In order to find time, the astronaut orbiting in an earth satellite should use

Acceleration due to gravity on moon is $\frac 16$ of the acceleration due to gravity on earth. If the ratio of densities of earth $({\rho _e})$ and moon $({\rho _m})$ is $\left( {\frac{{{\rho _e}}}{{{\rho _m}}}} \right) = \frac{5}{3}$ then radius of moon $R_m$ in terms of $R_e$ will be

If radius of the earth is $6347\, km ,$ then what will be difference between acceleration of free fall and acceleration due to gravity near the earth's surface ?

At what distance above and below the surface of the earth a body will have same weight, (take radius of earth as $R$.)

The radius of the earth is $6400\, km$ and $g = 10\,m/{\sec ^2}$. In order that a body of $5 \,kg$ weighs zero at the equator, the angular speed of the earth is

At what height above the surface of earth the value of $"g"$ decreases by $2 \%$ ..... $km$ [radius of the earth is $6400 \,km$ ]