Answer:
[tex]8.671 \mathrm{m} / \mathrm{s}^{2}[/tex] is the centripetal acceleration.
Explanation:
As per given values
Radius of earth (r) = 6371000 m
The "international space station" is orbiting with a velocity (v) = 7667 m/s.
"Centripetal acceleration" is the acceleration is equal to "the square of the velocity" divided by "the radius of the circular path".
[tex]\text { Centripetal acceleration a }=\frac{V^{2}}{R}[/tex]
V = velocity of the orbit
R = radius of the earth + height of the space station
R = 6,371,000 + 408,000
R = 6779000 m
The direction of the centripetal acceleration is always inwards along the radius vector of the circular motion.
[tex]a=\frac{7667^{2}}{6779000}[/tex]
[tex]a=\frac{58782889}{6779000}[/tex]
[tex]\mathrm{a}=8.671 \mathrm{m} / \mathrm{s}^{2}[/tex]
[tex]\text { The International Space Station's centripetal acceleration } 8.671 \mathrm{m} / \mathrm{s}^{2}.[/tex]
