Question

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?

Answer 1

`v=7.70 km.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 = (2piR)/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 = ((4pi^2)/(G M))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 km`.

`=> R = 6371 + 350 = 6721 km`

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= 5485.39/60 = 91 min`
`T= 5485.39/3600 = 1.5 hours`

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= 24/1.5= 16`