The mass and diameter of a planet have twice the value of the corresponding parameters of Earth. Acceleration due to gravity on the surface of the planet is ........ $m/s^2$.

What is the acceleration due to gravity at a distance of $2R$ from the surface of the Earth,where $R$ is the radius of the Earth?

The depth $d$ at which the acceleration due to gravity becomes $\frac{g}{n}$ is (where $R$ is the radius of the Earth,$g$ is the acceleration due to gravity at the surface,and $n$ is an integer).

$Assertion$ : In a free fall,weight of a body becomes effectively zero.
$Reason$ : Acceleration due to gravity acting on a body having free fall is zero.

$A$ $90 \,kg$ 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 $\,N$)

Acceleration due to gravity on the surface of Earth is $g$. If the diameter of Earth is reduced to one-third of its original value and mass remains unchanged,then the acceleration due to gravity on the surface of the Earth is . . . . . . $g$.