Answer:
a) v = 7.69 10³ m / s, b) T = 92.6 min
Explanation:
a) For this exercise we use the centripetal acceleration ratio, which in itself assumes a circular orbit, is equal to the acceleration of gravity
a = v² / r
v =
the distance to the ISS is
r = R_earth + d
r = 6400 10³ + 400 10³
r = 6800 10³ m
we calculate
v = Ra (8.69 6800 103)
v =
v = 7.687 10³ m / s
the result with the correct significant figures
v = 7.69 10³ m / s
b) The speed of the ISS is constant, so we can use the uniform motion relationships
v = d / t
if distance is the orbit distance
d = 2π r
time is called period
v = 2π r / T
T = 2π r / v
let's calculate
T = 2π 6800 10³ /7,687 10³
T = 5.558 10³ s
let's reduce the period to minutes
T = 5.558 10³ s (1 min / 60s)
T = 9.26 10¹ min
T = 92.6 min