The angular velocity of earth at present is `omega`. What should be its angular velocity so that the body lying at the equator flies off

Options

A. `17omega`
B. `8omega`
C. `2omega`
D. `289omega`

Answer

a

Explanation

`mRomega'^(2)=mg`
`therefore" "omega'^(2)=(g)/(R)`
`omega'=sqrt((g)/(R))=sqrt((9.8)/(6.4xx10^(6)))`
`omega'=(7sqrt2)/(8xx10^(-3))`
`omega=(2pi)/(T)=(6.28)/(86.400xx10^(3))`
`(omega')/(omega)=(7xxsqrt2)/(8xx10^(3))xx(86.4xx10^(3))/(6.28)`
`(omega')/(omega)=(7xx1.414xx86.4)/(8xx6.28)`
`(omega')/(omega)=17`
`omega'=17W.`