GRAVITATIONAL FIELD OF AN ELLIPTICAL SECTOR PLATE

GRAVITATIONAL FIELD OF AN ELLIPTICAL 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 Elliptical 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 an Elliptical Sector plate of major axis ‘a’ [m] and minor axis ‘b’ [m], angle ‘θ’ [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

We already determined the center of gravity of sector of regular ellipse in the previous post. The center of gravity of a regular Elliptical Sector is,

Figure .1 Sector of an ellipse

Points of interest and their distances from center

Consider an arbitrary point X ≡ (x,y) anywhere on the rim of the sector. We intend to determine the magnitude of gravitational field at point X. It is important to note that point X must lie on rim of the sector. If it lies inside the sector, we have to subtract the offset distance from the rim to determine the accurate value of gravitational field.

The distance between point X 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 = 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)

G field at point X

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

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

CONCLUSION

We determined the magnitude of gravitational field at one arbitrary point X. From the equation that we derived, we can determine the magnitude of g field at any point of interest as long as we know the coordinate and its distance from the center of gravity.