GRAVITATIONAL FIELD OF A SEMI-CIRCULAR PLATE

GRAVITATIONAL FIELD OF A SEMI-CIRCULAR 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 Semi-circular 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 Semi-circular plate of radius ‘r’ [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.

Figure .1 Semi-circular plate

Center of gravity

The center of gravity of a regular Semi-circle of radius ‘r’ is (x,y) ≡ (r, 4r/3π)

Points of interest and their distances from center

We will consider three points A, B and D 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 ≡ (2r, 0)

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

Point D ≡ (r, r)

The distance between Point D 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 D

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 a Semi-circular plate. Based on numerical calculation of the coefficient of the g terms we obtain point A as the point of strongest gravity field while point B is the weakest. Point D has a gravitational field 5 times less intense compared to point A.

Gravitational field of a Trapezium Plate

GRAVITATIONAL FIELD OF A TRAPEZIUM 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 Trapezium 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 Trapezium plate of upper side length ‘a’ [m], lower side length ‘b’ [m], height ‘h’ [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.

Figure.1 Trapezium Plate

Center of gravity
The center of gravity of a regular Trapezium with the dimensions as mentioned above is,

Points of interest and their distances from center

We will consider six points namely point O, A, B, D, E and F respectively.

Point O ≡ (0, 0) [The origin]

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

Point A ≡ (b/2, 0) [The midpoint of lower side]

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

Point B ≡ (b, 0) [The lower right corner of trapezium]

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

Point D ≡ ({b/2+a/2}, h) [The upper right corner of trapezium]

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

Point E ≡ (b/2, h) [The midpoint of upper side]

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

Point F ≡ ({b/2 - a/2}, h) [The upper left corner of trapezium]

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

From above calculations, we observe that distance OG = BG and distance DG = FG, which implies that the gravitational field at these points will be equal. Hence we will only calculate gravitational field at points OG and DG respectively.

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 O

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 A

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

G field at point D

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

G field at point E

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

CONCLUSION

We thus determined the magnitude of gravitational field at six points on a Trapezium plate. On performing numerical computation we obtain Point E with the maximum gravitational field followed by point A which is comparable to point E. Point D and O have very weak values with point O having the least value. This is evident from the fact that points E and A are closer to the center of gravity as compared to points like D and O.

Gravitational field of an Elliptic plate

GRAVITATIONAL FIELD OF AN ELLIPTIC 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 Elliptic 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 Elliptic plate of major axis length ‘2a’ [m], minor axis length ‘2b’ [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.

Fig 1 Ellipse plate

Center of gravity

The center of gravity of a regular Ellipse with the dimensions as mentioned above is,

(a, b) ≡ (a, 0)

Points of interest and their distances from center

We will consider two points namely the tip of major and minor axis.

Point A ≡ (2a, 0) [The tip of major axis]

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

 

Point B ≡ (a, b) [The tip of minor axis]

The distance between Point B 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,

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

CONCLUSION

We thus determined the magnitude of gravitational field at two points on an Elliptical plate. We observe that gravitational field is stronger at point B than at point A. This is because point B which is on the minor axis is closer to the center than point A which lies on the major axis. This implies that point B is attracted more toward the center because of its closer distance than point A which is far away from the center.

GRAVITATIONAL FIELD OF A TRIANGLULAR PLATE

GRAVITATIONAL FIELD OF A TRIANGLULAR 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 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 a 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

Fig 1 Triangle plate

The center of gravity of an Isosceles Triangle of base ‘b’ and height ‘h’ is (x,y) ≡ (b2, h/3)

Points of interest and their distances from center

We will consider two points namely the corner and midpoint of a side.

Point A ≡ (0, 0) or (b, 0)

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

 

Point B ≡ (b/2, 0)

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

Point D ≡ (b/2, h)

The distance between Point D 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 D

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 a Triangle plate. We observe that gravitational field is strongest at point B, less intense at point D and weakest at point A. This is because point B is closest to the centroid while point D and A are farther away. This implies that point B is attracted more toward the center because of its closer distance than point A or point D which is slightly far away from the center.