Gravitational Field of an Equilateral triangle plate

GRAVITATIONAL FIELD OF AN EQUILATERAL TRIANGLE PLATE

INTRODUCTION

Gravity is derived from the Latin word ‘gravitas’ meaning mass. The universal law of gravitation was coined by Sir Isaac Newton. According to the law, any two masses anywhere in the universe separated by a distance will attract each other. This force of attraction is proportional to the product of their masses and inversely proportional to square of the distance between them.

The distance between two masses can be finite or infinite, which is why gravitational force is referred to as long range force but is also the weakest force among all the other fundamental forces. All objects that have mass will attract other masses. This means that each mass has its own gravitational field just like Earth. So this implies that all objects will attract each other since they will have their own field. This is not evident on Earth since Earth’s gravitational field outweighs all other mass’s field and hence all objects no matter how massive are attracted toward the Earth. In this post we intend to determine the gravitational field of a two dimensional Equilateral Triangular plate and identify points where the plate’s own gravity is strong at some parts and weak at the other.  

ASSUMPTIONS

1. The thickness of the plate is negligible compared to its length and width.

2. The plate is not under the influence of an external gravitational field.

3. The plate is a homogeneous material.

4. All the mass is assumed to be concentrated at the center.

CALCULATION

Consider an Equilateral Triangle plate of length ‘a’ [m] and mass M [Kg]. We will first determine the center of gravity of this plate and then the magnitude of the gravitational field at points of interest.

Center of gravity

Fig1 Equilateral Triangle

The center of gravity of an Equilateral Triangle of side ‘a’ and height ‘h’ is (x,y) ≡ (a/2, h/3) ≡ [a/2, (a/2√3)]

Points of interest and their distances from center

We will consider three points namely Point A, B and D as shown in figure 1. Unlike a regular isosceles or right angled triangle, this one is an equilateral triangle, hence the distance AG must be equal to DG which will be evident from the calculations.

Point A ≡ (0, 0) or (a, 0)

The distance between Point A and C.G. can be calculated by the distance formula

Point B ≡ (a/2, 0)

The distance between Point B and C.G. can be calculated by the distance formula

Point D ≡ (b/2, h) ≡ [b/2, (√3/2)*a]

The distance between Point D and C.G. can be calculated by the distance formula

As we can observe, from the above equations, the distances AG and DG when solved are exactly the same.

Gravitational field

From Eq. 1, the force of attraction between a mass and its own surface is given by,

g = GM/R²

g – Acceleration due to gravity of the mass (m/s²)

G – Universal constant of gravitation

G = 6.67*10^-11 Nm²/Kg²

M – Mass of the object (Kg)

R – Distance between the centers of two masses (m)

Since distances AG and DG are same we will determine the gravitational field equations at points A and B.

G field at point A

The gravitational field beyond the surface is obtained by adding the additional distance,

d – Distance from the surface of the object to other object.

G field at point B

The gravitational field beyond the surface is obtained by adding the additional distance,

CONCLUSION

We thus determined the magnitude of gravitational field at various points on an Equilateral Triangle plate. We observe that gravitational field is strongest at point B and weak at points A and D. The gravitational field is exactly 4 times stronger at point B relative to the corner points like A and D. This is because point B is at a distance of h/3 from the centroid of the triangle while points A and D are located at 2h/3 from the centroid. This implies that point B is attracted more toward the center because of its closer distance than point A or point D which are far away from the center.