The mass of a planet and its diameter are three times those of the Earth. Then the acceleration due to gravity on the surface of the planet is ....... $m/s^2$.

If the Earth is assumed to be a sphere of radius $R$,and $g_{30}$ is the value of acceleration due to gravity at a latitude of $30^\circ$ and $g$ is the value at the equator,the value of $g - g_{30}$ is:

$A$ simple pendulum has a time period $T_1$ when on the earth's surface and $T_2$ when taken to a height $R$ above the earth's surface,where $R$ is the radius of the earth. The value of $T_2/T_1$ is

The ratio of the weights of a body on the Earth's surface to that on the surface of a planet is $9 : 4$. The mass of the planet is $\frac{1}{9}^{th}$ of that of the Earth. If $R$ is the radius of the Earth,what is the radius of the planet? (Assume the planets have the same mass density)

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).

The mass of the earth is $81$ times that of the moon and the radius of the earth is $3.5$ times that of the moon. The ratio of the acceleration due to gravity at the surface of the moon to that at the surface of the earth is