The mass of a planet and its diameter are three times those of earth's. Then the acceleration due to gravity on the surface of the planet is ....... $m/s^2$
$3.3$
$4.9$
$19.6$
$29.4$
The height at which the weight of the body become $\frac{1}{9}^{th}$. Its weight on the surface of earth (radius of earth $R$)
Find the magnitude of acceleration due to gravity at height of $10\, km$ from the surface of earth.
The acceleration due to gravity on the earth's surface at the poles is $g$ and angular velocity of the earth about the axis passing through the pole is $\omega .$ An object is weighed at the equator and at a height $h$ above the poles by using a spring balance. If the weights are found to be same, then $h$ is $:( h << R ,$ where $R$ is the radius of the earth)
The diameters of two planets are in ratio $4:1$ . Their mean densities have ratio $1:2$ . The ratio of gravitational acceleration on the surface of planets will be
The height ${ }^{\prime} h ^{\prime}$ at which the weight of a body will be the same as that at the same depth $'h'$ from the surface of the earth is (Radius of the earth is $R$ and effect of the rotation of the earth is neglected):