Q. A man is standing on an international space station, which is orbiting earth at an altitude $520\ km$ with a constant speed $7.6\ km/s$. If the man’s weight is $50\ kg$, his acceleration is
A) $7.6km/s^{2}$
B) $7.6m/s^{2}$
C) $8.4m/s^{2}$
D) $10m/s^{2}$
Explanation
(c) Acceleration due to gravity at an altitude $h$ is
$g_{h} = \frac{gR_{e}^{2}}{\left( R_{e} + h \right)^{2}}$; where $R_{e}$ is the radius of the earth
$g_{h} = \frac{9.8m/s^{2} \times \left( 6400 \times 10^{3}m \right)^{2}}{\left( 6400 \times 10^{3}m + 520 \times 10^{3}m \right)^{2}} = 8.4m/s^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.