Abstract
π appears often in the formulas of gravity. There is the 8π/3 in the equation for the expansion of the universe, the 3π/32 in the time for a star to collapse, the 4π² in the formula for the revolution of a planet. This paper places the same formulas side by side in a form written with density and a form written with mass, and counts where π enters and how it can be written away. No new theorem or law is claimed.
First, the π of the volume of a ball appears when one writes with density and vanishes when one writes with mass. Eight formulas for a uniform sphere (surface gravity, the angular speed and the period of a grazing orbit, the free-fall time, the escape speed, the expansion equation, the central pressure, the gravitational binding energy) were written in both ways. In the density form every one contains π. In the mass form π disappears from five of them. The π of the density form entered when the mass was rewritten as “density × (4π/3) × radius cubed”.
Second, the π that remains in the mass form is the angle of a revolution. A 2π remains in the period of an orbit and a π in the free-fall time. Free fall is half a revolution of a degenerate orbit whose semi-major axis is half the radius; the free-fall time multiplied by the angular speed of that orbit is exactly π. Rewriting the period as an angular speed removes this π as well. The free-fall time divided by the period of the grazing orbit is $\mathbf{0.176776695}$, which contains no π.
Third, the π of pressure is the π of an area. Only the central pressure shows a π in the denominator when written with mass. Pressure is force divided by area, and the surface area of the sphere, 4π times the radius squared, enters.
Fourth, numbers for the Sun. Using only the GM and the radius of the Sun, the time for the Sun to collapse if it lost its support is $\mathbf{29.487015}$ minutes, and the period of an orbit grazing its surface is $\mathbf{2.780062}$ hours. The value agreed with a numerical integration of the equation of motion to a relative difference of 2.220×10⁻¹⁶. Nowhere in this calculation does the gravitational constant G enter by itself.
Fifth, and this is the core of the paper: the GM of the Sun is 4π² in the astronomers’ units. The period of a circular orbit of radius 1 astronomical unit is $\mathbf{365.2568984}$ days, which agrees to a relative difference of 1.580×10⁻¹⁰ with the year fixed by the constant Gauss defined in 1809 (the Gaussian year), 365.2568983 days. Measuring length in astronomical units and time in Gaussian years, the GM of the Sun is 4π² = $\mathbf{39.478418}$, and Kepler’s third law reads “period = semi-major axis to the power 1.5” with no π. Measured with a year of 365.25 days it is $\mathbf{39.476926}$. In the choice used in this corpus (G = 4π, M = 4π, Paper 28), GM = 16π² and the period is half the semi-major axis to the power 1.5.
Sixth, the mass form has no π in any dimension. Writing the expansion equation in two to six dimensions, the mass form contains no π in any of them, and the power of π in the density form is 1, 1, 2, 2, 3, the same as the power of π in the volume of the unit ball. The angle 2π of a revolution remains 2π whatever the dimension.
To be candid, no meaning is read into the disappearance of π. It disappears because the volume of a ball, or the angle of a revolution, has been stored inside another quantity. The Sun is not a uniform sphere, so its mean density and the time for the passage through a tunnel are the values for a uniform sphere.
Keywords: gravitation, free-fall time, Kepler’s third law, Gaussian gravitational constant, choice of unit, density, dimension.
Scope (what is not claimed)
No new theorem or law is claimed. The gravity of a uniform sphere, the free-fall time, Kepler’s third law, the Jeans length and the Gaussian gravitational constant are all known (Binney and Tremaine 2008, Gauss 1809). No meaning is read into the disappearance of π. π goes away in the mass form because the volume of the ball has been stored inside the mass, not because anything has cancelled. The Sun is not said to be uniform. The mean density of the Sun and the time for the passage through a tunnel are the values for a uniform sphere. The free-fall time is the value for the outer shell when the pressure is removed. The precision of measurement is not discussed. Only the uncertainty of the gravitational constant (CODATA 2022) is quoted; how precisely GM is measured is not treated. The numbers put in are nominal values. The GM and the radius of the Sun are the nominal values fixed by the International Astronomical Union in 2015 (Prša et al. 2016), and the astronomical unit is the value of the 2012 definition. No claim is made that gravity exists in higher dimensions. The formula is only listed against the dimension. Relation to earlier papers: Paper 28 showed that putting M = 4π and G = 4π into Kepler’s third law makes the period half the semi-major axis to the power 1.5. Here it is placed beside the fact that in the units astronomers have long used GM is 4π² and the same law is written without the factor of a half. Paper 38 pointed out that π disappears from Kepler’s formula because the forced 4π² is divided by the chosen 16π², and that no meaning should be read into its disappearance. That caution is followed here as well. Paper 35 showed that of the two 4π in 4πG, the 4π of the Poisson equation is forced and the 4π put into G is a choice. Here G is fixed by the form of the force law, so the 4π of the Poisson equation does not come to the surface. Paper 907 showed that the 8π/3 of the Friedmann equation is the product of the 1/2 of the motion of a shell and the 4π/3 of the volume of a ball. Here it is seen that the same equation written with mass is 2GM divided by the radius cubed, with no π. Paper 1085 took a census of the origins of the 2 in 8π. Here it is not the 2 but the origin of π that is sorted. What is added: placing eight formulas side by side in density form and mass form and counting the power of π, sorting the π that remains into the angle of a revolution and an area, computing times from the GM and the radius of the Sun alone, and checking with present nominal values that the GM of the Sun is 4π² in astronomical units and Gaussian years.
1. The question
$\pi$ appears often in the formulas of gravity. The equation that fixes the expansion of the universe contains $(8\pi/3)G\rho$, the time for a cloud of gas to collapse contains $\sqrt{3\pi/(32G\rho)}$, and the formula for the revolution of a planet contains $4\pi^2a^3/(GM)$. Are these $\pi$ a property of gravity itself?
The way to check is simple. The same formula is placed side by side in a form written with the density $\rho$ and a form written with the mass $M$. A $\pi$ that goes away when the way of writing is changed is not a property of gravity but something the way of writing brought in. If a $\pi$ remains when the way of writing is changed, we look at which $\pi$ it is.
2. Writing with density, writing with mass
Consider a uniform sphere of radius $R$ and mass $M$. Its density is $\rho=M/((4\pi/3)R^3)$. Eight formulas for the sphere were written in both ways (symbolic computation).
| Quantity | Form with mass | Form with density | Power of $\pi$ (mass) | Power of $\pi$ (density) |
|---|---|---|---|---|
| Surface gravity | $GM/R^2$ | $(4\pi/3)G\rho R$ | 0 | 1 |
| Squared angular speed of a grazing orbit | $GM/R^3$ | $(4\pi/3)G\rho$ | 0 | 1 |
| Period of a grazing orbit | $2\pi\sqrt{R^3/(GM)}$ | $\sqrt{3\pi/(G\rho)}$ | 1 | $1/2$ |
| Free-fall time | $(\pi/2)\sqrt{R^3/(2GM)}$ | $\sqrt{3\pi/(32G\rho)}$ | 1 | $1/2$ |
| Squared escape speed | $2GM/R$ | $(8\pi/3)G\rho R^2$ | 0 | 1 |
| $H^2$ of the expansion equation | $2GM/R^3$ | $(8\pi/3)G\rho$ | 0 | 1 |
| Central pressure | $3GM^2/(8\pi R^4)$ | $(2\pi/3)G\rho^2R^2$ | $-1$ | 1 |
| Gravitational binding energy | $(3/5)GM^2/R$ | $(16\pi^2/15)G\rho^2R^5$ | 0 | 2 |
All eight contain $\pi$ in the density form. In the mass form $\pi$ disappears from five of them. The $(8\pi/3)G\rho$ of the expansion equation is the same thing as $2GM/R^3$. Paper 907 showed that this $8\pi/3$ is the product of the $1/2$ of the motion of a shell and the $4\pi/3$ of the volume of a ball. In the mass form only the 2, the reciprocal of the $1/2$, remains, and the $4\pi/3$ of the volume is stored inside $M$.
The $\pi$ of the density form entered when the mass was rewritten as “density $\times(4\pi/3)R^3$”. The binding energy contains $M$ twice, so its density form has two $\pi$. In each of the eight formulas, the difference between the power of $\pi$ in the density form and in the mass form equals the power of the mass in the formula.
3. The π that remains is the angle of a revolution
Three formulas keep a $\pi$ even in the mass form. They are the period, the free-fall time and the central pressure. The first two are examined here.
The $2\pi$ in the period of an orbit is the angle of one revolution. Writing the angular speed $2\pi/T$ instead of the period $T$ gives $\sqrt{GM/R^3}$, with no $\pi$. It is the difference between counting one revolution as 1 and counting it as an angle.
The $\pi$ in the free-fall time is the same angle. The path of a body that starts at rest at radius $R$ and falls straight to the centre is half of an infinitely flattened elliptical orbit whose semi-major axis is $R/2$. The angular speed of that orbit multiplied by the free-fall time is exactly $\pi$, the angle of half a revolution. Dividing the free-fall time by the period of the grazing orbit therefore gives
$$\frac{t_{\rm ff}}{T}=\frac{\sqrt{2}}{8}=0.176776695,$$which contains no $\pi$. The time to fall and the time to go around are both times that count an angle, so the angle drops out of their ratio.
The density form $\sqrt{3\pi/(32G\rho)}$ is a mixture of two $\pi$. There is one $\pi$ of the angle of half a revolution in the numerator and one $\pi$ of the volume of the ball in the denominator under the square root, which together make $\pi$ to the power $1/2$. The $\pi$ in the size at which a cloud of gas begins to collapse under its own weight (the Jeans length), $c_s\sqrt{\pi/(G\rho)}$, is built in the same way: the $2\pi$ of one cycle of a wave divided by the square root of the $4\pi$ of the equation of gravity.
4. The π of pressure is the π of an area
The central pressure in the mass form is $3GM^2/(8\pi R^4)$, with a $\pi$ in the denominator. This is neither the $\pi$ of a volume nor the $\pi$ of an angle. Pressure is force divided by area, and the surface area $4\pi R^2$ of the sphere enters the denominator. In the density form $(2\pi/3)G\rho^2R^2$, two $\pi$ of volume enter for the two masses and one $\pi$ of area divides, leaving one.
To sum up so far, the $\pi$ in the formulas of gravity has three origins. The volume of a ball (when mass is written as density), the angle of a revolution (when time is written as a period), and the area of a sphere (when force is written as pressure). None of them is a number that fixes the strength of gravity. They are numbers that record which variables were chosen.
5. Numbers for the Sun
The formulas are now evaluated for the Sun. The nominal values fixed by the International Astronomical Union are used: $GM=1.3271244\times10^{20}$ m$^3$/s$^2$ and a radius of $6.957\times10^{8}$ m (Prša et al. 2016).
| Quantity | Value |
|---|---|
| Period of a grazing orbit | $10008.225$ s ($2.780062$ hours) |
| Free-fall time | $1769.221$ s ($29.487015$ minutes) |
| Free-fall time $\div$ period | $0.176776695$ |
| Free-fall time (numerical integration of the equation of motion) | $1769.220925$ s |
| Passage there and back through a tunnel in a uniform sphere (numerical) | $10008.225$ s |
If the pressure supporting the Sun suddenly vanished, the outer shell would reach the centre in $29.487015$ minutes. The relative difference between the value of the formula and the numerical integration of the equation of motion was $2.220\times10^{-16}$. The time for a body dropped into a tunnel through a uniform sphere to go there and back equals the period of the grazing orbit to a relative difference of $3.586\times10^{-13}$ (a value for the Sun taken as a uniform sphere).
Nowhere in the calculation of this table is the gravitational constant $G$ used by itself. Only the product $GM$ and the radius enter. To do the same calculation in the density form, $G$ and the density are needed separately. From $G=6.67430\times10^{-11}$ (CODATA 2022) the mass of the Sun is $1.988410\times10^{30}$ kg and its mean density $1409.7798$ kg/m$^3$; putting these into the density form returns the same $1769.220925$ s and $10008.224908$ s. But the relative uncertainty of $G$ is $2.247\times10^{-5}$, and it enters the mass and the density unchanged. What the International Astronomical Union fixed as a nominal value is not the mass of the Sun but $GM$. The density form writes one precisely known quantity as two less precisely known quantities with a $\pi$ between them.
6. The $GM$ of the Sun is $4\pi^2$ in the astronomers’ units
Kepler’s third law relates the period $T$ and the semi-major axis $a$ of an orbit by $T=2\pi\sqrt{a^3/(GM)}$. It contains $(2\pi)^2=4\pi^2$. For a long time astronomers used units in which this $\pi$ need not be seen.
In 1809 Gauss fixed a constant $k=0.01720209895$, taking the mass of the Sun, the size of the orbit of the Earth and one day as units. $k$ is $\sqrt{GM}$ measured in these units. If $2\pi/k$ days is taken as one year (the Gaussian year, $365.2568983$ days), $GM$ is exactly $4\pi^2$ and the third law reads $T=a^{3/2}$. The astronomical unit was redefined as a length in 2012, and $k$ was removed from the table of constants. This was checked with present nominal values.
| Quantity | Value |
|---|---|
| Period of an orbit of radius 1 astronomical unit, $2\pi\sqrt{a^3/(GM)}$ | $365.2568984$ days |
| Gaussian year $2\pi/k$ | $365.2568983$ days |
| Relative difference of the two | $1.580\times10^{-10}$ |
| $GM$ of the Sun (astronomical units cubed $\div$ Gaussian years squared) | $39.478418$ |
| $GM$ of the Sun (astronomical units cubed $\div$ years of 365.25 days squared) | $39.476926$ |
| $4\pi^2$ | $39.478418$ |
The period obtained from the present $GM$ and astronomical unit agrees with the year fixed by Gauss’s constant to the tenth digit. The ratio of $GM$ measured in Gaussian years to $4\pi^2$ is $0.9999999997$. With 365.25 days as one year the ratio is $0.99996223$, slightly short of $4\pi^2$.
This corpus has a choice of the same kind. Paper 28 showed that putting in $G=4\pi$ and $M=4\pi$ gives $GM=16\pi^2$ and turns the third law into $T=a^{3/2}/2$. For $a=1$, 4 and 9, $GM=4\pi^2$ gives $T=1.000000$, $8.000000$ and $27.000000$, and $GM=16\pi^2$ gives $T=0.500000$, $4.000000$ and $13.500000$. In either choice $\pi$ becomes invisible. As Paper 38 cautioned, this is merely the $4\pi^2$ of the angle of a revolution divided by a chosen number. Gauss’s choice makes one year equal to 1, and the choice of Paper 28 matches the mass and $G$ to the number of the solid angle. Different choices leave different numbers behind, 1 or $1/2$.
7. Changing the dimension
If it is right to read the $\pi$ of the volume of a ball as being stored inside the mass, the same should happen when the dimension is changed. $G$ was fixed so that the force law has the form $F=GMm/r^{d-1}$, and the expansion equation was written in $d$ dimensions.
| $d$ | Volume of the unit ball | Power of $\pi$ | $H^2$ (with mass) | $H^2$ (with density) |
|---|---|---|---|---|
| 2 | $\pi$ | 1 | $2GM/R^2$ | $2\pi G\rho$ |
| 3 | $4\pi/3$ | 1 | $2GM/R^3$ | $(8\pi/3)G\rho$ |
| 4 | $\pi^2/2$ | 2 | $2GM/R^4$ | $\pi^2G\rho$ |
| 5 | $8\pi^2/15$ | 2 | $2GM/R^5$ | $(16\pi^2/15)G\rho$ |
| 6 | $\pi^3/6$ | 3 | $2GM/R^6$ | $(\pi^3/3)G\rho$ |
The mass form contains no $\pi$ in any dimension. The power of $\pi$ in the density form is 1, 1, 2, 2, 3, the same as the power of $\pi$ in the volume of the unit ball. The $2\pi$ in the period of a circular orbit, $2\pi\sqrt{R^d/(GM)}$, on the other hand, remains $2\pi$ in every dimension. An orbit goes around in a single plane, so the angle of a revolution does not know the dimension of space. When the dimension is moved, the $\pi$ of the volume moves and the $\pi$ of the angle does not. They are two different $\pi$.
8. Relation to earlier papers
Paper 35 separated the two $4\pi$ inside the product $4\pi G$. The $4\pi$ of the Poisson equation is a solid angle forced by Gauss’s law, and the $4\pi$ put into $G$ is a chosen number. This paper fixed $G$ by the form of the force law, so the $4\pi$ of the Poisson equation did not come to the surface. What did appear were three: the volume of a ball, the angle of a revolution and the area of a sphere. Only in the Jeans length does the $4\pi$ of the Poisson equation show itself, under a square root.
Paper 1085 took a census of the 2 in $8\pi$ and counted the $8\pi/3$ of the Friedmann equation as “the 2 of motion” times the volume of a ball. This paper looked at the same equation from the side of $\pi$ and counted that the $4\pi/3$ can be stored inside the mass. The 2 remains even in the mass form.
Closing
Closing: the π in the formulas of gravity is not a number that fixes the strength of gravity but a number that records which variables were chosen. Placing eight formulas for a uniform sphere side by side in density form and mass form, every density form contains π, and π disappears from five of the mass forms. The π that remains in the mass form is the angle of a revolution in the period and the free-fall time, and the area of a sphere in the pressure. The free-fall time divided by the period is $\mathbf{0.176776695}$, with no π. From the GM and the radius of the Sun alone, the free-fall time comes out as $\mathbf{29.487015}$ minutes and the period of a grazing orbit as $\mathbf{2.780062}$ hours. The GM of the Sun is 4π² = $\mathbf{39.478418}$ when measured in astronomical units and Gaussian years, and Kepler’s third law can then be written without π. In the choice of this corpus it is 16π², and a 1/2 attaches to the period. When the dimension is changed, the π of the volume moves and the π of the angle of a revolution does not. To be candid, it is not that the way of writing in which π disappears is the right one. What could be said is that when a π appears, one may look at whether it is counting a volume, an angle or an area. The separator is whether the formula takes in how dense the content is or how much is enclosed, and whether one counts revolutions or angular speed.
References
- Binney, J., & Tremaine, S. (2008). Galactic Dynamics. Princeton University Press.
- Gauss, C. F. (1809). Theoria Motus Corporum Coelestium in Sectionibus Conicis Solem Ambientium. Perthes et Besser.
- International Astronomical Union (2012). Resolution B2 on the Re-definition of the Astronomical Unit of Length. IAU.
- National Institute of Standards and Technology (2024). CODATA Recommended Values of the Fundamental Physical Constants: 2022. NIST.
- Prša, A., Harmanec, P., Torres, G., Mamajek, E., et al. (2016). Nominal Values for Selected Solar and Planetary Quantities: IAU 2015 Resolution B3. The Astronomical Journal.
- Yamagishi, Y. (2026). The Yamagishi-Kepler Equation [Paper 28]. YOI: https://www.yamagishiyuuki.com/yoi/28; DOI: 10.5281/zenodo.21430531
- Yamagishi, Y. (2026). The Yamagishi Gravitational Constant Unpacking Theory [Paper 35]. YOI: https://www.yamagishiyuuki.com/yoi/35; DOI: 10.5281/zenodo.21430118
- Yamagishi, Y. (2026). The Yamagishi Master Table of Classical-Physics Geometric Unpacking [Paper 38]. YOI: https://www.yamagishiyuuki.com/yoi/38; DOI: 10.5281/zenodo.21430234
- Yamagishi, Y. (2026). Cosmic Expansion Follows Exactly from a Newtonian Shell for Dust, but for Light the Deceleration Is Twice the Shell’s [Paper 907]. YOI: https://www.yamagishiyuuki.com/yoi/907; DOI: 10.5281/zenodo.22804787
- Yamagishi, Y. (2026). A Census of the 2 in $8\pi$: It Has Ten Origins, and Changing the Dimension Sorts Them into Three Ways of Moving [Paper 1085]. YOI: https://www.yamagishiyuuki.com/yoi/1085
Reference
Yamagishi, Y. (2026). The π in the Formulas of Gravity Appears with Density and Vanishes with Mass; What Remains Is the Angle of a Revolution [Paper 1086]. YOI: https://www.yamagishiyuuki.com/yoi/1086
ORCID: 0009-0006-8587-3450
PDF (English) · PDF (Japanese) · Date proof of the English PDF (OpenTimestamps .ots) · SHA256SUMS
The record page for this paper (in Japanese, with the PDFs in each language and the date proof): YOI 1086. This paper is a preprint and has not been peer reviewed.