The gravitational attraction between two objects is proportional to the product of the masses of the two objects divided by the square of the distance between the two objects.
Assume, to avoid having to do triple integrals, all mass of the earth is concentrated at a very dense point at the center of the earth, surrounded by a massless, 4000-mile-radius shell that we stand on.
We are thus separated from the mass of the earth by 4000 miles, which, squared, has a magnitude of 16000000.
The orbit of the ISS has an average altitude of about 225 miles.
It, then, is about 4225 miles from the mass of the earth; 4225 squared is about 17850000.
Therefore, the gravitational pull of the earth at the altitude of the space station is 16000000/17850000 relative to that on earth, or roughly 90%.
Even if you assume the far half of the earth doesn’t exist because the “1 over r-squared” term makes it less important and use a radius of 2000 miles rather than 4000, that still puts the earth’s gravitational attraction on the ISS equal to about 80% of that on the surface of the earth.
LOL I’ll take your word for it. I love the science but really suck at the math.