What is the change in mass of a body, when taken $64 \ km$ below the surface of the earth? $[$Take radius of the earth as $6400 \ km]$

What should be the angular velocity of the Earth due to rotation about its own axis so that the weight at the equator becomes $\left(\frac{3}{5}\right)$ of its initial value? (Radius of Earth at the equator $R = 6400 \ km$,$g = 10 \ m/s^2$,$\cos 0^{\circ} = 1$)

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:

The radius of the Earth is about $6400 \; km$ and that of Mars is $3200 \; km$. The mass of the Earth is about $10$ times the mass of Mars. An object weighs $200 \; N$ on the surface of the Earth. Its weight on the surface of Mars will be .......... $N$.

The maximum vertical distance through which a fully dressed astronaut can jump on the earth is $0.5\, m$. If the mean density of the moon is two-thirds that of the earth and the radius is one-quarter that of the earth,what is the maximum vertical distance through which he can jump on the moon and the ratio of the duration of the jump on the moon to that on the earth?

At what height above the earth's surface does the acceleration due to gravity fall to $1\%$ of its value at the earth's surface (in $,R$)?