
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
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Hint : We will use the relation of force between two objects using the universal law of gravitation to determine the gravitational force between the Earth and an object on the surface of the sun. The radius of the Earth is larger on the equator and smaller on the poles so the weight of an object will be different on the equator of the Earth than on the poles.
Formula used: In this solution, we will use the following formulae
$\Rightarrow F = \dfrac{{GMm}}{{{r^2}}} $ where $ F $ is the gravitational force between the Earth and an object, $ M $ is the mass of the Earth, $ m $ is the mass of the object, and $ r $ is the distance of the object from the center of the Earth.
Complete step by step answer
We know that gravitational force between the Earth an object is given as
$\Rightarrow F = \dfrac{{GMm}}{{{r^2}}} $
An object on the surface of the earth will experience a gravitational acceleration due to this law. We can find this gravitational acceleration using Newton’s second law which tells us that the net acceleration on a body is due to the net force acting on the object will be equal to mass times its acceleration or mathematically:
$\Rightarrow F = \dfrac{{GMm}}{{{r^2}}} = ma $
Dividing both sides by $ m $ , we get
$\Rightarrow a = \dfrac{{Gm}}{{{r^2}}} $ which is the gravitational acceleration on the surface of the Earth.
The weight of the object is actually the force it exerts on the Earth i.e. $ {\text{Weight}} = ma $ . And we know that the radius of the Earth is smaller on the poles and larger on the equator. Since the gravitational acceleration is inversely proportional to $ r $ i.e. the distance between the object and the surface of the Earth $ (a \propto 1/{r^2}) $ , we can say that the gravitational acceleration will be larger on the poles and smaller on the equator. And as the weight of a body is directly proportional to the gravitational acceleration of the object, so the weight will be larger on the poles and smaller on the equator.
Note
The gravitational acceleration of an object on the surface of the Earth is independent on if its mass which implies that the gravitational acceleration acting on a large object like a truck will be the same as the gravitational acceleration acting on a human. The value of gravitational acceleration on Earth is approximately $ 9.8\,m/{s^2} $ .
Formula used: In this solution, we will use the following formulae
$\Rightarrow F = \dfrac{{GMm}}{{{r^2}}} $ where $ F $ is the gravitational force between the Earth and an object, $ M $ is the mass of the Earth, $ m $ is the mass of the object, and $ r $ is the distance of the object from the center of the Earth.
Complete step by step answer
We know that gravitational force between the Earth an object is given as
$\Rightarrow F = \dfrac{{GMm}}{{{r^2}}} $
An object on the surface of the earth will experience a gravitational acceleration due to this law. We can find this gravitational acceleration using Newton’s second law which tells us that the net acceleration on a body is due to the net force acting on the object will be equal to mass times its acceleration or mathematically:
$\Rightarrow F = \dfrac{{GMm}}{{{r^2}}} = ma $
Dividing both sides by $ m $ , we get
$\Rightarrow a = \dfrac{{Gm}}{{{r^2}}} $ which is the gravitational acceleration on the surface of the Earth.
The weight of the object is actually the force it exerts on the Earth i.e. $ {\text{Weight}} = ma $ . And we know that the radius of the Earth is smaller on the poles and larger on the equator. Since the gravitational acceleration is inversely proportional to $ r $ i.e. the distance between the object and the surface of the Earth $ (a \propto 1/{r^2}) $ , we can say that the gravitational acceleration will be larger on the poles and smaller on the equator. And as the weight of a body is directly proportional to the gravitational acceleration of the object, so the weight will be larger on the poles and smaller on the equator.
Note
The gravitational acceleration of an object on the surface of the Earth is independent on if its mass which implies that the gravitational acceleration acting on a large object like a truck will be the same as the gravitational acceleration acting on a human. The value of gravitational acceleration on Earth is approximately $ 9.8\,m/{s^2} $ .
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