Different satellites orbit the earth with a vast range of altitudes, from just a couple hundred km, all the way to tens of thousands of km above the surface. The international space station (ISS) is in a low earth orbit, just 400km above the surface (you can see it with the naked eye at sunset and sunrise as a bright, moving dot). At this altitude, the acceleration due to gravity has a value of 8.69m/s Assuming that the radius of the earth is 6400km).
1. What is the speed of the ISS? Express your answer to three significant figures and include the appropriate units.
2. What is the orbital period (T) of the ISS in minutes?

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moya1316 moya1316 · Answer 1 · 2021-02-16T02:08:14+01:00

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 = [tex]\sqrt{a r}[/tex]

the distance to the ISS is

r = R_earth + d

r = 6400 10³ + 400 10³

r = 6800 10³ m

we calculate

v = [tex]\sqrt{8.69 \ 6800 \ 10^3}[/tex]Ra (8.69 6800 103)

v = [tex]\sqrt{59.09 \ 10^6}[/tex]

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