Questions & Answers

Question

Answers

A. Maximum at poles

B. Maximum at equator

C. Same everywhere

D. Minimum at poles

Answer
Verified

Gravitational attractive force felt by a body on the earth’s surface is given as;

$\overset{\to }{\mathop{F}}\,=\dfrac{GMm}{{{r}^{2}}}$

G is the gravitational constant

r is the distance from the earth’s center

M is the mass of the earth

M is the mass of the body.

We can observe from the above image that earth is not a perfect sphere. It is slightly bulged at the equator, and flattened at the poles. Thus, from the above diagram we can observe that;

$R_e > R_p$

where, $R_e$ is the diameter at the equator, $R_p$ is the diameter at the poles

We know from the gravitational attraction and thus the acceleration due to gravity ‘g’ becomes:

$\begin{align}

& g=\dfrac{GM}{{{r}^{2}}} \\

& \Rightarrow g\propto \dfrac{1}{{{r}^{2}}} \\

\end{align}$

Thus, for poles:

${{g}_{p}}\propto \dfrac{1}{R_{p}^{2}}$

Thus, for equator:

${{g}_{p}}\propto \dfrac{1}{R_{e}^{2}}$

So, since ${{R}_{e}}>{{R}_{p}},{{g}_{e}}$ will be less than ${{g}_{p}}$. That is, gravitational force and the ‘g’ will be maximum at the poles.

Therefore, the correct answer to this question is option A. That is ‘g’ is maximum at poles.

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