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Gravitational field due to a ring

Perpendicular components cancel each other.

E = E cos θ

E = G m cos θ a 2 + r 2

The trigonometric ratio “cosθ” is a constant for all points on the ring. Taking out cosine ratio and other constants from the integral,

E = G cos θ a 2 + r 2 m

Integrating for m = 0 to m = M, we have :

E = G M cos θ a 2 + r 2

From triangle OAP,

cos θ = r a 2 + r 2 1 2

Substituting for “cosθ” in the equation ,

E = G M r a 2 + r 2 3 2

For r = 0, E = 0. The gravitation field at the center of ring is zero. This result is expected also as gravitational fields due to two diametrically opposite equal elemental mass are equal and opposite and hence balances each other.

Position of maximum gravitational field

We can get the maximum value of gravitational field by differentiating its expression w.r.t linear distance and equating the same to zero,

E d r = 0

This yields,

r = a 2

Substituting in the expression of gravitational field, the maximum field strength due to a circular ring is :

E max = G M a 2 1 2 a 2 + a 2 2 3 2 = G M a 3 3 a 2

The plot of gravitational field with axial distance shows the variation in the magnitude,

Gravitational field due to a ring

The gravitational field along the axial line.

Gravitational field due to thin spherical shell

The spherical shell of radius “a” and mass “M” can be considered to be composed of infinite numbers of thin rings. We consider one such ring of infinitesimally small thickness “dx” as shown in the figure. We derive the required expression following the sequence of steps as outlined here :

Gravitational field due to thin spherical shell

The gravitational field is measured on axial point "P".

(i) Determine mass of the elemental ring in terms of the mass of shell and its surface area.

m = M 4 π a 2 X 2 π a sin α x = M a sin α x 2 a 2

From the figure, we see that :

x = a α

Putting these expressions,

m = M a sin α x 2 a 2 = M a sin α a α 2 a 2 = M sin α α 2

(ii) Write expression for the gravitational field due to the elemental ring. For this, we employ the formulation derived earlier for the ring,

E = G m cos θ A P 2

Putting expression for elemental mass,

E = G M sin α α cos θ 2 y 2

(v) Set up integral for the whole disc

We see here that gravitational fields due to all concentric rings are directed towards the center of spherical shell along the axis.

E = G M sin α cos θ α 2 y 2

The integral expression has three varibles "α","θ" and "y".Clearly, we need to express variables in one variable “x”. From triangle, OAP,

y 2 = a 2 + r 2 2 a r cos α

Differentiating each side of the equation,

2 y y = 2 a r sin α α

sin α α = y y a r

Again from triangle OAP,

a 2 = y 2 + r 2 2 y r cos θ

cos θ = y 2 + r 2 a 2 2 y r

Putting these values in the integral,

E = G M y y 2 + r 2 a 2 4 a r 2 y 2

E = G M y 4 a r 2 1 a 2 r 2 y 2

We shall decide limits of integration on the basis of the position of point “P” – whether it lies inside or outside the shell. Integrating expression on right side between two general limits, initial ( L 1 ) and final ( L 2 ),

E = G M L 1 L 2 y 4 a r 2 1 a 2 r 2 y 2

E = G M 4 a r 2 [ y + a 2 r 2 y ] L 1 L 2

Evaluation of integral for the whole shell

Case 1 : The point “P” lies outside the shell. The total gravitational field is obtained by integrating the integral from y = r-a to y = r+a,

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Source:  OpenStax, Physics for k-12. OpenStax CNX. Sep 07, 2009 Download for free at http://cnx.org/content/col10322/1.175
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