The International Space Station (ISS) experiences an acceleration due to the Earth's gravity of 8.83 m/s2. What is the orbital period of the ISS?

Question

So, I used 2pi r/v inputting 2 pi (6.37 x 10^6 + 380 x 10^3) / 529.6. 529.5 I got by muliplying 8.83 by 60 to hopefully translate to minutes. I got roughly 9065.9 and its telling me I'm wrong by a multiple of 10.

Answer 1

You almost had it, but there is a small mistake in your calculation method. Instead of directly converting the acceleration due to gravity into the orbital speed, we need to use the following formula for the orbital speed (v) first:

v^2 = GM/r, where G is the gravitational constant (6.674 * 10^-11 N m²/kg²), M is the mass of Earth (5.972 * 10^24 kg), r is the distance from the center of Earth to the ISS, which is the addition of the Earth's radius (6.37 * 10^6 m) and the ISS altitude (380 * 10^3 m).

r = 6.37 * 10^6 + 380 * 10^3 = 6.75 * 10^6 m

Now, we can find the orbital speed:

v^2 = (6.674 * 10^-11 N m²/kg²) * (5.972 * 10^24 kg) / (6.75 * 10^6 m)
v = √((6.674 * 10^-11 N m²/kg²) * (5.972 * 10^24 kg) / (6.75 * 10^6 m))
v ≈ 7,660 m/s

Now we can use the formula you mentioned for calculating the orbital period (T):

T = 2πr/v
T = 2π * (6.75 * 10^6 m) / (7,660 m/s)

T ≈ 5,548.8 sec or 92.48 min

So, the orbital period of the ISS is approximately 92.48 minutes.