The acceleration due to gravity at the surface of the planet is the same as that at the surface of the earth,but the density of the planet is thrice that of the earth. If $R$ is the radius of the earth,the radius of the planet will be:

If the value of $g$ at the surface of the earth is $9.8 \, m/s^2$,then the value of $g$ at a place $480 \, km$ above the surface of the earth will be .......... $m/s^2$. (Radius of the earth is $6400 \, km$)

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)

$A$ body of mass $m$ is taken to the bottom of a deep mine. Then

The dependence of acceleration due to gravity $g$ on the distance $r$ from the centre of the earth, assumed to be a sphere of radius $R$ of uniform density, is as shown in which of the following figures?

Find the angular speed of rotation of the Earth,so that the apparent $g$ at the equator becomes $(1/6)$th of its original value. $(R = 6.4 \times 10^6 \ m)$