The radius of Earth is about $6400 \mathrm{~km}$ and that of Mars is $3200 \mathrm{~km}$, and mass of the…
The radius of Earth is about $6400 \mathrm{~km}$ and that of Mars is $3200 \mathrm{~km}$, and mass of the Earth is about 10 times mass of Mars. An object weight $200 \mathrm{~N}$ on the surface of Earth. Then, its weight on the surface of Mars will be
$80 \mathrm{~N}$
$40 \mathrm{~N}$
$20 \mathrm{~N}$
$8 \mathrm{~N}$
Solution
Given, radius of Earth and Mars are $6400 \mathrm{~km}$ and $3200 \mathrm{~km}$, 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
$g=\frac{G M}{R^2},$
where, $M=$ mass of planet
and $R=$ radius of planet.
Taking the ratio of acceleration due to gravity of Mars to Earth,
$\frac{g_m}{g_e}=\frac{\frac{G M_m}{R_m^2}}{\frac{G M_e}{R_e^2}}=\frac{M_m}{M_e} \times\left(\frac{R_e}{R_m}\right)^2$
By substituting the values, we get
$\begin{aligned} & \frac{g_m}{g_e}=\frac{1}{10} \times\left(\frac{6400}{3200}\right)^2=\frac{1}{10} \times 4 \\ & g_m=\frac{2}{5} g_e\end{aligned}$
Given, the weight on earth's surface,
$w_e=m g_e=200 \mathrm{~N}$
$\therefore$ Weight on the Mars surface,
$w_m=m g_m=\frac{2}{5}\left(m g_e\right)=\frac{2}{5} \times 200 \mathrm{~N}$
$=80 \mathrm{~N}$