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**Physics, Astronomy & Cosmology / Re: How to calculate the orbital speed of the ISS around the centre of the earth?**

« **on:**03/05/2021 17:34:04 »

Using the numbers provided, and assuming a circular orbit (it's not quite circular, but close enough for this level of analysis), an orbit of once every 93 minutes (1.55 hours) means that the ISS travels the a distance of one circumference every 1.55 hours. The circumference of a circle is equal to 2 π times the radius of the circle, and in this case the radius of the orbit is equal to the radius of the Earth plus the altitude of the orbit.

so a distance of 2 times π times 6810 km ≈ 42790 km

if it can cover 42790 km in 1.55 hours it must be going 42790/1.55 km/hour = 27600 km/hour

and there are 3600 seconds in an hour (60 second per minute times 60 minutes per hour)

so 27600/3600 = 7.67 km/s

The difference could be due to rounding or deviation from circular orbit (also because the orbit isn't exactly circular, the speed isn't exactly fixed either)

so a distance of 2 times π times 6810 km ≈ 42790 km

if it can cover 42790 km in 1.55 hours it must be going 42790/1.55 km/hour = 27600 km/hour

and there are 3600 seconds in an hour (60 second per minute times 60 minutes per hour)

so 27600/3600 = 7.67 km/s

The difference could be due to rounding or deviation from circular orbit (also because the orbit isn't exactly circular, the speed isn't exactly fixed either)

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