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.]

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^(@) )

What is the value of the acceleration due to gravity at a height equal to radius of earth?

The value of acceleration due to gravity at a depth of 1600 km is equal to [Radius of earth = 6400 km]

Acceleration due to gravity vanishes at an altitude equal to half the radius of the Earth.

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)

Acceleration due to gravity is g on the surface of the earth. Then the value of the acceleration due to gravity at a height of 32 km above earth's surface is : (Assume radius of earth to be 6400 km)