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 ek 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.
