Chapter 2, Problem 7C

To determine

Why the acceleration of a freely falling body is less in the Galapagos island at the equator and more a Oslo, Norway?

Answer to Problem 7C

The acceleration of free fall is different at different latitudes because of the following reasons:

(i) The radius of the earth is greatest at the equator and the acceleration of free fall at points close to the surface of the earth is inversely proportional to the square of the radius of earth at the point.

(ii) Due to rotation of the earth, the points on the surface of the earth experience an outward force depending on the latitude.

Explanation of Solution

Introduction:

Acceleration of free fall is the acceleration experienced by a body due to the force of gravitation acting on it. This is given by the expression,

$g=\frac{GM}{R^2}$

Here, G is the universal gravitational constant, M is the mass of the Earth and R is the radius of the earth.

(i) Earth is not a perfect sphere. It is bulged at the equator and flattened at the poles. The equatorial radius is estimated to be 6378 km and the polar radius is around 6357 km. The difference between the equatorial and polar radius is about 21 km.

Since $g\propto \frac{1}{R^2}$, the acceleration of free fall is less at equator and more at points close to the poles.

(ii) The earth rotates from west to east with an angular velocity ω of $7.3\times {10}^{-5}\text{rad/s}$.

Consider a point A located at a latitude $\theta$. A mass m located at A rotates along with the earth in a circle of radius r. If the earth were stationary, the acceleration of free fall at A is directed towards the center of Earth O. If the mass is located at a very short distance from the surface of the earth, the mass experiences a radially outward force due to inertia. This force is equal to the magnitude of the centripetal force required for the mass to move along the circular path of radius r. This is often referred to, mistakenly, as centrifugal force. The magnitude of the outward force is $mr{\omega }^2$, as shown in the diagram.

Resolve the gravitational force mg into two components, $mg\cos \theta$ along AB and $mg\sin \theta$ along AX.

The net force along AB is given by,

$F_{AB}=mg\cos \theta -mr{\omega }^2$

From the triangle OBA,

$r=R\cos \theta$

Therefore,

$F_{AB}=mg\cos \theta -mR{\omega }^2\cos \theta$........(1)

The force along AX is given by,

$F_{AX}=mg\sin \theta$........(2)

The net force acting on the mass is given by,

$F=\sqrt{F_{AB}^2+F_{AX}^2}$........(3)

From Newtons second law,

$F=m{g}_{\theta }$......(4)

Here, $g_{\theta }$ is the acceleration of free fall at a latitude $\theta$.

From equations (1),(2), (3) and (4),

$\begin{array}{c}m{g}_{\theta }=\sqrt{{\left(mg\cos \theta -mR{\omega }^2\cos \theta \right)}^2+{\left(mg\sin \theta \right)}^2}\\ =mg\sqrt{{\left(\cos \theta -\frac{R{\omega }^2}{g}\cos \theta \right)}^2+{\sin }^2\theta }\end{array}$

Simplify the expression.

$\begin{array}{c}{g}_{\theta }=g{\left[{\cos }^2\theta -\frac{2R{\omega }^2}{g}{\cos }^2\theta +\frac{R^2{\omega }^4}{g^2}{\cos }^2\theta +{\sin }^2\theta \right]}^{1/2}\\ =g{\left(1-\frac{2R{\omega }^2}{g}{\cos }^2\theta +\frac{R^2{\omega }^4}{g^2}{\cos }^2\theta \right)}^{1/2}\end{array}$

The value of $\frac{R^2{\omega }^4}{g^2}{\cos }^2\theta$ is very small, considering the value of ω of earth equal to $7.3\times {10}^{-5}\text{rad/s}$ therefore, set $\frac{R^2{\omega }^4}{g^2}{\cos }^2\theta \approx 0$.

Hence,

$g_{\theta }=g{\left(1-\frac{2R{\omega }^2}{g}{\cos }^2\theta \right)}^{1/2}$

Using binomial expansion and neglecting all higher powers of $\frac{2R{\omega }^2}{g}{\cos }^2\theta$, the he acceleration of free fall at a latitude $\theta$ is given by,

$g_{\theta }=g\left(1-\frac{R{\omega }^2}{g}{\cos }^2\theta \right)$

At equator the latitude is zero, $\theta =0$, therefore the value of acceleration of free fall is given by,

$g_{eq}=g-R{\omega }^2$

At the poles, the latitude is 90°, the acceleration of free fall at poles is given by,

$g_{pole}=g$

It can be seen that the value of $g_{eq}<g_{pole}$.

Conclusion:

Thus, the acceleration of free fall is different at different latitudes because of the following reasons:

(i) The radius of the earth is greatest at the equator and the acceleration of free fall at points close to the surface of the earth is inversely proportional to the square of the radius of earth at the point.

(ii) Due to rotation of the earth, the points on the surface of the earth experience an outward force depending on the latitude.