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?

1 Answer
Nov 12, 2015

Answer:

#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#