11. Physics | Gravitation | Gravitational Field Strength due to a Uniform Ring | Ashish Arora (GA)

11. Physics | Gravitation | Gravitational Field Strength due to a Uniform Ring | Ashish Arora (GA)


let us discuss gravitational field strength
due to a uniform ring . again we will discuss it in 2 cases , if we talk about the case
1 , at a point . at centre of ring , if we talk about the gravitational field, at the
centre of ring , we can say we are given with a uniform ring . that means the total mass
of the ring is distributed uniformly at its sircumference which is of radius, say , r
, and if we talk about the centre of ring we can directly state that gravitational field
at the centre of ring is equal to zero . because for any elemental mass dm if we wish
to find the gravitational field at the centre it will be directed towards the dm , say it
is dg, and corresponding to this dm there always exist an element which is opposite
to this which will produce gravitational field , that is dg , in opposite direction and they
will cancel out , so at all points in a ring there exist , an opposite point diametrically
opposite point on the ring which will cancel the gravitational field, due to one point
. so we can say the net effect of all, the point
masses , at centre will be equal to zero so , the net gravitational field strength at
the centre of ring will be zero, similarly if we talk about, its second case . at an
axial point of ring .
and in this situation if we draw a 3 dimensional picture , say we are having a ring . which
is of mass m, and, the ring is placed along x and y axis , here y axis is vertical and
x axis is horizontally outward . and along z axis, there exist a point p .
located, at a distance z from the centre of the ring and we are required to find the gravitational
field strength due to the ring at point p, then in this situation , we can state that
we will consider a small element of length d l on the ring, due to which we find the
gravitational field at p , and due to this it will be directed towards this elemental
mass , say it is dg , then if we find the mass of this element dm , mass its elemental
mass dm can be written as m by 2 pie r into d l .
as the total mass of ring is distributed throughout its length that is its sircumference of length
2 pie r, so in this situation we can write , gravitational field strength . at p , due
to dm is , in this situation this dg is given as, g dm by , here we have to divide by the
square of this distance which is root of r square + z square , so it is r square + , z
square , this is the value of , gravitational field at p due to dm and if this angle is
alpha we can have 2 components of this gravitational field strength , one is d g coz alpha, and
other is in perpendicular direction to z axis which is d g , sine alpha .
so in this situation, if we just integrate , this elemental gravitational field for this
whole length, here we can state this d g sine alpha will cancel out each other because corresponding
to each element , there will be diametrically opposite element for which the normal components
will cancel each other , so here we can state net gravitational field strength .
at point p is , this is given as g at point p which is integration of dg coz alpha, if
we substitute the values this will be integration of g , dm we write as m by 2 pie r , multiplied
by d l, this is, r square + z square , multiplied by , coz alpha we can substitute as z by root
of r square, + , z square . and we will integrate this for the whole length
of the sircumference that is from zero to 2 pie r , integrating this the only variable
here is dl , so this dl will be integrated and this 2 pie r will cancel out and the result
we get will be , g m z by , r square , + z square to power, 3 by 2 , this is the result
we get , as the final result of the gravitational field strength at the axial point of a ring
located at a distance z from, centre of the ring .

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