Reasoning for distinct acceleration due to gravity on Polygons

REASONING FOR DISTINCT ACCLERATION DUE TO GRAVITY VALUES ON POLYGONS

INTRODUCTION

The gravitational field of Polygons viz. Pentagon, Hexagon, Heptagon and Octagon were determined in previous articles. One of the profound observations was, the magnitude of gravitational field at corner and mid-point were mathematically same for all the above mentioned four polygons. Although the equations for gravitational field at corner and mid-point were mathematically same, it is worthy to note that they do not yield same results for every polygon. The two predominant metrics defining the gravitational field of polygon are the side length of the polygon ‘a’ and its height ‘h’. This article aims at analyzing the equations to prove that the magnitude of gravitational field for all polygons will be distinct.

ASSUMPTIONS

1. All polygons considered are regular.

2. The thickness of the polygon is negligible compared to its length and width.

3. The polygon is not under the influence of an external gravitational field.

4. The polygon is homogeneous in nature.

5. All the mass of the polygon is assumed to be concentrated at the center.

CALCULATION

Consider a Polygonal plate of side length ‘a’ [m] and mass M [Kg]. The types of polygons considered are Pentagon, Hexagon, Heptagon and Octagon respectively. From the calculations of center of gravity of polygons, it is evident that every polygon can be divided equally into isosceles triangles with the exception of Hexagon which can be divided equally into equilateral triangles.

Figure .1

Consider one such isosceles triangle of side length ‘a’, height ‘h’ and slant length ‘b’ as depicted in figure .1. Assume all polygons considered have same side length ‘a’.

In ∆ABC by simple trigonometry,

θ – Angle between side ‘b’ and side ‘a’

Angle ‘θ’ is also half of the interior angle of the polygon. Hence θ = ϕ/2 where ‘ϕ’ is the interior angle of the polygon. The interior angle of the polygon increases with the increase in the number of sides of the polygon. Hence a pentagon has smaller interior angle compared to an Octagon. From equation (2) height ‘h’ is directly proportional to side ‘a’ and tan (θ). As angle ‘θ’ increases, tan (θ) will also increase. This implies that height ‘h’ would change for every polygon despite the same side ‘a’ since each polygon has different interior angle. 

The magnitude of gravitational field at corner (Point A) and mid-point (Point B) are as follows:

The gravitational field magnitude only depends on side length ‘a’ and height ‘h’. Since it is proved that height ‘h’ cannot be the same for every polygon, it is clear that gravitational field magnitude for every polygon at both corner and mid-point will not be the same despite the same equation.

INSIGHTS

The two most important observations regarding the dependency of height ‘h’ on the gravitational field magnitude are as follows:

•     For a given polygon of side length ‘a’, the height ‘h’ will always be different for every type of polygon.
•     The height ‘h’ is proportional to tangent of the angle of the triangle. Hence ‘h’ will increase as the angle increases or in other words ‘h’ will be greater for multi sided polygons assuming constant length ‘a’ for every polygon.