Gravitational field of a Circular sector plate

GRAVITATIONAL FIELD OF A CIRCULAR SECTOR 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.

(Eq. 1)

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 Circular sector 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 a Circular Sector plate of radius ‘r’ [m], angle ‘α’ [radian] 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

The center of gravity of a regular Sector is (x,y) ≡ (2rsinα/3α, 0)

Figure .1 Sector of a Circle

Points of interest and their distances from center

We will consider three points A, B and O as shown in figure 1.

Point A ≡ (r, 0)

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

Point B ≡ (r, rsinα)

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

Point O ≡ (0, 0)

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

Gravitational field

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

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)

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,

G field at point O

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 the Circular Sector plate. On performing numerical calculation to solve the coefficients of g field equations, point A turns out to be the point of strongest gravitational field. Point O follows point A in terms of field strength while point B has the weakest gravitational field due to its large distance from the center of gravity.