$ is the eccentricity, $M$ is the Mean anomaly and $E$ is the Eccentric anomaly.

Now, the thing is- $M$ and $E$ are angles, which might have unit in radians or degrees. However, both $e$ and $\sin E$ are dimensionless. Then how can we subtract a dimensionless number from an angle?

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    $\begingroup$ Related question on Physics.SE: Are units of angle really dimensionless? $\endgroup$
    – PM 2Ring
    Commented Aug 1, 2023 at 9:42
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    $\begingroup$ I kind of like to think of $M$ and $E$ as measurements of the area of sectors of the hypothetical circular orbit and the osculating circle around the elliptical orbit respectively, divided by $a^2$. Basically the same as treating them as angles for the elliptical orbits, but makes the comparison t othe elliptical Eccentric Anomaly to the analogous Hyperbolic Anomaly and hyperbolic sectors when working with hyperbolic orbits more evident. $\endgroup$
    – notovny
    Commented Aug 2, 2023 at 18:37
  • $\begingroup$ (+1) for an excellent question! This has always seemed to me just plain weird, but as DH explains, yep this mixes an angle with the sin of the angle and if that weren't enough, then throws the eccentricity in just show us who's boss in this universe; math, not people. $\endgroup$
    – uhoh
    Commented Aug 3, 2023 at 4:51
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    $\begingroup$ A helpful approach in understanding these is to treat both radians and results of trig functions as not actually dimensionless, but occurring in units of [m/m] - length/length (and refuse to cancel them outright). One is length of arc over length of radius, the other is ratios of lengths of sides of a triangle, like sin(angle) is length of opposite leg to hypotenuse. Sure the moment you reduce these you get a dimensionless value, but up until then you have units that help you verify how it meshes with other units. $\endgroup$
    – SF.
    Commented Aug 4, 2023 at 10:27
  • $\begingroup$ @SF. that's really nice; yes that works for me at least. $\endgroup$
    – uhoh
    Commented Aug 5, 2023 at 22:56

1 Answer 1

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$$\newcommand{\sin}{\mathop{\rm sin}\nolimits} M = E - e \sin E$$

Note that this expression only works when $M$ and $E$ are expressed in radians. Radians are a measure of arc length divided by radius, which also is a length. This means that radians are unitless and also are consistent. In comparison, degrees are an inconsistent way to express angle. To use Kepler's equation where angles are expressed degrees one must use

$$\newcommand{\sind}{\mathop{\rm sind}\nolimits} M = E - \frac{180}{\pi} e\sind E$$

where $\sind E$ is the sine function but with $E$ expressed in degrees rather than radians.

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    $\begingroup$ FWIW, modern derivations of Kepler's equation (like this) use calculus, so it's "obvious" that the angles must be in radians. Of course, Kepler's own derivation predates Newton, and uses clever geometrical reasoning, but I can't remember the details. $\endgroup$
    – PM 2Ring
    Commented Aug 1, 2023 at 9:47
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    $\begingroup$ @PM2Ring Newton did not write $F=ma$. He, like Kepler, used clever geometric reasoning. A more general way to write Newton's second law, using algebra, is $F=kma$ where $k$ is a non-unitary conversion constant required as punishment for insisting on using inconsistent units such as pounds-force, pounds-mass, feet, and seconds. $\endgroup$
    – David Hammen
    Commented Aug 1, 2023 at 10:04
  • $\begingroup$ @DavidHammen To put it another way, the units in the metric system come from picking a few basic measurements, and defining the remaining units so that equations like F = ma work without an additional conversion constant. $\endgroup$
    – Barmar
    Commented Aug 2, 2023 at 14:44
  • $\begingroup$ @Barmar The metric system is a good but incomplete start. It still has non-unitary conversion constants analogous to the $k$ in $F=kma$ such as the speed of light $c$, the universal gravitation constant $G$, and the reduced Planck constant $\hbar$. $\endgroup$
    – David Hammen
    Commented Aug 4, 2023 at 11:05

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