Gravitation  (Page 3/3)

An extended body is considered to be continuous aggregation of elements, which can be treated as particles. This fact can be represented by an integral of all elemental forces due to all such elements of a body, which are treated as particles. The force on a particle due to an extended body, therefore, can be computed as :

$$\mathbf{F}=\int d\mathbf{F}$$

where integration is evaluated to include all mass of a body.

Examples

Problem 1: Three identical spheres of mass “M” and radius “R” are assembled to be in contact with each other. Find gravitational force on any of the sphere due to remaining two spheres. Consider no external gravitational force exists.

Solution : The gravitational forces due to pairs of any two speres are equal in magnitude, making an angle of 60° with each other. The resultant force is :

$$\Rightarrow R=\sqrt{\left(F^2+F^2+2F^2\cos {60}^0\right)}$$

$$\Rightarrow R=\sqrt{\left(2F^2+2F^{2}X\frac{1}{2}\right)}$$

$$\Rightarrow R=\sqrt{3}F$$

Now, the distance between centers of mass of any pair of spheres is “2R”. The gravitational force is :

$$F=\frac{G{M}^2}{{\left(2R\right)}^2}=\frac{G{M}^2}{4R^2}$$

Therefore, the resultant force on a sphere is :

$$F=\frac{\sqrt{3}G{M}^2}{4R^2}$$

Problem 2: Two identical spheres of uniform density are in contact. Show that gravitational force is proportional to the fourth power of radius of either sphere.

Solution : The gravitational force between two spheres is :

$$F=\frac{G{m}^2}{{\left(2r\right)}^2}$$

Now, mass of each of the uniform sphere is :

$$m=\frac{4\pi r^{3}X\rho }{3}$$

Putting this expression in the expression of force, we have :

$$\Rightarrow F=\frac{GX16{\pi }^2{r}^6{\rho }^2}{9X4r^2}=\frac{Gx16{\pi }^2{r}^4{\rho }^2}{36}$$

Since all other quantities are constants, including density, we conclude that gravitational force is proportional to the fourth power of radius of either sphere,

$$\Rightarrow F\propto r^4$$

Measurement of universal gravitational constant

The universal gravitational constant was first measured by Cavendish. The measurement was an important achievement in the sense that it could measure small value of “G” quite accurately.

The arrangement consists of two identical small spheres, each of mass “m”. They are attached to a light rod in the form of a dumb-bell. The rod is suspended by a quartz wire of known coefficient of torsion “k” such that rod lies in horizontal plane. A mirror is attached to quartz wire, which reflects a light beam falling on it. The reflected light beam is read on a scale. The beam, mirror and scale are all arranged in one plane.

The rod is first made to suspend freely and stabilize at an equilibrium position. As no net force acts in the horizontal direction, the rod should rest in a position without any torsion in the quartz string. The position of the reflected light on the scale is noted. This reading corresponds to neutral position, when no horizontal force acts on the rod. The component of Earth's gravitation is vertical. Its horizontal component is zero. Therefore, it is important to keep the plane of rotation horizontal to eliminate effect of Earth’s gravitation.

Two large and heavier spheres are, then brought across, close to smaller sphere such that centers of all spheres lie on a circle as shown in the figure above. The gravitational forces due to each pair of small and big mass, are perpendicular to the rod and opposite in direction. Two equal and opposite force constitutes a couple, which is given by :

$${\tau }_G=F_{G}L$$

where “L” is the length of the rod.

The couple caused by gravitational force is balanced by the torsion in the quartz string. The torque is proportional to angle “θ” through which the rod rotates about vertical axis.

$${\tau }_T=k\theta$$

The position of the reflected light is noted on the scale for the equilibrium. In this condition of equilibrium,

$$\Rightarrow F_{G}L=k\theta$$

Now, the expression of Newton’s law of gravitation for the gravitational force is:

$$F_G=\frac{GMm}{r^2}$$

where “m” and “M” are mass of small and big spheres. Putting this in the equilibrium equation, we have :

$$\Rightarrow \frac{GMmL}{r^2}=k\theta$$

Solving for “G”, we have :

$$\Rightarrow G=\frac{r^{2}k\theta }{MmL}$$

In order to improve accuracy of measurement, the bigger spheres are, then, placed on the opposite sides of the smaller spheres with respect to earlier positions (as shown in the figure below). Again, position of reflected light is noted on the scale for equilibrium position, which should lie opposite to earlier reading about the reading corresponding to neutral position.

The difference in the readings (x) on the scale for two configurations of larger spheres is read. The distance between mirror and scale (y) is also determined. The angle subtended by the arc “x” at the mirror is twice the angle through which mirror rotates between two configurations. Hence,

$$\Rightarrow 4\theta =\frac{x}{y}$$

$$\Rightarrow \theta =\frac{x}{4y}$$

We see here that beam, mirror and scale arrangement enables us to read an angle, which is 2 times larger than the actual angle involved. This improves accuracy of the measurement. Putting the expression of angle, we have the final expression for determination of “G”,

$$G=\frac{r^{2}kx}{4MmLy}$$