Question

Derive an expression for the acceleration due to gravity on the surface of the earth. How does the weight of an object change as the object moves from the equator to the pole of the earth?

Answer 1

Hint According to Newton’s universal law of gravitation, every object in this universe will attract every other object with a force that is directly proportional to the masses of the objects and inversely proportional to the square of the distance between the two objects.

Complete Step by step solution
The acceleration due to gravity can be defined as the acceleration experienced by a body falling freely from a height to the surface of the earth.
According to Newton's law of gravitation, we can write the force between the freely falling body and earth as,
$F = G\dfrac{{Mm}}{{{r^2}}}$
Where $F$ stands for the force of attraction, $G$ is a constant called the gravitational constant, $M$stands for the mass of the earth, $m$stands for the mass of the object, and $r$stands for the distance between the centres of the earth and the object.
We know that,
$F = ma$
Where $F$ stands for the force, $m$ stands for the mass of the object and $a$stands for the acceleration
Let $g$ be the acceleration due to gravity, then
$F = mg$
Since the L.H.S is the same we can equate the two expressions
$mg = G\dfrac{{Mm}}{{{r^2}}}$
Since the mass of the object $m$is common on both sides, we cancel it and the acceleration due to gravity will be.
$g = \dfrac{{GM}}{{{r^2}}}$
On the surface of the earth,$r = R$
Where $R$ stands for the radius of the earth.
$g = \dfrac{{GM}}{{{R^2}}}$
This is the expression for acceleration due to gravity on the surface of the earth.
The weight of a body can be taken as,
$w = mg$
Where $w$ stands for the weight of the object, $m$stands for the mass of the object, and $g$ stands for the acceleration due to gravity.
The value of acceleration due to gravity varies as we go to the poles from the equator. This is due to the nonspherical shape of the earth and the rotation of the earth. Since earth is not a perfect sphere, the radius is different in different locations. The radius will be minimum at the poles. Therefore the acceleration due to gravity will be maximum at the poles. This will cause an increase in the weight of the object at the poles.

Note
The value of acceleration due to gravity is taken to be $9.8m/{s^2}$. This will vary slightly from place to place. As the height from the surface of the earth increases, the value of acceleration due to gravity decreases. The value will decrease with an increase in depth from the surface of the earth too.