$A$ $90 \,kg$ body placed at $2R$ distance from the surface of the Earth experiences a gravitational pull of: ($R=$ Radius of Earth,$g=10 \,ms^{-2}$) (in $\,N$)

Since the Earth is not a perfect sphere,what is the effect on the acceleration due to gravity $(g)$?

The speed with which the earth would have to rotate about its axis so that a person on the equator would weigh $\frac{1}{6}$ th as much as at present is ($g=$ gravitational acceleration,$R=$ equatorial radius of the earth).

What is the time period of a seconds pendulum on a planet whose mass and radius are twice that of the Earth?

The radius of the Earth is $6400 \,km$ and the acceleration due to gravity is $g=10 \,ms^{-2}$. For the weight of a body of mass $5 \,kg$ to be zero at the equator,the rotational angular velocity of the Earth must be (in $rad/s$):

The escape velocities of two planets $A$ and $B$ are in the ratio $1:2$. If the ratio of their radii respectively is $1:3$,then the ratio of acceleration due to gravity of planet $A$ to the acceleration due to gravity of planet $B$ will be: