For a satellite moving in an orbit around the earth, the ratio of kinetic energy to potential energy is

Solution :
Potential energy, $$U = - \frac{{G{M_e}m}}{{{R_e}}}$$
where,
$${M_e} =$$  mass of the earth
$$m =$$  mass of satellite
$${R_e} =$$  radius of the earth
$$G =$$  gravitational constant
or $$\,\left| U \right| = \frac{{G{M_e}m}}{{{R_e}}}$$
Kinetic energy, $$K = \frac{1}{2}\frac{{G{M_e}m}}{{{R_e}}}$$
Thus, $$\frac{K}{{\left| U \right|}} = \frac{1}{2}\frac{{G{M_e}m}}{{{R_e}}} \times \frac{{{R_e}}}{{G{M_e}m}} = \frac{1}{2}$$
NOTE
The total energy, $$E = K + U = - \frac{{G{M_e}m}}{{2r}}$$

This energy is constant and negative, i.e. the system is closed. To farther the satellite from the earth, the greater is its total energy.