Question

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Why the value of g is minimum at equator?

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Answer 1

Explanation:

In the above question, the central idea is the factors affecting the accleration due to gravity of earth. Here, the factor is Shape of earth.

We know that Earth is geoid shape i.e. not perfectly spherical in shape. The radius of the earth at poles is more than that of the radius of the earth at equator.

✏️ Refer to the attachment....

We know,

Acceleration due to gravity is given by:

$\large{ \boxed{ \sf{g = \frac{GM}{ {R}^{2} } }}}$

Where,

• g = Accleration due to gravity
• G = Gravitational constant
• M = Mass of the planet
• R = Radius

Let,

• $\sf{g_p}$ = acceleration due to gravity at poles.
• $\sf{g_e}$ = accleration due to gravity at equator.

Then,

$\sf{ \longrightarrow{g_p = \dfrac{GM}{R_p {}^{2} } - - - - - - (1) }}$

And,

$\sf{ \longrightarrow{g_e = \dfrac{GM}{R_e {}^{2} } - - - - - - (2) }}$

Dividing equation (1) by equation (2),

$\sf{ \longrightarrow{ \dfrac{g_p}{g_e} = \dfrac{GM / R_p {}^{2} }{GM / R_e {}^{2} } }}$

GM will cancel from numerator and denominator, and we wil get that,

$\sf{ \longrightarrow{ \dfrac{g_p}{g_e} = \dfrac{R_e {}^{2} }{R_p {}^{2} } }}$

So here we can see a inverse relation between them. And we know that Re > Rp (Radius at equator is more than that of radius at poles of earth). So,

$\sf{ \longrightarrow{g_p > g_e}}$

As the radius at the equator of the earth is minimum is case of all the dimension of earth, acceleration due to gravity at equator is minimum.

And we are done !!