The acceleration due to gravity becomes $\left(\frac{g}{2}\right)$ ($g =$ acceleration due to gravity on the surface of the earth) at a height equal to

$R$ is the radius of the Earth,$\omega$ is its angular velocity,and $g_p$ is the value of acceleration due to gravity at the poles. The effective value of $g$ at latitude $\lambda = 60^\circ$ will be equal to:

If a particle takes $t$ seconds less and acquires a velocity of $v \text{ m/s}$ more in falling through the same distance on two planets,where the accelerations due to gravity are $2g$ and $8g$ respectively,then:

The acceleration due to gravity on a planet is $1.96 \, m/s^2$. If it is safe to jump from a height of $3 \, m$ on the Earth,the corresponding height on the planet will be ........ $m$.

Find the angular speed of rotation of the Earth,so that the apparent $g$ at the equator becomes $(1/6)$th of its original value. $(R = 6.4 \times 10^6 \ m)$

The approximate height from the surface of earth at which the weight of the body becomes $\frac{1}{3}$ of its weight on the surface of earth is $.......... \, km$ : [Radius of earth $R = 6400 \, km$ and $\sqrt{3} = 1.732$]