According to Kepler’s law the time period of a satellite varies with its radius as
${T^2} \propto {R^3}$
${T^3} \propto {R^2}$
${T^2} \propto (\frac{1}{R^3})$
${T^3} \propto (\frac{1}{R^2})$
A body weighs $700\,gm\,wt.$ on the surface of the earth. How much will it weigh on the surface of a planet whose mass is $\frac {1}{7}$ and radius half of that of the earth ....... $gm\, wt$
A spherical planet far out in space has a mass ${M_0}$ and diameter ${D_0}$. A particle of mass m falling freely near the surface of this planet will experience an acceleration due to gravity which is equal to
Time period of simple pendulum increases by an amount $\sqrt 2 $ times at height $'h'$ from the surface of earth. Then the value of $h$ is
A body of mass $m$ is kept at a small height $h$ above the ground. If the radius of the earth is $R$ and its mass is $M$, the potential energy of the body and earth system (with $h = \infty $ being the reference position ) is
A body of mass $m$ is lifted up from the surface of the earth to a height three times the radius of the earth. The change in potential energy of the body is
where $g$ is acceleration due to gravity at the surface of earth.