The Earth is assumed to be a sphere of radius $R$. A platform is arranged at a height $R$ from the surface of the Earth. The escape velocity of a body from this platform is $fv$, where $v$ is its escape velocity from the surface of the Earth. the value of $f$ is
$\sqrt 2 $
$\frac{1}{{\sqrt 2 }}$
$\frac{1}{3}$
$\frac{1}{2}$
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 figure below
If the radius of the earth were to shrink by $1\%$ its mass remaining the same, the acceleration due to gravity on the earth's surface would
If $M$ is mass of a planet and $R$ is its radius then in order to become black hole [ $c$ is speed of light]
In order to make the effective acceleration due to gravity equal to zero at the equator, the angular velocity of rotation of the earth about its axis should be $(g = 10\,m{s^{ - 2}}$ and radius of earth is $6400 \,kms)$
The radius of a planet is $R$. A satellite revolves around it in a circle of radius $r$ with angular velocity $\omega _0.$ The acceleration due to the gravity on planet’s surface is