If earth has a mass nine times and radius twice to the of a planet $P$. Then $\frac{v_e}{3} \sqrt{x}\; ms ^{-1}$ will be the minimum velocity required by a rocket to pull out of gravitational force of $P$, where $v_e$ is escape velocity on earth. The value of $x$ is

At what depth below the surface of the earth, acceleration due to gravity $g$ will be half its value $1600 \,km$ above the surface of the earth

A body weights $49\,N$ on a spring balance at the north pole. ..... $N$ will be its weight recorded on the same weighing machine, if it is shifted to the equator?

(Use $g=\frac{G M}{R^{2}}=9.8 \,ms ^{-2}$ and radius of earth, $R =6400\, km .]$

Weight of a body of mass m decreases by $1\%$ when it is raised to height $h$ above the earth’s surface. If the body is taken to a depth h in a mine, change in its weight is

At what height over the earth's pole, the free fall acceleration decreases by one percent ........ $km$. (assume the radius of earth to be $6400 \,km$)