The international space station makes 15.65 revolutions per day in its orbit around the earth.

Answer:- The radius of orbit is given by the expresson

R = ∛(μT^2)/(4π^2 )

Where μ=398.6 .〖10〗^3 〖km〗^3/ s^2 , is the geocentric gravitational constant and T =〖(15.65rdp)〗^(-1) =〖(1.81 .〖10〗^(-4))〗^(-1)   is the period of revolution. Substituting values, obtain:

R =  ∛(398.6 .〖10〗^3 〖5521〗^2)/(4π^2 ) ≈6752 km

Subtracting the Earth’s equatorial radius, 6378 kilometers, obtain the height of this satellite above the surface of the Earth:

h = 6752 – 6378 = 374 km