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Question: What is the moment of inertia of the earth? Given that the radius of the earth is \(6.38 \times {10^...

What is the moment of inertia of the earth? Given that the radius of the earth is 6.38×106m6.38 \times {10^6}m and the mass of the earth is 5.98×1024kg5.98 \times {10^{24}}kg .

Explanation

Solution

A quantity expressing a body’s tendency to resist angular acceleration. This can be calculated by the sum of the products of the mass of each particle in the body with the square of its distance from the axis of rotation. Earth is a type of solid sphere. To calculate the moment of inertia of the earth simply put the values in the formula of moment of inertia of the solid sphere according to earth to get the answer.

Complete answer:
Given,
Radius of the earth is 6.38×106m6.38 \times {10^6}m
Mass of earth is 5.98×1024kg5.98 \times {10^{24}}kg
Now to find the moment of inertia of a solid sphere can be find by the formula =25mr2 = \,\dfrac{2}{5}m{r^2}
Where mm is the mass of the object, and rr is the radius of the sphere
Now put the given values in the formula to find the value of moment of inertia of earth.
Therefore, moment of inertia of earth =25mr2 = \,\dfrac{2}{5}m{r^2}
25×(5.98×1024)×(6.38×106)2\Rightarrow \dfrac{2}{5} \times (5.98 \times {10^{24}})\, \times {(6.38 \times {10^6})^2}
97.36×1036kgm2\Rightarrow {97.36} \times {\text{1}}{{\text{0}}^{{\text{36}}}}kg{m^2}
Thus, the required answer is 97.36×1036kgm2{97.36} \times {\text{1}}{{\text{0}}^{{\text{36}}}}kg{m^2}

Note: The moment of inertia of a figure about a line is the sum of the products formed by multiplying the magnitude of each element (of area or of mass) by the square of its distance from the line. A relation between the area of a surface or the mass of a body to the position of a line. The moment of inertia of a figure is the sum of moments of its parts.