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

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

$$\Rightarrow E=\frac{GM}{4a{r}^2}[r+a+\frac{a^2-r^2}{r+a}-r+a-\frac{a^2-r^2}{r-a}]$$

$$\Rightarrow E=\frac{GM}{4a{r}^2}[2a+\left(a^2-r^2\right)\left(\frac{1}{r+a}-\frac{1}{r-a}\right)]$$

$$\Rightarrow E=\frac{GM}{4a{r}^2}X4a$$

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

This is an important result. We have been using this result by the name of Newton’s shell theory. According to this theory, a spherical shell, for a particle outside it, behaves as if all its mass is concentrated at its center. This is how we could calculate gravitational attraction between Earth and an apple. Note that radius of the shell, “a”, does not come into picture.

Case 2 : The point “P” lies outside the shell. The total gravitational field is obtained by integrating the integral from x = a-r to x = a+r,

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

$$\Rightarrow E=\frac{GM}{4a{r}^2}[a+r+\frac{a^2-r^2}{a+r}-a+r-\frac{a^2-r^2}{a-r}]$$

$$\Rightarrow E=\frac{GM}{4a{r}^2}[2r+\left(a^2-r^2\right)\left(\frac{1}{a+r}-\frac{1}{a-r}\right)]$$

$$\Rightarrow E=\frac{GM}{4a{r}^2}[2r-2r]=0$$

This is yet another important result, which has been used to determine gravitational acceleration below the surface of Earth. The mass residing outside the sphere drawn to include the point below Earth’s surface, does not contribute to gravitational force at that point.

The mass outside the sphere is considered to be composed of infinite numbers of thin shells. The point within the Earth lies inside these larger shells. As gravitational intensity is zero within a shell, the outer shells do not contribute to the gravitational force on the particle at that point.

A plot, showing the gravitational field strength, is shown here for regions both inside and outside spherical shell :

Gravitational field 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 spherical shell of infinitesimally small thickness “dx” as shown in the figure. The gravitational field strength due to thin spherical shell at a point outside shell, which is at a linear distance “r” from the center, is given by

$$dE=\frac{Gdm}{r^2}$$

The gravitational field strength acts along the line towards the center of sphere. As such, we can add gravitational field strengths of individual shells to obtain the field strength of the sphere. In this case, most striking point is that the centers of all spherical shells are coincident at one point. This means that linear distance between centers of spherical shell and the point ob observation is same for all shells. In turn, we can conclude that the term “ $r^2$ ” is constant for all spherical shells and as such can be taken out of the integral,

$$\Rightarrow E=\int \frac{Gdm}{r^2}=\frac{G}{r^2}\int dm=\frac{GM}{r^2}$$

We can see here that a uniform solid sphere behaves similar to a shell. For a point outside, it behaves as if all its mass is concentrated at its center. Note that radius of the sphere, “a”, does not come into picture. Sphere behaves as a point mass for a point outside.

Gravitational field at an inside point

We have already derived this relation in the case of Earth.

For this reason, we will not derive this relation here. Nevertheless, it would be intuitive to interpret the result obtained for the acceleration (field strength) earlier,

$$\Rightarrow g\prime =g_0(1-\frac{d}{R})$$

Putting value of “g0” and simplifying,

$$\Rightarrow g\prime =\frac{GM}{R^2}\left(1-\frac{d}{R}\right)=\frac{GM}{R^2}\left(\frac{R-d}{R}\right)=\frac{GMr}{R^3}$$

As we have considered “a” as the radius of sphere here – not “R” as in the case of Earth, we have the general expression for the field strength insider a uniform solid sphere as :

$$\Rightarrow E=\frac{GMr}{a^3}$$

The field strength of uniform solid sphere within it decreases linearly within “r” and becomes zero as we reach at the center of the sphere. A plot, showing the gravitational field strength, is shown here for regions both inside and outside :