Gravitational field lines of Hyperbola

GRAVITATIONAL FIELD LINES OF A HYPERBOLA

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 hyperbolic arc or a hyperbolic 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 hyperbola of major axis ‘a’ [m], minor axis ‘b’ [m] and mass ‘m’ [Kg].

Fig .1 A hyperbola

There is a 3 step procedure to determine the field lines

Step1 Determine differential equation of Hyperbola and its slope

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

Rearranging the above equation,

Differentiating equation (2) with respect to x, 

Equation (3) represents the slope of equation (1) [Hyperbola]. Thus an orthogonal trajectory to the Hyperbola 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 a Hyperbola. The orthogonal trajectories which represent hyperbolas are pictorially represented in figure 2. It is interesting to note that both ellipse and hyperbola have the same orthogonal trajectory which is the hyperbola.

REPRESENTATION

The graph of a Hyperbola and its orthogonal trajectories are represented in Figure 2. The hyperbolas itself which are orthogonal trajectories to the hyperbola appear emanating outward and perpendicular to the parent hyperbola. Thus a hyperbola or a hyperbolic plate will have its gravitational field directed as hyperbolas.

Fig .2 Gravitational field lines of Hyperbola

EXPLANATION

Since Hyperbola 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 perimeter of Hyperbola. From figure 2, it is evident that the field lines are non-uniform since they are not parallel to each other. The non-uniformity stems from the curvature of the curve [Hyperbola]. The field lines in general represent the direction of the field.

CONCLUSION

The final equation (5) implies that the orthogonal trajectory of Hyperbola 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 Hyperbola.