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# Using GIS tools & Mathematics to Improve Kepler's law of Planetary Motion

Rohit Kumar

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Rohit Kumar. Using GIS tools & Mathematics to Improve Kepler's law of Planetary Motion. 2020. [hal-02906058](https://hal.science/hal-02906058)

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# Using GIS tools & Mathematics to Improve Kepler's law of Planetary Motion

Rohit Kumar¹

AT+PO - Matihani, Dist.- Begusarai, Bihar

Corresponding author: Rohiit@live.in

**Abstract:**

Kepler's law can be considered as one of the foundational astrophysics' theory, which laid
foundation of modern theories of astrophysics, especially Newtonian mechanics. It informs the
planets motion around sun and its orbital period. But the planet-sun system is in constant
equilibrium at the center of mass. The movement of planet around the sun should be measured
as per its distance from the center of mass instead of the distance from parent star. In this
paper it is shown that equal area can be covered in equal time interval only when considered
from the center of mass. The area differs between the major and minor axis, when
measurement is done from star center which would simply mean one of the foci will lie at the
center of mass instead of center of sun.

**Keyword:**

Kepler's Law, linear acceleration, center of mass, foci, rotational period, GIS, Spatial analysis

**Introduction:**

Between 1609 and 1619 John Kepler gave three laws based on observation having scientific approach. Observations made over the movement of Mars was used to assume the motion of other planetary bodies, which was indeed a right assumption. Kepler's theory was an improvement over Copernicus theory of circular motion of planets around sun. It was perhaps the first theory which showed how the motion of planets very over time and distance. The second law helps to establish that when a planet is closer to the Sun, it travels faster. The third law expresses that the farther a planet is from the Sun, the longer its orbit, and vice versa [2].

There are three different Kepler's Laws namely Law of Orbits, Areas, and Periods.

**Three Laws of Kepler:**

* Kepler's Law of Orbits – The Planets moves in elliptical orbit around sun, with the sun at one foci [2,3,4,6].

* Kepler's Law of Areas – The line joining the center of planet to the center of Sun sweeps out equal areas in equal interval of time [2,3,4,6].

* Kepler's Law of Periods – The square of the time period is directly proportional to the cube of the semimajor axis of orbit of the planet [2,3,4,6].

But now an improvement can be made in the Kepler's law as stated above. It is known since
ancient Greek 'Archimedes' times that two bodies remains at equilibrium at their center of
mass [1]. This basic concept was not used by him while creating the laws. So, while creating the laws, assuming the planets rotation around the sun instead the center of mass [1] is incorrect. In stating the planets rotation around the sun, the area covered by the planets between any two identical time period will not be same.

Sun being at one foci as suggested by Kepler will get changed when comparison is made considering the rotation of Jupiter size planet around sun. as in this case the center of mass [1] is outside the circumference of sun. it will be more exaggerated in case of planet of super Jupiter size revolving around a white dwarf, or in case of rotation of two stars or white dwarf or neutron star or black hole as shown by the Australian National University (ANU) [5].

Newtons gravitational theory which was devised using Kepler's work also has distance between the two rotating bodies. In this paper it is shown that Kepler's law of planetary motion can be improved.

Harmonic oscillation has been favorite way to demonstrate and analyze the movement of planets around sun. Indeed, it can be one of the correct approach to consider. But this can provide correct result only when center of mass [1] approach is included instead of considering the distance between the object like sun and planets or planets and their satellites. The ratio of area coverage in equal time interval can be constant for same. The oscillation frequency will remain same for each time[5], but if consideration is made using center of star approach as is suggested in [7,8,9], there will be overlap of area coverage in equal time interval, ie; start and end area will get overlapped. Entire elliptical area can be converted into equal ratio for same time interval in case of considering the center of mass. In this case there can't be any overlap thus perfect display of oscillation can be observed.

The fact that planets moves around its parent star and these star revolves around their galactic center will remain true. The effect of this revolution over star-planet system in the movement over planets around their host star as suggested by Fatima [10] can be one of the possibility. But overall there is stability in the movement of planets around their center of mass, as the fundamental balance of mass-distance equilibrium can't be overruled at any point of time.

Methodology:

Steps to prove inaccuracy in Kepler's law:

Area covered by planets in 1 day while revolving around sun: Example earth [6]

$$ \text{Area of ellipse}/365 = \frac{\pi(ab)}{365} \text{ km}^2 $$

Area covered by planet (earth) in 1 day:

$$ \text{Total area covered} / 365 = \frac{\pi(ab)}{365} \text{ m}^2 $$
$$ \frac{\pi * 1.47 * 1.52 * 10^{10}}{365} m^2 = 1.92 * 10^{18} m^2 $$

But this area corresponds to the ellipse created by the movement of planet (earth) from the center of mass, while Kepler's law indicates about area swept by the movement of planets around center of parent star.

<figure><img src="image_2.png" /><figcaption>**Figure1:** Planets elliptical orbital area coverage from center of mass</figcaption></figure>

Comparison of the area swept by the planet around their parent star center [2,3,4,6] will have

* Decrease in area for minor axis.

* Increased in area for major axis.

<figure><img src="image_3.png" /><figcaption>**Figure 2A &2B:** Planets elliptical orbital area coverage from center of parent star</figcaption></figure>

Difference in the area swept by the planet while revolving around the star can be calculated by taking two different approach.

* GIS method

* Mathematical modelling

**GIS Method:**

GIS is a powerful tool to visualize and analyze data in spatial reference. The best part to have knowledge of GIS tools is that without having great knowledge of mathematics, the result can be derived. The mathematical part is done in the background by tools.

Over here movement of planet is just an example of spatial positioning of planet or relatively lighter object compared to heavier at any moment of time. Comparison using GIS tools can be quick and easy. Method used to make comparison using this approach is:
<figure><img src="image_4.png" /></figure>

A hypothetical scenario is created identical to that of star-planet system in GIS environment with projection WGS-84 near equator where distortion in area calculation due to earth's curvature is minimum. Calculation is done in square meter to calculate the difference in area. Output of the difference between area coverage of major and minor axis between center of mass and star center is shown in below image.

Minor glitch in understanding the difference in spatial location can create the rise of incorrect understanding of fundamentals. The has been with Kepler's law of Planetary motion. In below image shaded violet area with identifiable thickness represents minor axis, while violet area with very little thickness refers to major axis.
<figure><img src="image_5.png" /><figcaption>Figure3: Difference in area coverage between major and minor axis between center of mass and star center</figcaption></figure>

**Statistical details:**

<table><thead><tr><td>Sl no.</td><td>Axis type</td><td>Total area (square meter)</td><td>Difference</td></tr></thead><tbody><tr><td>1</td><td>Area for center of mass for Major Axis</td><td>31893.1784497206</td><td>~ 5 m²</td></tr><tr><td>2</td><td>Area for center of mass for Minor Axis</td><td>31888.820116431</td><td></td></tr><tr><td>3</td><td>Area from Sun's center for Major Axis</td><td>33029.8413142413</td><td>~ 4649 m²</td></tr><tr><td>4</td><td>Area from Sun's center for Minor Axis</td><td>28380.1139471648</td><td></td></tr></tbody></table>

**Mathematical modelling:**

Considering the most basic situation in which the difference in coverage area swept by the planets can be explained.

<figure><img src="image_6.png" /><figcaption>Figure3: Overlay of area covered by planet in same time interval.</figcaption></figure>

In the above figure 3,
* Red lines show the area created when considerations are made using the center of mass, while in blue it shows the area coverage from center of sun.

* Sun is displayed in thick purple colour circle.

* Earth is in blue circle at the ellipse circumference.

<figure><img src="image_7.png" /><figcaption>Fig4: Area coverage comparison</figcaption></figure>

In the above figure4 considering AOB be the equilateral triangle where 'O' is the position of center of mass and ABO' is the isosceles right-angled triangle with "O" being the Sun's center. Let's consider, the length of each red colored arm of triangle created equilateral triangle with side length 10cm and blue colour arm created isosceles right triangle with equal arm of length 6cm. This 10 cm and 6 cm can be equal to or replaced with 10¹⁰m or 6¹⁰m to represent real scenarios present in universe. Then,

$$ \text{Area of equilateral triangle: } \frac{\sqrt{3}}{4} * a^2 = \frac{\sqrt{3}}{4} * OA^2 = \frac{\sqrt{3}}{4} * 10^2 = 25\sqrt{3} = 43.301 \text{ cm}^2 $$

$$ \text{Area of isosceles right triangle: } \frac{1}{2} * a^2 = \frac{1}{2} * O'A^2 = \frac{1}{2} * 6^2 = 18 \text{ cm}^2 $$

Area comparison due to change in position from center of mass (that governs the rotation) to the center of sun's mass.

$$ \text{Difference in area: } (43.301 - 18) \text{ cm}^2 = 25.301 \text{ cm}^2 \text{ for minor axis.} $$

For the decrease in area of minor axis, there is increase in area for major axis. If considered equal, total difference in area will become twice between the major and minor axis for same time period. Thus, proving the claim made in Kepler's law incorrect.
Explanation for the Law of motion of planets:

Equilibrium of a system can be obtained using the linear acceleration of mutually rotating bodies instead of considering the centers of mass of each bodies separately. So, on considering linear acceleration motion of planets can be explained:

$$m_P * v_P^2 * r_P = m_S * v_S^2 * r_S \text{ ------------------------ } 1$$

$$\text{Considering velocity } 'v' = \left(\frac{2\pi r}{T}\right)$$

$$m_P * r_P * \left(\frac{2\pi r_P}{T_P}\right)^2 = m_S * r_S * \left(\frac{2\pi r_S}{T_S}\right)^2 \text{ ------------------ } 2$$

$$\Rightarrow \left(\frac{T_S * 2\pi r_P}{T_P * 2\pi r_S}\right)^2 * \frac{r_P}{r_S} = \frac{m_S}{m_P} \text{ --------------------- } 3$$

From above equation following conditions can be obtained:

$$(1) \text{ When } \frac{m_S}{m_P} = \frac{m_S}{m_{NS}} = 1:$$

In the above case mass of planet - m<sub>P</sub> is replaced by stars (neutron, white dwarf or blackhole) revolving around another star of identical mass. Or the condition can be converted if two identical star (neutron, white dwarf or blackhole) are revolving around each other.

$$\left(\frac{T_S * 2\pi r_P}{T_P * 2\pi r_S}\right)^2 * \frac{r_P}{r_S} = 1, \text{ or } \frac{T_S^2 * 4\pi^2 * r_P^3}{T_P^2 * 4\pi^2 * r_S^3} = 1 \text{ or } \frac{T_S^2 * r_P^3}{T_P^2 * r_S^3} = 1 \text{ ------------------ } 4$$

$$\text{or, } \frac{T_S^2}{T_P^2} = \frac{r_S^3}{r_P^3} \text{ -------------------------- } 5$$

$$\text{Or, } \left(\frac{T_2}{T_1}\right)^2 = \frac{4\pi^2}{4\pi^2} \left(\frac{r_2}{r_1}\right)^3 \text{ -------------------------- } 6$$

On multiplying and dividing the numerator by 3 and cancelling one π

$$\left(\frac{T_2}{T_1}\right)^2 = \frac{\text{Volume of 2nd Object}}{\text{Volume of 1st Object}} \text{ -------------------------- } 7$$

Orbital period between the centers:

⇒ Equation 5 indicates square of orbital period is directly proportional to the cube of distance from the center of mass and not by distance between the centers of rotating object 'r', as suggested in Kepler's law.

⇒ Orbital period and distance or volume covered by the two rotating objects will always be in same proportion from their center of mass.
⇒ One single object cannot rotate on its own around any center, there must be presence of other object.

$$ (2) \text{ When } \frac{m_P * v_P^2}{m_S * v_S^2} = \frac{r_S}{r_P} : ....................................34 $$

⇒ Total energy contained by mutually rotating body within a system, is inversely proportional to distance from their center of mass.

$$ (3) \text{ When } \frac{m_P * v_P}{m_S * v_S} = \frac{P_P}{P_S} = \frac{v_S * r_S}{r_P * v_P} = \frac{r_S^2 * t_P}{r_P^2 * t_S} ....................................35 $$

⇒ Momenta of the planet (P<sub>P</sub>) & star (P<sub>S</sub>) are inversely proportional to VeDi (Ve - velocity & Di - distance) or inversely proportional to surface area swept per unit time, or surface area covered since 't<sub>P</sub>' will always be equal to 't<sub>S</sub>'.

$$ (4) \text{ When } \frac{m_P * r_P}{m_S * r_S} = \frac{v_S^2}{v_P^2} $$

⇒ m<sub>P</sub> * r<sub>P</sub> = m<sub>S</sub> * r<sub>S</sub>, from equivalence principle, so v<sub>S</sub><sup>2</sup> = v<sub>P</sub><sup>2</sup>

$$ \text{Or, } \left(\frac{T_S * 2\pi r_P}{T_P * 2\pi r_S}\right)^2 = 1 \Rightarrow \left(\frac{T_S}{T_P}\right)^2 = \left(\frac{r_S}{r_P}\right)^2 ....................................36 $$

$$ \left(\frac{T_2}{T_1}\right)^2 = \frac{\text{Surface Area of 2nd Object}}{\text{Surface Area of 1st Object}} $$

⇒ Square of the orbital period is directly proportional to the surface area covered.

Rotation of single object can only be in artificially created objects. In universe, it is highly unlikely that the object will move away from its center of mass, or there would be any change in either of the condition by any object. These objects will continue to move in its original path till an external force is applied to displace it from its radius. Once moved, it will continue to move till it again obtains the equilibrium and follows the same condition.

It can be said that the same condition exists at microscopic level, with only difference that these microscopic objects are responsible for maintaining the equilibrium of the universe. Over here its certain that all the atoms are in hydrostatic equilibrium. Movement of electron around nucleus can also be attributed to this. Photon ejected should be considered an attempt to maintain the hydrostatic equilibrium by electron, proton or nucleus. In the same way neutron ejecting the electron is an attempt to maintain its hydrostatic equilibrium with surrounding.

## Result & Discussion:

Modification in the Kepler's law of motion are:
1. Kepler's Law of Orbits – The Planets moves in elliptical orbit around sun along their center of gravity, with one of the foci at center of gravity. In other words, since equal area is covered in equal time interval from center of gravity so one foci will be at center of mass. Location of other foci can be obtained by reversing it.

2. Kepler's Law of Areas – The line joining the center of planet to the center of mass sweeps out equal areas in equal interval of time.

3. Kepler's Law of Periods – The square of the time period is directly proportional to the cube of the semimajor axis of orbit of the planet. This law can remain unchanged.

Other understandings from the derived equation:

1) Square of the orbital period is directly proportional to the surface area covered.

2) Momenta of the planets & star are inversely proportional to VeDi (Ve - velocity & Di - distance) or inversely proportional to surface area swept per unit time, or surface area covered.

3) Total energy contained by mutually rotating body within a system, is inversely proportional to distance from their center of mass.

4) Square of orbital period is directly proportional to the cube of distance from the center of mass and not by distance between the centers of rotating object as suggested in Kepler's law.

5) Orbital period and distance, volume or surface area covered by the two rotating objects will always be in same proportion from their center of mass.

6) One single object cannot rotate on its own around any center, there must be presence of other object, else equilibrium can't get maintained.

7) Area swept by the planet around their parent star center will have

a) Decrease in area for minor axis.

b) Increased in area for major axis.

c) Equal area can be covered in equal time interval only when considered from the center of mass and not from the center of parent star.

**Conclusion:**

Kepler's law being incorrect, its modified version discussed here is part of the center of mass theory discussed in previous paper with inclusion of time period. Law governing the motion of planets are not different from the law governing motion of other particles around each other.

**Acknowledgement:**

I am thankful to my uncle Dr. M. K. Roy, Physics dept., Mohan Lal Sukhadia College-Udaipur who is always there when I need help, my, as well to all those who taught me GIS. Special thanks to Navin Nayan - Manager GIS Data Engineering, HERE technology with whom I am
working since last 5 years. His efforts to share knowledge with his colleague helped us to
understand the depth of spatial thinking. The strength obtained has been used to look in details
given by Kepler, spatially and logically.

References:

1. https://en.wikipedia.org/wiki/Center_of mass

2. https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary-motion

3. https://www.physicsclassroom.com/mmedia/circmot/ksl.cfm

4. NCERT Class 11 book, Physics

5. https://phys.org/news/2018-12-scientists-biggest-black-hole-collision.html

6. https://www.texasgateway.org/resource/71-keplers-laws-planetary-motion

7. P. A. Horvathy, “The laws of planetary motion, derived from those of a harmonic oscillator (following Arnold)”, April 2014, arXiv:1404.2265v1

8. V. I. Arnold: Huygens & Barrow, Newton & Hook. Birkhäuser (1990).

9. P. Horvathy, *The laws of planetary motion, derived from those of an oscillator,* (in Hungarian) Fizikai Szemle 54/6 June (2004) pp. 180-184.

10. Galuh Titi Fatimah, “Analysis of Kepler’s law to the movement of celestial bodies”, IOP Publishing, 0.1088/1742-6596/1282/1/012037, Conf. Series 1282 (2019) 012037