The angular velocity of the Earth at present is $\omega$. With what angular velocity should it rotate so that the weight of a body at the equator appears to be zero (in $\omega$)? (Express the answer as a multiple of $\omega$)

When a ball is dropped from a height $h$,it takes $t \ s$ to reach the ground. If the same experiment is done on a different planet having a mass $100$ times the earth's mass and a radius $10$ times the earth's radius,then the time it will take to cover the same height on the new planet is:

If a particle takes $t$ seconds less and acquires a velocity of $v \text{ m/s}$ more in falling through the same distance on two planets,where the accelerations due to gravity are $2g$ and $8g$ respectively,then:

What should be the angular velocity of the Earth due to rotation about its own axis so that the weight of an object at the equator becomes $3/5$ of its initial value? (Radius of Earth at the equator is $R = 6400 \, km$,take $g = 10 \, m/s^2$)

The mass of a planet is $\frac{1}{10}$ that of the earth and its diameter is half that of the earth. The acceleration due to gravity on that planet is: (in $m \ s^{-2}$)

The height in terms of radius of the earth $(R)$,at which the acceleration due to gravity becomes $g/9$,where $g$ is acceleration due to gravity on earth's surface,is . . . . . . .