Question

A man is standing on an international space station, which is orbiting earth at an altitude $520\, km$ with a constant speed $7.6 \, km\,s^{-1} $ . If the mans weight is $50\, kg$, his acceleration is

• A. $7.6\, km\,s^{-2} $
• B. $7.6 \, km\,s^{-2} $
• C. $8.4 \, km\,s^{-2} $
• D. $10 \, km\,s^{-2} $

Answer 1

Answer: (C)

A man of mass $50\, kg$ is standing on an international space station.
Then, distance of station from centre of earth
$r=R+h=6400+520$
$[\because$ radius of earth $R =6400 \,km ]$
$=6920\, km$
Now, centripetal acceleration $a=\frac{v^{2}}{r}$
$=\frac{\left(7.6 \times 10^{3}\right)^{2}}{6920 \times 10^{3}}=8.4 \,ms ^{-2}$