Let's say we're interested in finding how fast the International Space Station has to move in order not to fall back to earth (i.e. to stay in orbit). The earth has an equatorial radius of about 6384km and the satellite is about 422km above the equator when it's overhead on its inclined orbital path. For the sake of simplicity we will ignore the oblate spheroid shape of the earth, and the replace the elliptical orbit with a circular one, 6806km from the center of the earth.

On Earth's surface the acceleration due to gravity is \(9.8 m \cdot s^{-2}\). The further you go away, the lower the force of gravity. 422km above the surface we're told that the acceleration is 89% of surface gravity - it's \(8.722 m \cdot s^{-2}\).

Using the SUVAT equation \(s = ut + {1 \over 2}at^2\) we can see that an object dropped from 422km (i.e. \(g = 8.772 m \cdot s^{-2}\)) would fall 4.361m in the first second.

In the same second, we know that the ISS traces out the circular orbit (i.e. it doesn't crash into the Earth). The angular distance it travels (we could use the unit circle for visual confirmation) is \(\arccos({{r - s} \over r})\) or \(\arccos({{6806000 - 4.361} \over 6806000})\) or \(\arccos({6805995.639 \over 6806000})\) or simply 0.001132041 radians.

If it travels 0.001132041 radians in a second, it will take 5550.316851 seconds to perform a complete revolution of 2π. 5550 seconds is 92½ minutes, a figure that's very close to the published figure (on Wikipedia) and that's quite amazing given the rough estimates we've made to get this far. A circle of radius 6806km has a circumference of 13612km. Divide that circumference by the orbital period in seconds and you get \(7704 m \cdot s^{-1}\). You could also try \(r \sin(\theta)\) or \(6806000 \sin(0.001132041)\) which gives the same result. It's moving pretty quickly, but would have to be even quicker if the orbit was any closer to the surface!

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## 1 comment:

The circumference needs to be multiplied by pi, which gives 42763

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