At what height above the Earth's surface does the acceleration due to gravity become $1\%$ of its value at the Earth's surface (in $R$)? (Radius of Earth $= R$)

The mass and diameter of a planet have twice the value of the corresponding parameters of Earth. Acceleration due to gravity on the surface of the planet is ........ $m/s^2$.

The speed with which the earth would have to rotate about its axis so that a person on the equator would weigh $\frac{3}{5}$ th as much as at present weight is ($g=$ gravitational acceleration,$R=$ equatorial radius of the earth).

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:

If a body takes time $t$ to reach the ground when dropped from a height $h$ on Earth,how much time will it take to reach the ground when dropped from the same height $h$ on the Moon?

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 $R = 6400 \ km$,$g = 10 \ m/s^2$,$\cos 0^{\circ} = 1$)