F = g(m2*m2)/d^2. Earth surface gravity using Earth's radius as d; iss using that radius plus orbital height of 408km. Earth's surface gravity = 9.81m/s2, ISS gravity = 8.66m/s2.

So the answer is, if you could hold the ISS stationary without falling, people in it would feel almost as much gravity. You would have to go much higher than the ISS to feel weightless without orbiting/falling.

Assuming people feel weightless at 1m/s2 or less, you can solve for that! And the answer is 13,600 km.

Gravity is proportional to 1/r^2 (r being distance), so you could take a stab at computing your ISS question with that info.

The centripetal force question is a little more challenging, but still in the realm of algebra. The complicated part is setting up the problem and understanding where the forces are coming from and if/how they cancel each other.

Still, 19 miles vs 250 miles (and static not 17,500 mph) for the ISS - maybe the balloon passengers feel 0.98g?

Since the inverse square law applies and you're only 100 miles away from a planet 25,000 miles in circumference, I think it would be in the high 90's percentage of your weight on earth.

You have to get pretty far away to experience true "weightlessness".

You can use Newton's equations for gravity to calculate this.

https://www.quora.com/What-is-the-acceleration-due-to-gravit...

At the ISS, the extra distance from the center of the earth is minimal compared to the radius of the earth itself. Therefore, there wouldn't be a great difference.

Thanks to all the replies about the (very small) extra weight at the height of the ISS.

As for the centrifugal force from the Earth's rotation, Wikipedia puts it at about 0.3% (which, added to the effect of the Earth's bulge, means you feel a total of 0.5% less weight at the equator than at the poles).

https://en.m.wikipedia.org/wiki/Gravity_of_Earth#Latitude