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A body suspended on a spring balance in ...

A body suspended on a spring balance in a ship weighs `W_(0)` when the ship is at rest. When the ship begins to move along the equator with a speed `v`, show that the scale reading is very close to `W_(0) (1+-2omegaV//g`), where `omega` is the angular speed of the earth.

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Let `W_(0)` be weight of the body when ship is at rest then
`W_(0)=F-(mv_(0)^(2))/R`
Where `F=` gravitational pull of the earth when it is at rest and `v_(e)=` velocity of the earth at the equator when both move due east, the net speed of the ship.
`v'+v_(e)+v`
Therefore the apparent weight `W=F=(m(v_(e)+v)^(3))/R`
`W-W_(0)=(mv_(e)^(2))/R=(m(v_(e)+v)^(2))/R=(mv_(e)^(2))/R=(mv^(2))/R[1-(1+v/(v_(e))^(2))]`
As `vlt ltv_(e)^(')`
`W-W_(0)=mv_(e)^(2)[1-(1+v/(2v_(0)))]=-(mv_(e)v)/(2R)`
As `=(W_(0))/g` and `omega=(v_(3))/RimpliesW=W_(0)(1-(2omegav)/g)`
Similarly if the ship moes due west, then the net velocity of the ship is
`v'=v_(e)-v`
`W=W_(0)-=(m(v_(e)-v)^(2))/R=W=W_(0)(1/(2omegav)/g)`
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