If the mass and radius of a planet are double those of the Earth,what will be the acceleration due to gravity on this planet compared to the acceleration due to gravity on Earth?

If the change in the value of $g$ at a height $h$ above the surface of the earth is the same as at a depth $x$ below it,then (both $x$ and $h$ being much smaller than the radius of the earth)

The ratio of the height $h$ above the surface of the Earth to the depth $d$ below the surface of the Earth,for which the gravitational accelerations are the same (assuming small heights),is:

The time period of a simple pendulum on the surface of the earth is $T$. The height above the surface of the earth at which the time period of the pendulum becomes $2T$ is (Radius of the earth $= 6400 \text{ km}$) (in $\text{ km}$)

If the angular speed of the earth is doubled,the value of acceleration due to gravity $(g)$ at the north pole:

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 $rad/s$):