For a satellite moving in an orbit around the earth at height ' \(h\) ' the ratio of kinetic energy to potential energy is

Let \(\mathrm{M}_{\mathrm{e}}=\) Mass of earth
\(\mathrm{m}=\) Mass of satellite
\(\mathrm{R}_{\mathrm{e}}=\) Radius of earth
\(\mathrm{G}=\) Gravitational constant
\(\therefore \quad\) Potential Energy,
\(\mathrm{U}=\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{~m}}{\mathrm{R}_{\mathrm{c}}}\)
Kinetic Energy,
\(\begin{aligned}
& \mathrm{K}=\frac{1}{2} \frac{\mathrm{GM}_{\mathrm{e}} \mathrm{~m}}{\mathrm{R}_{\mathrm{e}}} \\
& \frac{\mathrm{~K}}{\mathrm{U}}=\frac{\frac{1 \cdot \mathrm{GM}_{\mathrm{e}} \mathrm{~m}}{2}}{\frac{\mathrm{R}_{\mathrm{e}}}{\mathrm{GM}_{\mathrm{e}} \mathrm{~m}}} \\
& \mathrm{U} = 2 \mathrm{K}
\end{aligned}\)