The radii of two planets are respectively $R_1$ and $R_2$ and their densities are respectively ${\rho _1}$ and ${\rho _2}$. The ratio of the accelerations due to gravity at their surfaces is
${g_1}:{g_2} = \frac{{{\rho _1}}}{{R_1^2}}:\frac{{{\rho _2}}}{{R_2^2}}$
${g_1}:{g_2} = {R_1}{R_2}:{\rho _1}{\rho _2}$
${g_1}:{g_2} = {R_1}{\rho _2}:{R_2}{\rho _1}$
${g_1}:{g_2} = {R_1}{\rho _1}:{R_2}{\rho _2}$
Radius of the earth is $R$. If a body is taken to a height $3R$ from the surface of the earth than change in potential energy will be
If $R$ is the radius of earth and $g$ is the acceleration due to gravity on the earth's surface. Then mean density of earth is ..........
The weight of a body on the surface of the earth is $63\, N$. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth ? (in $N$)
If $M$ is mass of a planet and $R$ is its radius then in order to become black hole [ $c$ is speed of light]
$Assertion$ : The escape speed does not depend on the direction in which the projectile is fired.
$Reason$ : Attaining the escape speed is easier if a projectile is fired in the direction the launch site is moving as the earth rotates about its axis.