Question

Amit buys a few grams of gold at the poles as per the instruction of one of his friends. He hands over the same when he meets him at the equator. Will the friend agree with the weight of gold bought? If not, why? [Hint: The value of g is greater at the poles than at the equator.]

Answer 1

Hint
In this question, the mass of the gold bought by Amit at the poles will remain the same at the equator, but the weight of the gold bought changes. This is because the acceleration due to gravity at the pole is greater than the acceleration due to gravity at the equator (${g_{pole}} > {g_{equator}}$)
$\Rightarrow g = \dfrac{{Gm}}{{{r^2}}}$, gravity at the surface of earth.
$\Rightarrow$ Weight =$m \times g$
Where $G,M,m,r,g$ represent universal gravitation constant, the mass of the earth, the mass of the object, the distance between the object and center of the earth, and gravity at the surface of earth respectively.

Complete step by step answer
We know that the shape of the earth is slightly ellipsoidal, which means that the distance from the poles to the center of the earth is lesser than the distance from the equator to the center of the earth.
From the equation of acceleration due to gravity, that is, $g = \dfrac{{Gm}}{{{r^2}}}$ , we know the gravitational force is inversely proportional to the square of the distance between the two objects, $g \propto \dfrac{1}{{{r^2}}}$. By which we can say that with an increase in distance the value of g decreases. Because the distance from poles to the center of the earth is greater, the gravity is smaller compared to the equator. Now using the formula for gravity at the surface of the earth. The weight of gold at the poles will be greater than the weight of gold at the equator.
$\Rightarrow {g_{equator}} = \dfrac{{G{m_{gold}}}}{{{r_{equator}}^2}}$
$\Rightarrow {g_{pole}} = \dfrac{{G{m_{gold}}}}{{{r_{pole}}^2}}$
$\Rightarrow {r_{pole}} < {r_{equator}}$
$\Rightarrow {g_{pole}} > {g_{equator}}$
$\Rightarrow {W_{pole}} = {m_{gold}} \times {g_{pole}}$
$\Rightarrow {W_{equ}} = {m_{gold}} \times {g_{equ}}$
$\Rightarrow {g_{pole}} > {g_{equator}}$
$\therefore {W_{pole}} > {W_{equator}}$
Hence Amit’s friend will not agree with the weight if he doesn’t know the concept of gravitation.

Note
This is the same reason why an object weighs 1/6 of its actual weight on the moon. The mass of the moon is 1/100 times and its radius 1/4 times that of earth. As a result, the gravitational attraction on the moon is about one-sixth when compared to earth. Hence, the weight of an object on the moon is 1/6th of its weight on the earth.