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 rocket of mass $M$ is launched vertically from the surface of the earth with an initial speed $V.$ Assuming the radius of the earth to be $R$ and negligible air resistance, the maximum height attained by the rocket above the surface of the earth is
The potential energy of a satellite of mass $m$ and revolving at a height $R_e$ above the surface of earth where $R_e =$ radius of earth, is
A satellite of mass $m$ is at a distance $a$ from $a$ star of mass $M$. The speed of satellite is $u$. Suppose the law of universal gravity is $F = - G\frac{{Mm}}{{{r^{2.1}}}}$ instead of $F = - G\frac{{Mm}}{{{r^2}}}$, find the speed of the statellite when it is at $a$ distance $b$ from the star.
A geo-stationary satellite is orbiting the earth at a height of $6 R$ above the surface of earth, $R$ being the radius of earth. The time period of another satellite at a height of $2.5 R$ from the surface of earth is
The two planets have radii $r_1$ and $r_2$ and their densities $p_1$ and $p_2$ respectively. The ratio of acceleration due to gravity on them will be