Name: Gravitational Potential Energy
Uploaded: 2019-07-19
Duration: 9 min 10 s
Description: The energy possessed by a body because of its position in a gravitational field is known as Gravitational Potential Energy. Let’s learn about it in detail in...

Question

Explain the variation of g with latitude.

Accepted Answer

Let us consider the Earth to be a spherical ball of mass $M$ and radius $R$ . An object of mass $m$ is at point $P$ at latitude $λ$ , when Earth is not rotating the weight of the object is $m g$ but Earth is rotating with angular velocity $ω$ . So object is moving in a circular path of radius $r$ as shown in the figure, The object experiences a centrifugal force. Due to rotation of earth its apparent weight can be given by

The object is being acted by two forces $m g$ and $F_{c}$ . The resultant of these two forces gives the apparent weight of the object $(m g^{'})$ where

$F_{c} = m ω^{2} r = m ω^{2} R \cos λ [r = R \cos λ]$

Now from parallelogram of vector addition

$m g^{'} = \sqrt{(m g)^{2} + {(F_{c})}^{2} + 2 m g F_{c} \cos (180 - λ)}$

${(m g')}^{2} = {(m g)}^{2} + {(m ω^{2} R \cos λ)}^{2} + 2 m g m ω^{2} R \cos λ (- \cos λ) m^{2} g'^{2} = m^{2} g^{2} + m^{2} ω^{4} R^{2} \cos^{2} λ - 2 m^{2} g ω^{2} R \cos^{2} λ g'^{2} = g^{2} (1 + \frac{ω^{4} R^{2} \cos^{2} λ}{g^{2}} - \frac{2 ω^{2} R \cos^{2} λ}{g})$

Now value of $ω$ is of the order of $10^{- 4}$ . So the term containing $ω^{4}$ is much smaller than $1$ and $ω^{2}$ so term containing $ω^{4}$ can be neglected.

$g'^{2} = g^{2} (1 - \frac{2 ω^{2} R \cos^{2} λ}{g}) g' = g {(1 - \frac{2 ω^{2} R \cos^{2} λ}{g})}^{\frac{1}{2}}$

Applying binomial expansion and neglecting higher powers we get,

$g' = g (1 - \frac{1}{2} \times \frac{2 ω^{2} R \cos^{2} λ}{g}) g' = g (1 - \frac{ω^{2} R \cos^{2} λ}{g}) g' = g - ω^{2} R c o s^{2} λ$

This gives the acceleration due to gravity due to rotation of Earth.

Explain the variation of g with latitude.

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