The value of $‘g’$ at a particular point is $9.8\,m/{s^2}$. Suppose the earth suddenly shrinks uniformly to half its present size without losing any mass. The value of $‘g’$ at the same point (assuming that the distance of the point from the centre of earth does not shrink) will now be ......... $m/{\sec ^2}$.

The weight of a body on the surface of the earth is $100\,N$. The gravitational force on it when taken at a height, from the surface of earth, equal to onefourth the radius of the earth is $..........\,N$

If the density of the earth is doubled keeping its radius constant, then acceleration due to gravity will be ........ $m/s^2$. $(g = 9.8\,m/sec^2)$

The angular speed of earth, so that the object on equator may appear weightless, is $(g = 10\,m/{s^2}$, radius of earth $6400\, km$)

The value of acceleration due to gravity at Earth’s surface is $9.8\, m\,s^{-2}$. The altitude above its surface at which the acceleration due to gravity decreases to $4.9\, m\,s^{-2}$, is close to: (Radius of earth $= 6.4\times10^6\, m$)

Figure shows variation of acceleration due to gravity with distance from centre of a  uniform spherical planet, Radius of planet is $R$. What is $r_2 -r_1.$