# The ISS orbits around Earth, at an altitude of 350. km. What is the speed? What is the period? How many orbits will complete per day?

Nov 12, 2015

$v = 7.70 k m . {s}^{- 1}$
$T = 91 \min$
and it will complete 16 orbits per day.

#### Explanation:

For this solution we will make the (reasonable) assumption that the ISS exhibits uniform circular motion.

In that case we can use this equation for the speed: $v = \frac{2 \pi R}{T}$
where $R$ is the radius of the orbit and $T$ is the period.

But two of the variables in that equation are unknowns ($v$ and $T$). We can use the following equation to solve for the time period:
${T}^{2} = \left(\frac{4 {\pi}^{2}}{G M}\right) {R}^{3}$
where $G$ is the universal gravitational constant, 6.67×10^(-11) N m^2 kg^(-2),
and $M$ is the mass of the object being orbited (Earth in this case), 5.972 × 10^24kg.

First work out the radius of orbit. R≠350km. The ISS orbits at an altitude of 350 km, i.e. 350 km above the Earth's surface. So $R = {R}_{E} + 350 k m$.

$\implies R = 6371 + 350 = 6721 k m$

Now calculate the time period.
T = sqrt(((4pi^2)/(6.67×10^(-11) × 5.972 × 10^24)) (6.721× 10^6)^3) = 5485.39 … s

We can use that in the first equation above to calculate the speed:
v = (2pi × 6.721× 10^6)/(5485.39 …) = 7698 m.s^(-1) = 7.70 km.s^(-1)

We can write the period in minutes or hours to make it more practical:
$T = \frac{5485.39}{60} = 91 \min$
$T = \frac{5485.39}{3600} = 1.5 h o u r s$

Lastly, we want to know how many orbits the ISS will make per day. Divide the total time of one day (24 hours) by the time for one orbit in hours:
$n = \frac{24}{1.5} = 16$