4.8 Gravitational potential due to rigid body  (Page 2/2)

(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 potential due to the elemental ring. For this, we employ the formulation derived earlier,

$$dV=-\frac{Gdm}{y}$$

Putting expression for elemental mass,

$$\Rightarrow dV=-\frac{GM\sin \alpha d\alpha }{2y}$$

(i) Set up integral for the whole disc

$$\Rightarrow V=-GM\int \frac{\sin \alpha d\alpha }{2y}$$

Clearly, we need to express variables in one variable “x”. From triangle, OAP,

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

Differentiating each side of the equation,

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

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

Replacing expression in the integral,

$$\Rightarrow V=-GM\int \frac{dy}{2ar}$$

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 ( $L_2$ ),

$$\Rightarrow V=-\frac{GM}{2ar}[y\underset{L_1}{\overset{L_2}{]}}$$

Case 1: gravitational potential at a point outside

The total gravitational field is obtained by integrating the integral from y = r-a to y = r+a,

$$\Rightarrow V=-\frac{GM}{2ar}[y\underset{r-a}{\overset{r+a}{]}}$$

$$\Rightarrow V=-\frac{GM}{2ar}\left[\right[r+a-r+a\left]\right]=-\frac{GM}{2ar}2a$$

$$\Rightarrow V-\frac{GM}{r}$$

This is an important result. It again brings the fact that a spherical shell, for a particle outside it, behaves as if all its mass is concentrated at its center. In other words, a spherical shell can be considered as particle for an external point.

Check : We can check the relationship of potential, using differential equation that relates gravitational potential and field strength.

$$E=-\frac{dV}{dr}=\frac{d}{dr}\frac{GM}{r}$$

$$\Rightarrow E=-GMX-1X{r}^{-2}$$

$$\Rightarrow E=-\frac{GM}{r^2}$$

The result is in excellent agreement with the expression derived for gravitational field strength outside a spherical shell.

Case 2: gravitational potential at a point inside

The total gravitational field is obtained by integrating the integral from y = a-r to y = a+r,

$$\Rightarrow V=-\frac{GM}{2ar}[y\underset{a-r}{\overset{a+r}{]}}$$

We can see here that "a-r" involves mass of the shell to the right of the point under consideration, whereas "a+r" involves mass to the left of it. Thus, total mass of the spherical shell is covered by the limits used. Now,

$$\Rightarrow V=-\frac{GM}{2ar}[a+r-a+r]=-\frac{GM}{2ar}2r=-\frac{GM}{a}$$

The gravitational potential is constant inside the shell and is equal to the potential at its surface. The plot of gravitational potential for spherical shell is shown here as we move away from the center.

Check : We can check the relationship of potential, using differential equation that relates gravitational potential and field strength.

$$E=-\frac{dV}{dr}=-\frac{d}{dr}\frac{GM}{a}$$

$$\Rightarrow E=0$$

The result is in excellent agreement with the result obtained for gravitational field strength inside a spherical shell.

Gravitational potential due to uniform solid sphere

The uniform solid sphere of radius “a” and mass “M” can be considered to be composed of infinite numbers of thin spherical shells. We consider one such thin spherical shell of infinitesimally small thickness “dx” as shown in the figure.

Case 1 : the point lies outside the sphere

In this case, potential due to elemental spherical shell is given by :

$$dV=-\frac{Gdm}{r}$$

In this case, most striking point is that the centers of all spherical shells are coincident at center of sphere. This means that linear distance between centers of spherical shells and the point of observation is same for all shells. In turn, we conclude that the term “r” is constant for all spherical shells and as such can be taken out of the integral,

$$\Rightarrow V=-\frac{G}{r}\int dm$$

$$\Rightarrow V=-\frac{GM}{r}$$

Case 2 : the point lies inside the sphere

We calculate potential in two parts. For this we consider a concentric smaller sphere of radius “r” such that point “P” lies on the surface of sphere. Now, the potential due to whole sphere is split between two parts :

$$V=V_S+V_R$$

Where " $V_S$ " denotes potential due to solid sphere of radius “r” and " $V_S$ " denotes potential due to remaining part of the solid sphere between x = r and x = a. The potential due to smaller sphere is :

$$\Rightarrow V_S=-\frac{GM\prime }{r}$$

The mass, “M’” of the smaller solid sphere is :

$$M\prime =\frac{3M}{4\pi a^3}X\frac{4\pi r^3}{3}=\frac{M{r}^3}{a^3}$$

Putting in the expression of potential, we have :

$$\Rightarrow V_S=-\frac{GM{r}^2}{a^3}$$

In order to find the potential due to remaining part, we consider a spherical shell of thickness “dx” at a distance “x” from the center of sphere. The shell lies between x = r and x =a. The point “P” is inside this thin shell. As such potential due to the shell at point “P” inside it is constant and is equal to potential at the spherical shell. It is given by :

$$\Rightarrow d{V}_R=-\frac{Gdm}{x}$$

We need to calculate the mass of the thin shell,

$$\Rightarrow M\prime =\frac{3M}{4\pi a^3}X4\pi x^{2}dx=\frac{3M{x}^{2}dx}{a^3}$$

Substituting in the expression of potential,

$$\Rightarrow d{V}_R=-\frac{G3M{x}^{2}dx}{a^{3}x}$$

We integrate the expression for obtaining the potential at “P” between limits x = r and x =a,

$$\Rightarrow V_R=-\frac{3GM}{a^3}\underset{r}{\overset{a}{\int }}xdx$$

$$\Rightarrow V_R=-\frac{3GM}{a^3}[\frac{x^2}{2}\underset{r}{\overset{a}{]}}$$

$$\Rightarrow V_R=-\frac{3GM}{2a^3}\left(a^2-r^2\right)$$

Adding two potentials, we get the expression of potential due to sphere at a point within it,

$$\Rightarrow V_R=-\frac{GM{r}^2}{a^3}-\frac{3GM}{2a^3}\left(a^2-r^2\right)$$

$$\Rightarrow V_R=-\frac{GM}{2a^3}\left(3a^2-r^2\right)$$

This is the expression of gravitational potential for a point inside solid sphere. The potential at the center of sphere is obtained by putting r = 0,

$$\Rightarrow V_C=-\frac{3GM}{2a}$$

This may be an unexpected result. The gravitational field strength is zero at the center of a solid sphere, but not the gravitational potential. However, it is entirely possible because gravitational field strength is rate of change in potential, which may be zero as in this case.

The plot of gravitational potential for uniform solid sphere is shown here as we move away from the center.