Answer
The ratio of the kinetic energy of satellite B to satellite A is $~~\frac{1}{2}$
Work Step by Step
Let $R_E$ be the Earth's radius
The kinetic energy of an orbiting satellite at a height $h$ above the Earth's surface is $K = \frac{GMm}{2(R_E+h)}$
We can find the ratio of the kinetic energy of satellite B to satellite A:
$\frac{K_B}{K_A} = \frac{\frac{GMm}{(2)(6370~km+19,110~km)}}{\frac{GMm}{(2)(6370~km+6370~km)}}$
$\frac{K_B}{K_A} = \frac{(2)(6370~km+6370~km)}{(2)(6370~km+19,110~km)}$
$\frac{K_B}{K_A} = \frac{12,740~km}{25,480~km}$
$\frac{K_B}{K_A} = \frac{1}{2}$
The ratio of the kinetic energy of satellite B to satellite A is $~~\frac{1}{2}$
