AP EAPCET 23-Aug-2021 Shift 2 Paper

Given, radius of Earth and Mars are $6400 \text{km}$ and $3200 \text{km}$ , respectively. Mass of Earth $=10$ (Mass of Mars) $M_e = 10 M_m$ We know that, acceleration due to gravity at surface of a planet is $g = \frac{GM}{R^2},$ where, $M =$ mass of planet and $R =$ radius of planet. Taking the ratio of acceleration due to gravity of Mars to Earth, $\frac{g_m}{g_e} = \frac{ \frac{GM_m}{R_m^2} }{ \frac{GM_e}{R_e^2} }$ $= \frac{M_m}{M_e} \times \left( \frac{R_e}{R_m} \right)^2$ By substituting the values, we get $\frac{g_m}{g_e} = \frac{1}{10} \times \left( \frac{6400}{3200} \right)^2 = \frac{1}{10} \times 4$ $\Rightarrow g_m = \frac{2}{5} g_e$ Given, the weight on earth's surface, $w_e = m g_e = 200 \text{N}$ ∴ Weight on the Mars surface, $w_m = m g_m = \frac{2}{5} (m g_e)$ $= \frac{2}{5} \times 200 \text{N} = 80 \text{N}$