(N/A) Consider a satellite of mass $m$ moving around the Earth in a circular orbit of radius $R$. The orbital speed $v_0$ is given by $v_0 = \sqrt{\frac{GM}{R}}$,where $M$ is the mass of the Earth.
$(a)$ Kinetic Energy $(KE)$: $K = \frac{1}{2}mv_0^2 = \frac{GMm}{2R}$. Thus,$K \propto \frac{1}{R}$. The graph is a rectangular hyperbola in the first quadrant,showing $KE$ decreases as $R$ increases.
$(b)$ Potential Energy $(PE)$: $U = -\frac{GMm}{R}$. Thus,$U \propto -\frac{1}{R}$. The graph lies in the fourth quadrant,where $U$ is negative and its magnitude decreases as $R$ increases,approaching zero from below.
$(c)$ Total Energy $(TE)$: $E = K + U = \frac{GMm}{2R} - \frac{GMm}{R} = -\frac{GMm}{2R}$. Thus,$E \propto -\frac{1}{R}$. Similar to $PE$,the graph lies in the fourth quadrant,representing a bound state where total energy is negative.