Q. The ratio of radii of earth to another planet is $\frac{2}{3}$ and the ratio of their mean densities is $\frac{4}{5}.$ If an astronaut can jump to a maximum height of 1.5 m on the earth, with the same effort, the maximum height he can jump on the planet is
Explanation
(c) $Given\ \ \ \ \ \ \ \frac{R_{e}}{R_{p}} = \frac{2}{3}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{d_{e}}{d_{p}} = \frac{4}{5}$
$As\ \ \ \ \ \ \ \ \ \ MG = gR_{e}^{2}$
$and\ \ \ \ \ \ \ \ \ M = d_{e} \times \frac{4}{3}\pi R_{e}^{3}$
$\ \ \ \ \ \ \ \ \ \ \ \ d_{e} \times \frac{4}{3}\pi R_{e}^{3} \times G = g_{e}R_{e}^{2}$
$or\ \ \ \ \ \ \ \ \ d_{e} \times \frac{4}{3}\pi R_{e} \times G = g_{e}\ \ \ \ \ \ \ \ \ \ \ \ \ \ldots.(i)$
Similarly for planet
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ d_{p} \times \frac{4}{3}\pi R_{p}G = g_{p}\ \ \ \ \ \ \ \ \ \ \ \ \ \ldots.(ii)$
Dividing Eq. (i) by Eq. (ii), we get
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{g_{e}}{g_{p}} = \frac{R_{e}}{R_{p}} \times \frac{d_{e}}{d_{p}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{g_{e}}{g_{p}} = \frac{2}{3} \times \frac{4}{5} = \frac{8}{15} = 0.5$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.