Question

What should be the velocity of earth due to rotation about its own axis so that the weight at equator becomes `(3/5)^"th"` of initial value?

(Radius of earth on equator = 6400 km, `g=10m/s^2,cos0^circ=1`)

Options

• `7.91xx10^-4"rad"/"s"`

• `2.5xx10^-4"rad"/"s"`

• `3.5xx10^-4"rad"/"s"`

• `6.5xx10^-4"rad"/"s"`

Answer 1

`7.91xx10^-4"rad"/"s"`

Explanation:

g' = g - ω² r cosλ = g - ω² r [λ = 0]

`3/5g=g-omega^2r`

`omega^2r=g-3/5g=2/5g=2/5xx10=4`

`omega^2=4/r=4/(6400xx1000)`

`omega=2/(8xx100xxsqrt10)`

`omega=1/4xx(10sqrt10)/10000=7.91xx10^-4"rad"/"s"`