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)`.

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^(@) )

At the centre of the earth , the value of the acceleration due to gravity becomes zero.

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 imaginary angular velocity of the earth for which the effective acceleration due to gravity at the equator shall be zero is equal to [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.]

In order to make the effective acceleration due to gravity at the equator to zero, the numerical value of the angular velocity of rotation of the earth should be [Radius of the earth=6400 km, g=10m/s^(2) ]

The acceleration due to gravity at the latitude 45^(@) on the earth becomes zero if the angular velocity of rotation of the earth is

Calculate the centripetal acceleation of a point on the equator of earth due to the rotatin of earth about its wn axis . Radius of earth = 6400 km .

The height at which the acceleration due to gravity becomes g//9 in terms of R the radius of the earth is

What should be the angular velocity of earth, if the apparent value of acceleration due to gravity at earth's surface on equatorial plane is zero ? Radius of earth is 6400 km and g at earth's surface is 10m//s^(2) .