My bad, I read "satellite" and "power" and my brain jumped to deep-space communication.
According to Wikipedia, the power of the GPS at your receiver is a miniscule 8 attowatts. That is 8 billionths of a billionth of a watt. If we take the 50kW number and transmit it to the Continental US with a surface area of 86,977 billion square feet, that 50kW transmission is roughly 0.574 pico-watts per square foot, 718 million times more power than the 8 attowatts the receiver is looking for.
The math above is VERY rough, back-of-an-envelope stuff and should not be considered accurate by any means, but for order-of-magnitude purposes it will serve. So, that 50kW figure seems a bit high.
That's okay. Everyone makes mistakes.
A better calculation would be the area of half of Earth, and the area of a GPS reciever.
I found one that is 15 mm by 15 mm.
That means the area of a receiver is 0.000225 m^2.
The area of half the Earth (cross sectional) is 1.278E+14 m^2.
Also, the GPS reciever I found only needs 1.6 aW.
Thus using the 2kW figure, that gives us 1.56E-11 W/m^2
so our gps receiver takes 3.52E-15 W of power.
That is roughly 3520 attowatts. So that 2 kW gives plenty.
Using other math, we treat the GPS satallite as a point source.
Greater than the maximum distance to the reciever we have a distance of 20992383.37 m.
This gives us a sphere area of 5.53775E+15 m^2.
So for 2kW, this gives us 3.61157E-13 W/m^2.
So our GPS reciever would pick up 81 aW.
Still far more than the 1.6aW needed.
So yeah, it looks like 2kW of power works fine for GPS.