Question

At what altitude will a satellite complete a circular orbit of the Earth in 2.6 h ?

Answer 1

Kepler's 3rd law says: Orbital period T = 2*pi*R^1.5/SQRT(GM)
where R is the radius (in the case of a circular orbit), G is the universal gravitational constant and M is the mass of the central body. In orbit, centrifugal acceleration CA balances gravitational acceleration GA. CA=w^2R (where w is omega, the orbital angular rate which = 2pi/T). So CA = (2pi/T)^2*R. From Newton we know GA = GM/R^2. Put 'em together,
CA = GA
R*(2pi)^2/T^2 = GM/R^2
T^2 = R^3*(2pi)^2/GM

Putting the values:

T = 2.6*60*60 Sec.

G = 6.67 X 10^-11

M = 5.98 X 10^24

We get, R = 8224.43 KM.

This is the required altitude.