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Question Number 31766 by NECx last updated on 14/Mar/18

Calculate the moment of inertia  of the earth abouth its axis.If  the mass of the earth is 6.0×10^(34) kg  and radius 6.4×10^6 m.  (Assume the earth to be a perfect  sphere)

$${Calculate}\:{the}\:{moment}\:{of}\:{inertia} \\ $$$${of}\:{the}\:{earth}\:{abouth}\:{its}\:{axis}.{If} \\ $$$${the}\:{mass}\:{of}\:{the}\:{earth}\:{is}\:\mathrm{6}.\mathrm{0}×\mathrm{10}^{\mathrm{34}} {kg} \\ $$$${and}\:{radius}\:\mathrm{6}.\mathrm{4}×\mathrm{10}^{\mathrm{6}} {m}. \\ $$$$\left({Assume}\:{the}\:{earth}\:{to}\:{be}\:{a}\:{perfect}\right. \\ $$$$\left.{sphere}\right) \\ $$

Commented by Tinkutara last updated on 14/Mar/18

(2/5)MR^2 =9.8304×10^(47)  kg m^2

$$\frac{\mathrm{2}}{\mathrm{5}}{MR}^{\mathrm{2}} =\mathrm{9}.\mathrm{8304}×\mathrm{10}^{\mathrm{47}} \:{kg}\:{m}^{\mathrm{2}} \\ $$

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