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I'm not sure if it's ever measured, but it could be approximated by calculating the surface gravity of a spherical asteroid of equal mass and dimensions.

Assuming the mass of the International Space Station is 450 000 kg (M) and its mean radius is about 30 meters (r) , the surface gravity would be g = MG/r2 = about 0,00000003337 m/s2 (about 30 nanometers/square second).

For comparison, Earth's surface gravity is about 9.81 m/s2, so the gravity you would experience standing on the surface of the International Space Station is about 0.3 millionth of a percent compared to earth. It's certainly too small a gravity to hold you attached to the station if you were standing on it.

If you're inside the space station, in the center of the station, there is zero gravity because you are in the center of gravity because the mass of the station is situated around you.

Panu, M.Sc.

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Q: What is the value of 'g' where has International Space Station?
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