What is the Height of the International Space Station Above the Earth's Surface?

[Chan M]

Homework Statement

The International Space Station, launched in 1998, makes 15.65 revolutions around the Earth each day in a circular orbit. Find the height of the space station (in kilometers) above the earth’s surface.

Homework Equations

T = (2*pi*r) / v
r - radius, distance between center of Earth and the station
r = Re + h
Re - radius of earth
h = height above ground
Find h

The Attempt at a Solution

So there is only one force acting on the station, which is gravity. Acceleration due to gravity is also centripetal acceleration. Gravity is the centripetal force.

15.65 revolution each day is one revolution every 5520.77 seconds

Fg = m*Ac
mg = m * (v^2 / r)
g = v^2 / r
g = [ (2*pi*r) / T ]^2 / r
g = 4*pi*pi*r / T^2
h = [(T^2)*g / 4*pi*pi ] - Re
h = [(5520.77^2)*9.8 / 4*pi*pi ] - (6.37 * 10^6)
h =

(this is what I put in the calculator)
h = 1195987.99 meters
h = 1196 km

This is not the correct answer, I know the correct answer, I just don't know how to get it or what I'm doing wrong. Thanks!

 

[NFuller]

Gravity is weaker that high up so $g=9.8$m/s won't work. You need
$$g=G\frac{M_{Earth}}{r^{2}}$$

 

[andrevdh]

can one use Kepler's 3rd law?