Gravitational field lines of an Ellipse

GRAVITATIONAL FIELD LINES OF AN ELLIPSE

INTRODUCTION

The force of gravity as described by Sir Isaac Newton is a force of attraction. The gravitational force acting on a body of mass ‘m’ is equal to the product of its mass ‘m’ and the acceleration due to gravity acting on that body. Now Force is a vector quantity since it is a product of mass [scalar] and acceleration [vector]. By definition a scalar quantity can be represented by magnitude alone whereas a vector quantity must be represented by magnitude and direction. Thus while solving a vector quantity, it is important to obtain both magnitude and direction to obtain complete information of the vector. Acceleration due to gravity being a vector quantity has both magnitude and direction. The direction of gravitational force is termed as gravitational field lines or orthogonal trajectories which are always orthogonal [perpendicular] to the surface as explained in detail in the calculation section. This article intends to determine the gravitational field lines of a two dimensional elliptical plate.

ASSUMPTIONS

1. The plate has negligible thickness hence is assumed to be two dimensional

2. The plate is homogeneous in nature meaning composed of only one material

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

4. The whole mass of the plate is assumed to be concentrated in the center

CALCULATION

Consider a two dimensional elliptical plate of major axis ‘a’ [m], minor axis ‘b’ [m] and mass ‘m’ [Kg]. Let the eccentricity of ellipse be around 1.2 which implies an ellipse with major axis along x axis and a minor axis along y axis.

Fig .1 An Ellipse

There is a 3 step procedure to determine the field lines

Step1 Determine differential equation of ellipse and its slope

The equation of a regular Ellipse with major and minor axes ‘a’ and ‘b’ is,

Rearranging the above equation,

Differentiating equation (1) with respect to x, 

Equation (3) represents the slope of equation (1) [Ellipse]. Thus an orthogonal trajectory to the Ellipse must have a slope that is negative inverse of the slope in equation (3).

Step2 Determine the slope of the orthogonal trajectory

Thus the new slope or the slope of the orthogonal trajectory which is a negative inverse of equation (3) is

Step3 Determine equation of the orthogonal trajectory by integration

To obtain the equation of the orthogonal trajectory, integrate equation (4) by separating the variables. Rearranging equation (4)

{‘k’ is a constant of integration}

Equation (5) represents the family of orthogonal trajectories of an ellipse. The orthogonal trajectories which represent hyperbolas are pictorially represented in figure 2.

REPRESENTATION

The graph of an Ellipse and its orthogonal trajectories are represented in Figure 2. The hyperbolas which are orthogonal trajectories to the Ellipse appear emanating from the Ellipse radially outward thereby satisfying the mathematical condition. Thus an elliptical plate will have its gravitational field directed as hyperbolas. Since an Ellipse is only symmetric about one axis x or y but not both, the gravitational field lines are not equidistant from the center and hence the magnitude of gravitational field is not the same at all points on the circumference of Ellipse. From figure2, the field lines can be interpreted as non-uniform since they are not parallel to each other. The non-uniformity stems from the curvature of the curve [Ellipse].

Fig .2 Gravitational field lines of Ellipse

EXPLANATION

The field lines to the ellipse which are represented by hyperbolas do not appear to pass through the center unlike the circle where straight lines did pass through the center. In fact asymptotes of hyperbolas which are tangents to the hyperbolas do pass through the center of ellipse. Asymptotes which are straight lines intersect the hyperbolas at only one point. Since they intersect hyperbolas, their slope is 0. Although hyperbolas themselves do not pass through the Ellipse’s center of gravity their asymptotes do pass.

CONCLUSION

The final equation (5) implies that the orthogonal trajectory of an Ellipse is family of hyperbolas. It is important to note that the constant ‘2k’ can have infinite values hence there are infinite orthogonal trajectories for a given shape in this case the Ellipse.

Gravitational field lines of a circle

GRAVITATIONAL FIELD LINES OF A CIRCLE

INTRODUCTION

The force of gravity as described by Sir Isaac Newton is a force of attraction. The gravitational force acting on a body of mass ‘m’ is equal to the product of its mass ‘m’ and the acceleration due to gravity acting on that body. Now Force is a vector quantity since it is a product of mass [scalar] and acceleration [vector]. By definition a scalar quantity can be represented by magnitude alone whereas a vector quantity must be represented by magnitude and direction. Thus while solving a vector quantity, it is important to obtain both magnitude and direction to obtain complete information of the vector. Acceleration due to gravity being a vector quantity has both magnitude and direction. The direction of gravitational force is termed as gravitational field lines or orthogonal trajectories which are always orthogonal [perpendicular] to the surface as explained in detail in the calculation section. This article intends to determine the gravitational field lines of a two dimensional circular plate.

ASSUMPTIONS

1. The plate has negligible thickness hence is assumed to be two dimensional

2. The plate is homogeneous in nature meaning composed of only one material

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

4. The whole mass of the plate is assumed to be concentrated in the center

CALCULATION

Consider a two dimensional circular plate of radius ‘r’ [m] and mass ‘m’ [Kg].

There is a 3 step procedure to determine the field lines

Step1 Determine differential equation of circle and its slope

The equation of circle with radius ‘r’ is   

Differentiating equation (1) with respect to x, 

Equation (3) represents the slope of equation (1) [Circle]. Thus an orthogonal trajectory to the Circle must have a slope that is negative inverse of the slope in equation (3).

Step2 Determine the slope of the orthogonal trajectory

The new slope or the slope of the orthogonal trajectory which is a negative inverse of equation (3) is

Step3 Determine equation of the orthogonal trajectory by integration

To obtain the equation of the orthogonal trajectory, integrate equation (4) by separating the variables. Rearranging equation (4)

On solving the above integral, the solution is  

Raising exponential to above equation, 

It is of the form  

REPRESENTATION

The graph of a circle and its orthogonal trajectories are represented in Figure 1. The family of straight lines which are orthogonal trajectories to the circle appear emanating from the circle radially thereby satisfying the mathematical condition. Thus a circular plate will have its gravitational field directed as infinite straight lines.

EXPLANATION

A circle is symmetric about both x and y axes, thereby the gravitational field lines are equidistant from the center and hence the magnitude of gravitational field is same at all points on the circumference of circle. This statement is made assuming that field lines are uniform. Although this is an assumption, it is not necessarily true. The field lines of circle being straight lines are parallel to every other neighboring lines in close proximity of the circle. The farther the lines are away from the circle, the more spread out they are, which means the field lines are non-uniform at larger distances.

Fig .1 Gravitational field lines of circle

CONCLUSION

The final equation (6) implies that the orthogonal trajectory of a circle is a straight line of slope ‘M’ where M is equal to e^k which is a constant. It is important to note that the constant ‘k’ can have infinite values hence there are infinite orthogonal trajectories for a given shape in this case the circle.

Circle as an infinite sided polygon

CIRCLE AS AN INFINITE SIDED POLYGON

INTRODUCTION

A circle is defined as a closed loop curve belonging to the conic family. It is a set of points in a plane that are at equal distances from a fixed point which is the center of the circle. This equal distance is called as the radius of the circle. A circle in fact is a special case of an ellipse i.e. an ellipse with equal major and minor axis is a circle or more mathematically an ellipse with zero eccentricity is a circle. In contrast a regular polygon is again a closed loop plane figure that is bounded by finite chain of straight line segments. The segments are the sides of the polygon. The number of segments depend on the type of polygon in question. For example a pentagon has 5 sides, hexagon has 6 sides and so on. This article aims to prove that a circle can be approximated as a polygon with large number of segments such that the center of gravity of circle can be determined by using the polygon and trigonometry without the need for the equation of the circle.

ASSUMPTIONS

1. The polygon considered is a regular polygon

2. The number of sides is a large finite value

CALCULATION

From the previous posts, it is evident that an ‘n’ sided regular polygon can always be divided into equal number of isosceles triangles. As the number of sides of the polygon increases, the number of triangles required to equally divide the polygons also increases. So this implies that an infinite sided polygon can be literally divided into infinite triangles. Thus the individual triangle as represented in figure 2 is very thin with the base angles approaching 90°.

Consider an infinite sided polygon. Infinite side signifies large number of sides. Let the number of sides be 1000. Hence this polygon can be equally divided into 1000 isosceles triangles. Figure 1 represents how a circle can be discretized into large sided polygon.

Fig .1 Circle as an infinite sided polygon

Fig .2 Thin isosceles triangle

The sum of angles of the polygon can be determined using the equation,

Hence individual angle (θ) of the polygon is equal to S/1000 which is 179.64°

Now the isosceles side of the triangle bisects individual angle of the polygon. Thus the individual angle (ϕ) of the triangle is half of the individual angle (θ) of the polygon.

In ∆ABC from figure 2, sum of interior angles is equal to 180°

Where, φ - top angle or the third angle of the triangle

Side BC = a - Base of triangle (m)

AM = h – Height of triangle (m)

AB = AC = b – Isosceles sides of triangle (m)

In order to prove that the polygon can be approximated as a circle, the height and isosceles sides of the triangle must be proved equal.

In ∆ABC, ⦤ABC = ⦤ACB = ϕ = 89.82°, ⦤BAC = φ = 0.36°

Consider ∆AMB,

Thus from above equation it is clearly evident that h is almost equal to b or in other words length AM is almost equal to length AB or AC. Therefore ‘h’ or ‘b’ is indistinguishable up to 5 decimal places.

Since it is proved that quantity b ≈ h, every point on the side of the polygon is located at a distance of ‘h’ from the center of polygon.

Consider the same ∆ABC inscribed inside the polygon. Point M has coordinates M ≡ (h, 0) while origin is located at O ≡ (0, 0). Since AB ≈ AM, BM ≪ AM, Point B will have the same coordinates as point M. Thus B ≡ (h, 0)

Test for Polygon to qualify as a circle

The equation of circle is,

r – Radius of the circle (m)

Since we assumed that a 1000 sided polygon must approximately represent a circle, we will solve the LHS of equation of circle to check if RHS is valid at point B

Neglecting negative value, we get radius r = h

Hence the polygon satisfies the condition of circle not only at point B but on any point on its perimeter.

INSIGHTS

1    

1    1. As the number of sides of the polygon increase, the individual angle of the polygon approach 180° but never equal to that.

2    2. The straight edges of the polygon form the curvature of the circle.

CONCLUSION

Thus an infinite sided polygon can be approximated as a circle with reasonable accuracy with the radius of circle equaling to height of the polygon. It is to be noted that in this example, a 1000 sided polygon was considered whereas a higher sided polygon would give more accuracy and thus best simulate a circle.

Gravitational field of an Octagonal plate

GRAVITATIONAL FIELD OF AN OCTAGONAL 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.

F = GM₁M₂/R² (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 article the gravitational field of a two dimensional Octagonal plate will be determined and points on the plate would be identified where the plate’s own gravity is strong or weak.   

ASSUMPTIONS

1. The Octagon is assumed to be a regular polygon.

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

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

4. The plate is a homogeneous material.

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

CALCULATION

Consider an Octagonal plate of side length ‘a’ [m] and mass M [Kg]. First the center of gravity of this plate will be determined, followed by the magnitude of the gravitational field at points of interest.

Center of gravity

 

Fig 1 Octagonal plate

The center of gravity of Octagon as determined from previous article is (x, y) ≡ (h, h)

Points of interest and their distances from center

Consider two points namely Point A and Point B as shown in figure 1. Point A represents the corner point while point B represents the mid-point of a side. All the other points depicted in figure 1 are at the same distances as points A and B are from the center.

Point A ≡ (a/√2, 0)

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

{Since a/√2 + a/2 = h; h – a/√2 = a/2}

Point B ≡ (a/√2 + a/2, 0) ≡ (h, 0)

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 = 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 A

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

d – Distance between point A and any point in space

G field at point B

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

d – Distance between point B and any point in space

CONCLUSION

Thus the magnitude of gravitational field at two points on an Octagonal plate was found successfully. On comparing the gravitational acceleration values at point A and B it is important to note that gravitational field is stronger at point B due to its closer distance to the center while it is relatively weaker at point A due to its greater distance from the center. This is true for every mid-point and corner point on the Octagon.

Gravitational field of a Heptagonal plate

GRAVITATIONAL FIELD OF A HEPTAGONAL 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.

F = GM₁M₂/R² (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 article the gravitational field of a two dimensional Heptagonal plate will be determined and points on the plate would be identified where the plate’s own gravity is strong or weak.   

ASSUMPTIONS

1. The Heptagon is assumed to be a regular polygon.

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

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

4. The plate is a homogeneous material.

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

CALCULATION

Consider a Heptagonal plate of side length ‘a’ [m] and mass M [Kg]. First the center of gravity of this plate will be determined, followed by the magnitude of the gravitational field at points of interest.

Center of gravity

Fig 1 Heptagonal plate

The center of gravity of Heptagon as determined from previous article is (x, y) ≡ (1.123a, h)

Points of interest and their distances from center

Consider two points namely Point A and Point B as shown in figure 1. Point A represents the corner point while point B represents the mid-point of a side. All the other points depicted in figure 1 are at the same distances as points A and B are from the center.

Point A ≡ (0.623a, 0)

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

Point B ≡ (1.123a, 0)

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 = 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 A

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

d – Distance between point A and any other point in space.

G field at point B

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

d – Distance between point B and any other point in space.

CONCLUSION

Thus the magnitude of gravitational field at two points on a Heptagonal plate was found successfully. On comparing the gravitational acceleration values at point A and B it is important to note that gravitational field is stronger at point B due to its closer distance to the center while it is relatively weaker at point A due to its greater distance from the center. This is true for every mid-point and corner point on the Heptagon.