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Home  >>  EAMCET  >>  Physics

Assume the earths orbit around the sun as circular and the distance between their centers as D. Mass of the earth is M and its radius is R. If earth has an angular velocity $\omega _0$ with respect to its center and $\omega$with respect to the center of the sun, the total kinetic energy of earth is :

\[\begin {array} {1 1} (a)\;\frac{MR^2\omega^2_0}{5}\bigg[1+\bigg(\frac{\omega}{\omega_0}\bigg)+\frac{5}{2}\bigg(\frac{D \omega}{R\omega_0}\bigg)^2\bigg] \\ (b)\;\frac{MR^2\omega^2_0}{5}\bigg[1+\frac{5}{2}\bigg(\frac{D \omega}{R\omega_0}\bigg)\bigg] \\ (c)\;\frac{2}{5}MR^2 \omega^2 \bigg[1+\frac{5}{2}\bigg(\frac{D \omega}{R\omega_0}\bigg)^2\bigg] \\ (d)\;\frac{2}{5}MR^2w_0^2 \bigg[1+\bigg(\frac{\omega}{\omega_0}\bigg)+\frac{5}{2}\bigg(\frac{D \omega}{R\omega_0}\bigg)^2\bigg] \end {array}\]

1 Answer

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$(b)\;\frac{MR^2\omega^2_0}{5}\bigg[1+\frac{5}{2}\bigg(\frac{D \omega}{R\omega_0}\bigg)\bigg]$
answered Nov 7, 2013 by pady_1
 

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