4.1 Gravity  (Page 4/6)

• Pseudo force ( $m{\omega }^{2}r$ )
• Normal force (N)
• gravitational force ( $m{g}_0$ )

The particle is subjected to normal force against the net force on the particle or the weight as measured. Two forces are equal in magnitude, but opposite in direction.

$$N=W=mg$$

It is worthwhile to note here that gravitational force and normal force are different quantities. The measured weight of the particle is equal to the product of mass and the measured acceleration. It is given by the expression “mg”. On the other hand, Gravitational force is given by the Newton’s equation, which considers Earth at rest. It is equal to “ $m{g}_0$ ”.

$$\Rightarrow F_G=\frac{GM}{R^2}=m{g}_0$$

Since particle is stationary on the surface of Earth, three forces as enumerated above constitute a balanced force system. Equivalently, we can say that the resultant of pseudo and gravitational forces is equal in magnitude, but opposite in direction to the normal force.

$$N=F_G+F_P$$

In other words, resultant of gravitational and pseudo forces is equal to the magnitude of measured weight of the particle. Applying parallelogram theorem for vector addition of gravitational and pseudo forces, the resultant of the two forces is :

$$N^2={\left(m{g}_0\right)}^2+{\left(m{\omega }^{2}R\right)}^2+2m^2{\omega }^2{g}_{0}R\cos \left(180-\phi \right)$$

Putting N = mg and rearranging, we have:

$$\Rightarrow m^2{g}^2=m^2{g}_0^2+m^2{\omega }^4{R}^2-2m^2{\omega }^2{g}_{0}R\cos \phi$$

Angular velocity of Earth is quite a small value. It may be interesting to know the value of the term having higher power of angular velocity. Since Earth completes one revolution in a day i.e an angle of “2π” in 24 hrs, the angular speed of Earth is :

$$\Rightarrow \omega =\frac{2\pi }{24X60X60}=7.28X{10}^{-5}\mathrm{rad}/s$$

The fourth power of angular speed is almost a zero value :

$$\Rightarrow {\omega }^4=2.8X{10}^{-17}$$

We can, therefore, safely neglect the term “ $m^2{\omega }^4{r}^2$ ”. The expression for the measured weight of the particle, therefore, reduces to :

$$\Rightarrow g^2=g_0^2-2{\omega }^2{g}_{0}R\cos \phi$$

$$\Rightarrow g=g_0{\left(1-\frac{2{\omega }^{2}R\cos \phi }{g_0}\right)}^{1/2}$$

Neglecting higher powers of angular velocity and considering only the first term of the binomial expansion,

$$\Rightarrow g=g_0\left(1-\frac{1}{2}X\frac{2{\omega }^{2}R\cos \phi }{g_0}\right)$$

$$\Rightarrow g=g_0-{\omega }^{2}R\cos \phi$$

This is the final expression that shows the effect of rotation on gravitational acceleration ( $m{g}_0$ ). The important point here is that it is not only the magnitude that is affected by rotation, but its direction is also affected as it is no more directed towards the center of Earth.

There is no effect of rotation at pole. Being a point, there is no circular motion involved and hence, there is no reduction in the value of gravitational acceleration. It is also substantiated from the expression as latitude angle is φ = 90° for the pole and corresponding cosine value is zero. Hence,

$$\Rightarrow g=g_0-{\omega }^{2}R\cos {90}^0=g_0$$

The reduction in gravitational acceleration is most (maximum) at the equator, where latitude angle is φ = 0° and corresponding cosine value is maximum (=1).

$$\Rightarrow g=g_0-{\omega }^{2}R\cos 0^0=g_0-{\omega }^{2}R$$

We can check approximate reduction at the equator, considering R = 6400 km = 6400000 m = $6.4X{10}^6$ m.

$$\Rightarrow {\omega }^{2}R={\left(7.28X{10}^{-5}\right)}^{2}X\left(6.4X{10}^6\right)=3.39X{10}^{-3}m/s^2=0.0339m/s^2$$