Respuesta :
Answer:
Moment of inertia of Earth about its own axis is given as
[tex]I = 9.7 \times 10^{37} kg m^2[/tex]
Explanation:
Since Earth is considered as solid sphere
So we will have
[tex]I = \frac{2}{5}M_eR_e^2[/tex]
so we will have
[tex]I = \frac{2}{5}(5.97 \times 10^{24})(6.371 \times 10^6)^2[/tex]
so we have
[tex]I = 9.7 \times 10^{37} kg m^2[/tex]
The moment of inertia of Earth = [tex]I_E[/tex] = 9.7 [tex]\times[/tex] 10^37 kg m^2
Find an expression for the moment of inertia
[tex]I_E= (2/5)\times M_ER_E^2[/tex]
The Moment of Inertia of solid sphere about the axis passing through center of mass of is defined by equation (1) as
[tex]I= 2/5 \times MR^2[/tex]........ (1)
[tex]I = Moment \; of \; Inertia[/tex]
[tex]M = Mass \; of\; the\; Sphere[/tex]
[tex]R = Radius \;of\; the\; Sphere[/tex]
According to the given question
Moment of Inertia of Earth = [tex]I_E[/tex]
Mass of Earth = [tex]M_E[/tex]
Radius of the Earth = [tex]R_E[/tex]
Considering Earth to be a Solid sphere
Putting these variables in equation (1) we get the expression
[tex]I_E= (2/5)\times M_ER_E^2[/tex]
According to the given data
[tex]M_E=[/tex] Mass of Earth= 5.97[tex]\times[/tex][tex]10^{24}[/tex] kg
[tex]R_E[/tex]= Radius of Earth = 6,371 km= 6371000 m
Hence
The moment of Inertia of earth = [tex]I_E[/tex] = (2/5) [tex]\times[/tex]5.97[tex]\times[/tex][tex]10^{24}[/tex] kg[tex]\times[/tex] 6371000 m[tex]\times[/tex] 6371000 m = 9.7 [tex]\times[/tex] 10^37 kg m^2
For more information please refer to the link below
https://brainly.com/question/14701082
