
What is the effect of the shape of the earth on the value of $ 'g' $ ?
Answer
572.1k+ views
Hint To answer this question we need to analyse the exact shape of the earth, which deflects from its ideal spherical shape. Then, using the basic formula of $ g $ we can predict its variation due to the shape of the earth.
Formula Used: The formula which is used in solving this question is given by
$ \Rightarrow g = \dfrac{{G{M_e}}}{{{r^2}}} $ , here $ g $ is the value of the acceleration due to gravity at the point on the earth which is at a distance of $ r $ from its centre, $ {M_e} $ is the mass of the earth and $ G $ is the universal gravitational constant.
Complete step by step answer
We usually think that the earth is perfectly spherical in shape. But this is not so. The radius of the earth is less at the poles, while it is slightly greater at the equator. So, instead of being perfectly spherical, it is ellipsoid in shape. The major axis of this ellipsoid is the equatorial axis, while the minor axis is the polar axis.
Now, we know that the value of $ g $ at a location on the earth is given by the expression
$ \Rightarrow g = \dfrac{{G{M_e}}}{{{r^2}}} $
The mass of the earth $ {M_e} $ is a constant. Also $ G $ is the universal constant. So we have
$ \Rightarrow g \propto \dfrac{1}{{{r^2}}} $
Now, as the distance of a point on the equator is greater at the equator than at the poles, so from the above proportionality we have the value of $ g $ greater at the equator than at the poles.
Thus, instead of being constant, the value of $ g $ is slightly greater at the points near the equator than at the points which are near to the poles.
Hence, we have found the effect of the shape of the earth on the value of $ g $ .
Note
We have slightly modified the formula of $ g $ which is used in this solution. In the denominator we have a square of the radius of the earth. But we have modified it to be square of the distance of the point from the centre of the earth. This is to take care of the fact that the earth is an ellipsoid and the radius of an ellipsoid is not defined.
Formula Used: The formula which is used in solving this question is given by
$ \Rightarrow g = \dfrac{{G{M_e}}}{{{r^2}}} $ , here $ g $ is the value of the acceleration due to gravity at the point on the earth which is at a distance of $ r $ from its centre, $ {M_e} $ is the mass of the earth and $ G $ is the universal gravitational constant.
Complete step by step answer
We usually think that the earth is perfectly spherical in shape. But this is not so. The radius of the earth is less at the poles, while it is slightly greater at the equator. So, instead of being perfectly spherical, it is ellipsoid in shape. The major axis of this ellipsoid is the equatorial axis, while the minor axis is the polar axis.
Now, we know that the value of $ g $ at a location on the earth is given by the expression
$ \Rightarrow g = \dfrac{{G{M_e}}}{{{r^2}}} $
The mass of the earth $ {M_e} $ is a constant. Also $ G $ is the universal constant. So we have
$ \Rightarrow g \propto \dfrac{1}{{{r^2}}} $
Now, as the distance of a point on the equator is greater at the equator than at the poles, so from the above proportionality we have the value of $ g $ greater at the equator than at the poles.
Thus, instead of being constant, the value of $ g $ is slightly greater at the points near the equator than at the points which are near to the poles.
Hence, we have found the effect of the shape of the earth on the value of $ g $ .
Note
We have slightly modified the formula of $ g $ which is used in this solution. In the denominator we have a square of the radius of the earth. But we have modified it to be square of the distance of the point from the centre of the earth. This is to take care of the fact that the earth is an ellipsoid and the radius of an ellipsoid is not defined.
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