Calculate that imaginary angular velocity of the Earth for which effective acceleration due to gravity at the equator becomes zero. In this condition, find the length (in hours) of a day? Radius of Earth `= 6400 km. g = 10 ms^(-2)`.

Calculate the angular velocity of the earth for which effective acceleration due to gravity at the equator becomes zero. In this condition, what will be the length of the day ? Take R_e =6400 km and g=10 m*s^(-2)

Find the imaginary angular velocity of the earth for which the effective acceleration due to gravity at the equator shall be zero. (take g = 10 m//s^(2) for the acceleration due to gravity, if the earth were at rest and radius of earth equal to 6400 km and phi = 60^(@) )

Find the imaginary angular velocity of the earth for which the effective acceleration due to gravity at the equator shall be zero. (take g = 10 m//s^(2) for the acceleration due to gravity, if the earth were at rest and radius of earth equal to 6400 km and phi = 60^(@) )

Find the imaginary angular velocity of the earth for which the effective acceleration due to gravity at the equator shall be zero. (take g = 10 m//s^(2) for the acceleration due to gravity, if the earth were at rest and radius of earth equal to 6400 km and phi = 60^(@) )

The angular velocity of the earth with which it has to rotate so that the acceleration due to gravity on 60^@ latitude becomes zero is

The angular velocity of the earth with which it has to rotate so that the acceleration due to gravity on 60^@ latitude becomes zero is

The angular velocity of the earth with which it has to rotate so that the acceleration due to gravity on 60^@ latitude becomes zero.