The value of g on the earth's surface is $980\,cm/{\sec ^2}$. Its value at a height of $64 \,km$ from the earth's surface is …….. $cm/{\sec ^2}$

(Radius of the earth $R = 6400$ kilometers) 

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

If $R$ is the radius of the earth and the acceleration due to gravity on the surface of earth is $g=\pi^2 \mathrm{~m} / \mathrm{s}^2$, then the length of the second's pendulum at a height $h=2 R$ from the surface of earth will be,:

The angular velocity of the earth with which it has to rotate so that acceleration due to gravity on $60^o$ latitude becomes zero is (Radius of earth $= 6400\, km$. At the poles $g = 10\,m{s^{ – 2}})$

The acceleration of a body due to the attraction of the earth (radius $R$) at a distance $2 \,R$ from the surface of the earth is ($g =$ acceleration due to gravity at the surface of the earth)