JEE Main · 2020 · 06 Sep 2020 · Shift 1 · Gravitation

Angular momentum about the planet is conserved along the orbit, so the speed is greatest where the satellite is closest and least where it is farthest, with $v_{\max }{r}_{\min }=v_{\min }{r}_{\max }$. The data state that the farthest-point speed is six times smaller than the closest-point speed, i.e. $v_{\max }=6v_{\min }$. Substituting, $6v_{\min }{r}_{\min }=v_{\min }{r}_{\max }$, so $r_{\max }=6r_{\min }$ and the ratio of closest to farthest distance is $\frac{r_{\min }}{r_{\max }}=\frac{1}{6}$, that is $1:6$.