4.5 Gravitational field due to rigid bodies  (Page 2/3)

$$\Rightarrow E=\int dE\cos \theta$$

$$\Rightarrow E=\int \frac{Gdm\cos \theta }{\left(a^2+r^2\right)}$$

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

$$\Rightarrow E=\frac{G\cos \theta }{\left(a^2+r^2\right)}\int dm$$

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

$$\Rightarrow E=\frac{GM\cos \theta }{\left(a^2+r^2\right)}$$

From triangle OAP,

$$\Rightarrow \cos \theta =\frac{r}{{\left(a^2+r^2\right)}^{\frac{1}{2}}}$$

Substituting for “cosθ” in the equation ,

$$\Rightarrow E=\frac{GMr}{{\left(a^2+r^2\right)}^{\frac{3}{2}}}$$

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,

$$\frac{dE}{dr}=0$$

This yields,

$$\Rightarrow r=\frac{a}{\sqrt{2}}$$

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

$$\Rightarrow E_{\max }=\frac{GMa}{2^{\frac{1}{2}}{\left(a^2+\frac{a^2}{2}\right)}^{\frac{3}{2}}}=\frac{GMa}{3\sqrt{3}{a}^2}$$

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

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 :

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

$$dm=\frac{M}{4\pi a^2}X2\pi a\sin \alpha dx=\frac{Ma\sin \alpha dx}{2a^2}$$

From the figure, we see that :

$$dx=ad\alpha$$

Putting these expressions,

$$\Rightarrow dm=\frac{Ma\sin \alpha dx}{2a^2}=\frac{Ma\sin \alpha ad\alpha }{2a^2}=\frac{M\sin \alpha d\alpha }{2}$$

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

$$\Rightarrow dE=\frac{Gdm\cos \theta }{{AP}^2}$$

Putting expression for elemental mass,

$$\Rightarrow dE=\frac{GM\sin \alpha d\alpha \cos \theta }{2y^2}$$

(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.

$$\Rightarrow E=GM\int \frac{\sin \alpha \cos \theta d\alpha }{2y^2}$$

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

$$\Rightarrow y^2=a^2+r^2-2ar\cos \alpha$$

Differentiating each side of the equation,

$$\Rightarrow 2ydy=2ar\sin \alpha d\alpha$$

$$\Rightarrow \sin \alpha d\alpha =\frac{ydy}{ar}$$

Again from triangle OAP,

$$\Rightarrow a^2=y^2+r^2-2yr\cos \theta$$

$$\Rightarrow \cos \theta =\frac{y^2+r^2-a^2}{2yr}$$

Putting these values in the integral,

$$\Rightarrow E=GM\int \frac{dy\left(y^2+r^2-a^2\right)}{4a{r}^2{y}^2}$$

$$\Rightarrow E=GM\int \frac{dy}{4a{r}^2}\left(1-\frac{a^2-r^2}{y^2}\right)$$

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 ( $L_2$ ),

$$\Rightarrow E=GM\underset{L_1}{\overset{L_2}{\int }}\frac{dy}{4a{r}^2}\left(1-\frac{a^2-r^2}{y^2}\right)$$

$$\Rightarrow E=\frac{GM}{4a{r}^2}[y+\frac{a^2-r^2}{y}\underset{L_1}{\overset{L_2}{]}}$$

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,