$A$ body weighs $700\,gm\,wt.$ on the surface of the earth. How much will it weigh on the surface of a planet whose mass is $\frac{1}{7}$ and radius is half of that of the earth?

The weight of a body on the surface of the earth is $100\,N$. The gravitational force on it when taken at a height,from the surface of earth,equal to one-fourth the radius of the earth is $..........\,N$.

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).

What is the height from the surface of earth,where acceleration due to gravity will be $\frac{1}{4}$ of that of the earth (in $km$)? $(R_E = 6400 \ km)$

The acceleration due to gravity is $g$ at a point distant $r$ from the centre of the Earth of radius $R$. If $r < R$,then:

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 $10\,m/s^2$).