Question

A spaceship of weight $39200 \,kg-W$, lands on planet mars whose radius is $3.4 \times 10^{6} \,m$ and mass is $6.42 \times 10^{23} \,kg$ The weight of space ship on mars will be (surface tension of water $\left.=7 \times 10^{-2} \,N / m \right)$

• A. 5000 kg-W
• B. 10000 kg-W
• C. 15000 kg-W
• D. 20000 kg-W

Answer 1

Answer: (C)

The mass of spaceship
$m=\frac{39200}{9.8}=4000\, kg$
On earth; $g=\frac{G M m}{R^{2}}\,\,\,...(i)$
On mars, $g'=\frac{G M'm}{R'^{2}}\,\,\,...(ii)$
$\therefore \frac{g'}{g}=\frac{M'}{M} \times \frac{R^{2}}{R'^{2}}\,\,\,...(iii)$
Here $: R=6400\, km =6.4 \times 10^{6} \,m$
$R'=3.4 \times 10^{6}\, m$
$R'=3.4 \times 10^{6} \,m $
$M=6.0 \times 10^{24} \,kg ,$
$ M'=6.42 \times 10^{23}\, kg$
Putting the given values in eq (i)
$\therefore g'=\frac{6.42 \times 10^{23} \times\left(6.4 \times 10^{6}\right)^{2}}{6 \times 10^{24} \times\left(3.4 \times 10^{6}\right)^{2}} \times 9.8$
$\Rightarrow g' \approx 3.7 \,m / sec ^{2}$
Hence, the weight of spaceship on mars
$w'=m g'$
$=4000 \times 3.7$
$=14800$
$ \approx 15000\, kg$