# Why does Kepler's Law not apply to geostationary satellites?

The radius of orbit of the earth around the sun is equal to 1 AU and the period is 365.25 days(a year). Kepler's Law, $T^2\propto r^3$ prop to $r^3$ then means that $(T/365.25)^2 = (r/AU)^3$ gives the time and radius of orbit of any satellite in terms of days and AU.

The period of orbit of a geostationary satellite is 24 hours(i.e. 1 day). The radius worked out from the above equations then becomes,$(1/365.25)^2)^{1/3}\text{ AU} = r = 2.93\times10^9\text{ m}$. The $4.22\times10^7\text{ m}$, several orders of magnitude lower. Why doesn't Kepler's law apply in this situation?

Kepler's law does work for Geostationary orbits. The reason is that the $T^2=kr^3$ is valid for k for only one particular system. Using the first data you calculated k for the earth-sun system(where k depends on the mass of the sun) not the earth-satellite system( where k depends on the mass of the earth). Thus the discrepancy in your calculation.
Note: The value of the constant $k$ in the above equation is given by $$k=(\frac{GM}{4\pi^2})^{-1}$$ where M is the mass of the body which the satellite is orbiting. Where to calculate period for the earth and sun, $M$ would be mass of sun while for the geostationary satellite $M$ would be the mass of the earth.
The equation you used is only valid for objects that orbit the Sun. Geostationary satellites orbit the Earth (hence the "geo"). To calculate the orbital radius, you need to use the general form of Kepler's law that works for any central planet or star. $$T^2\left(\frac{GM_{Earth}}{4\pi^2}\right) = R^3$$ For any other central body, replace $M_{Earth}$ with the mass of the body being orbited.