GRAVITATIONAL
FIELD LINES OF A PARABOLA
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 parabolic 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 parabolic plate of focal length ‘a’ [m] and mass ‘m’ [Kg].
Fig .1 A parabola |
There is a 3 step procedure to determine the field lines
Step1 Determine differential equation of
Parabola and its slope
The equation of a
regular Parabola with focal length ‘a’ is,
Rearranging the above
equation,
Differentiating equation
(2) with respect to x,
Equation (3) represents
the slope of equation (1) [Parabola]. Thus an orthogonal trajectory to the Parabola
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 Parabola. The orthogonal
trajectories which represent family of ellipses are pictorially represented in
figure 2.
REPRESENTATION
The graph of a Parabola
and its orthogonal trajectories are represented in Figure 2. The ellipses which
are orthogonal trajectories to the Parabola appear emanating outward and complete
the trajectory. Thus a parabolic plate will have its gravitational field directed
as ellipses.
Fig .2 Gravitational field lines of Parabola |
EXPLANATION
Since Parabola 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 Parabola. 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
[Parabola]. The field lines in general represent the direction of the field.
CONCLUSION
The final equation (5)
implies that the orthogonal trajectory of a Parabola is family of ellipses. It
is important to note that the constant ‘m’ can have infinite values hence there
are infinite orthogonal trajectories for a given shape in this case the Parabola.
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