For what value of the angular velocity of the earth, the acceleration due to gravity in the equatorial region would have been zero ? The average density of the earth's material `=5.5 g*cm^(-3) and G=6.67xx10^(-8)` CGS unit. Find the ratio of this calculated value and the present value of the angular velocity of the earth.

Explanation

The acceleration due to gravity at the equator
`g^.=g-omega^2R`
For `g^.` to be zero, the value of `omega_0` should be such that
`g-omega_0^2R=0 or, omega_0=sqrt(g/R)`.
As average density of the earth is `rho` and its radius is R,
mass of the earth `M=4/3pi R^3rho`
Also, `g=(GM)/(R^2)=G/(R^2)*4/3*piR^3rho =(4piGRrho)/(3)`
Hence, required angular velocity ,
`omega_0=sqrt(g/R)=sqrt((4piGrho)/3)`
`therefore omega_0=sqrt((4xxpixx6.67xx10^(-8)xx5.5)/3)=1.24xx10^(-3)rad *s^(-1)`
Present angular velocity of the earth
`omega=(2pi)/(24xx60xx60)=7.27xx10^(-5)rad *s^(-1)`
`therefore (omega_0)/(omega)=(1.24xx10^(-3))/(7.27xx10^(-5))=17` (approx.)
Hence , if the angular velocity of diurnal motion of the earth would have been 17 times its present value, the acceleration due to gravity at the equator would be zero.
ALTERNATIVE METHOD:
From known values of `omega` , R and g,
`(omega^2R)/g~~1/(289)or, omega=sqrt(g/(289R))`
If the acceleration due to gravity is zero at the equator, then the required angular velocity `omega_0=sqrt(g/R)`
`therefore (omega_0)/(omega)=(sqrt(g//R))/(sqrt(g//(289R)))=sqrt(289)=17`.