$A$ planet has the same density as that of the Earth,and the universal gravitational constant $G$ is twice that of the Earth. The ratio of the acceleration due to gravity on the planet to that on the Earth is:

If the Earth loses its gravity,then for a body:

$A$ body has a weight of $90 \ kgf$ on the earth's surface. The mass of the moon is $1/9$ of the earth's mass and its radius is $1/2$ of the earth's radius. On the moon,the weight of the body is .......... $kgf$.

At what depth $d$ below the surface of the Earth does the acceleration due to gravity become $\frac{1}{n}$ times its value at the surface? ($R$ = radius of the Earth)

Assume there are two identical simple pendulum clocks. Clock-$1$ is placed on the Earth and Clock-$2$ is placed on a space station located at a height $h$ above the Earth's surface. Clock-$1$ and Clock-$2$ operate with time periods of $4\,s$ and $6\,s$ respectively. Then the value of $h$ is $....\,km$ (consider the radius of the Earth $R_E = 6400\,km$ and $g$ on Earth $10\,m/s^2$).

The acceleration due to gravity at a height $1\, km$ above the earth is the same as at a depth $d$ below the surface of earth. Then $d = $ ......... $km$