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):

If earth suddenly stop rotating, then the weight of an object of mass $m$ at equator will $[\omega$ is angular speed of earth and $R$ is its radius]

A particle of mass $10\, g$ is kept of the surface of a uniform sphere of mass $100\, kg$ and a radius of $10\, cm .$ Find the work to be done against the gravitational force between them to take the particle far away from the sphere. (you make take $\left.G=6.67 \times 10^{-11} Nm ^{2} / kg ^{2}\right)$

Acceleration due to gravity is$ ‘g’ $on the surface of the earth. The value of acceleration due to gravity at a height of $32\, km$ above earth’s surface is …….. $g$. (Radius of the earth$ = 6400 \,km$)

Find the gravitational field at a distance of $2000\, km$ from centre of earth. (in $m / s ^{2}$)

(Given $\left.R_{\text {earth }}=6400 km , r =2000 km , M _{\text {earth }}=6 \times 10^{24} kg \right)$