Answer
The ratio of the potential energy of satellite B to satellite A is $~~\frac{1}{2}$
Work Step by Step
Let $R_E$ be the Earth's radius
The gravitational potential energy at a height $h$ above the Earth's surface is $U_h = -\frac{GMm}{R_E+h}$
We can find the ratio of the potential energy of satellite B to satellite A:
$\frac{U_B}{U_A} = \frac{-\frac{GMm}{6370~km+19,110~km}}{-\frac{GMm}{6370~km+6370~km}}$
$\frac{U_B}{U_A} = \frac{6370~km+6370~km}{6370~km+19,110~km}$
$\frac{U_B}{U_A} = \frac{12,740~km}{25,480~km}$
$\frac{U_B}{U_A} = \frac{1}{2}$
The ratio of the potential energy of satellite B to satellite A is $~~\frac{1}{2}$
