[dr_lew]

Yeah. v²/r = g(R/r)², so v = sqrt( g R ) sqrt( R/r ).

sqrt( g R ) represents the speed of an orbit at distance R, or about 7901 m/sec = 17674 mph and the speed of higher orbits are lower by the second factor.

Then we require r/R =( 17674/17000 )² ~= 1.08, giving an altitude of 0.08 R or about 317 miles .

[JRandomFreeper]

Leave it to a cook to point this out, and I should have known it, being a zoomie.... it's not a circular orbit.

So all bets are off.

BTW, I find your math correct to an order of magnitude for a cirular orbit.

/johnny

[Eltair]

For a circular orbit, the required speed is : v = sqrt(mu/r)
where mu is the gravitational parameter and r is the orbital radius. mu = GM, where G is the universal constant of gravitation, M is the mass of the central body (technically it should be M+m, but we assume m small relative to M).

so if v = 17000 mph, then r is 1.0820655*re (re = radius of Earth, I solved this using canonical earth units, it’s easier). This means the orbit is 324.98 miles up, which is very reasonable (I believe the space station is 230 miles).