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InterviewSolution
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The fastest possible rate of rotation of a planet is that for which the gravitational force on material at the equator just nearly provides the centripetal force needed for rotation. The corresponding shortest period of rotation is (If `rho` is the density of the earth)A. `sqrt((3pi)/(Grho))`B. `sqrt((3pirho)/(G))`C. `sqrt((3piG)/(rho))`D. `sqrt((Grho)/(3pi))` |
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Answer» Correct Answer - a `mRomega^(2)=(GMm)/(R^(2))` `omega^(2)=(GM)/(R^(3))` `omega=sqrt((GM)/(R^(3)))=sqrt((Grho)/(K^(3)))(4pi)/(3)R^(3)=sqrt((4pi)/(3)Grho)` `T=(2pi)/(omega)=(2pi)/(sqrt((4pi G rho)/(3)))=sqrt((3xx4pi^(2))/(4piG rho))=sqrt((3pi)/(Grho)).` |
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