Question

The earth is rotating with angular velocity $\omega$ about its own axis. $R$ is the radius of the earth. If $R{{\omega }^{2}}=0.03386m{{s}^{-2}}$, calculate the weight of a body of mass $100g$ at latitude ${{25}^{\circ }}$.
$\left( g=9.8m{{s}^{-2}},\text{ }\cos {{25}^{\circ }}=0.9063 \right)$

Answer 1

Due to the rotation of the planet earth about its axis, the weight of an object as well as the gravitational acceleration experienced by it differs at the poles as compared to the equator. We will calculate the value of acceleration due to gravity at the given latitude and determine the weight of the body at that particular point on the earth’s surface.

Complete answer:
Earth's rotation is defined as the rotation of the planet Earth around its own axis. Earth rotates in the eastward direction. The earth rotates about an imaginary line that passes through the North Pole and the South Pole of the planet earth. This line about which the planet rotates is known as the axis of rotation.
The value of acceleration of gravity experienced by different objects changes, as we move from the equator towards the pole of the earth, or vice versa. This value of acceleration due to gravity depends on the latitude angle that the object’s position makes with the equatorial line or the equatorial plane.

Radius of plane of object at latitude angle $\lambda$ is given as,
$r=R\cos \lambda$
Where,
$R$ is the radius of the earth at equatorial plane
Force due to rotation of the earth, centripetal force is given as,
$F=\dfrac{m{{v}^{2}}}{r}$
Where,
$m$ is the mass of the body
$v$ is the velocity of the earth
We know,
$v=r\omega$
Where,
$\omega$ is the angular velocity of the rotation

Therefore,
$\begin{aligned} & F=\dfrac{m{{\left( r\omega \right)}^{2}}}{r}=\dfrac{m{{r}^{2}}{{\omega }^{2}}}{r} \\\ & F=mr{{\omega }^{2}} \\\ \end{aligned}$

Tangential centripetal force,
${{F}_{c}}=mr{{\omega }^{2}}\cos \lambda$
Now,
$r=R\cos \lambda$
Therefore,
${{F}_{c}}=mR{{\omega }^{2}}{{\cos }^{2}}\left( \lambda \right)$
Net force on the body,
${{F}_{\text{net}}}={{F}_{g}}-{{F}_{c}}$
Gravitational force on the body is given as,
${{F}_{g}}=\dfrac{GMm}{{{r}^{2}}}$
We know,
$g=\dfrac{GM}{{{r}^{2}}}$

Therefore,
${{F}_{\text{net}}}=mg-mR{{\omega }^{2}}{{\cos }^{2}}\left( \lambda \right)$
Put,
${{F}_{\text{net}}}=m{{g}_{\lambda }}$
We get,
${{g}_{\lambda }}=g-{{\omega }^{2}}R{{\cos }^{2}}\left( \lambda \right)$
At the equator,
$\lambda =0$

Therefore,
${{g}_{\lambda }}=g-{{\omega }^{2}}R$
At the poles,
$\lambda ={{90}^{\circ }}$
Therefore,
${{g}_{\lambda }}=g$

Given that,
$\begin{aligned} & m=100g \\\ & {{\omega }^{2}}r=90.03386m{{s}^{-2}} \\\ \end{aligned}$

Also,
Latitude angle is given as,
$\lambda ={{25}^{\circ }}$

Now,
${{g}_{\lambda }}=g-{{\omega }^{2}}R{{\cos }^{2}}\lambda$
Where,
$g$ is the acceleration due to gravity
${{g}_{\lambda }}=9.8-0.03386{{\cos }^{2}}\left( {{25}^{\circ }} \right)$
We know,
$\cos {{25}^{\circ }}=0.9063$
Therefore,

$\begin{aligned} & {{g}_{\lambda }}=9.8-0.03386\times 0.9063 \\\ & {{g}_{\lambda }}=9.8-0.03068 \\\ & {{g}_{\lambda }}=9.77219m{{s}^{-2}} \\\ \end{aligned}$

Weight of body is given as,
$W=mg\lambda$

Putting values,
$\begin{aligned} & m=100g=0.1Kg \\\ & {{g}_{\lambda }}=9.77219m{{s}^{-2}} \\\ \end{aligned}$

We get,
$\begin{aligned} & W=0.1\times 9.77219 \\\ & W=0.977N \\\ \end{aligned}$

The weight of the body is $0.977N$

Note:
Due to the rotation of the planet earth about its axis, the weight of an object as well as the gravitational acceleration experienced by it differs at the poles as compared to the equator.
If we place an object on a scale at the pole of the planet, we will notice that the object is actually not rotating and therefore the normal force that the scale exerts on the object will be exactly equal to the gravitational force exerted by the Earth on the object, expressed as mass times the gravitational constant. On the other hand, if we place an object at the equator, it creates a centripetal acceleration that will act to decrease the normal force as read by the scale and also decreases the apparent gravitational constant. Therefore, the object weighs less at the equator than what they weigh at the poles.