Let us consider a body of mass $m$ is revolving around a star $S$ of mass $M$ in circular path of radius $r$.
(i) Orbital velocity of the body
$v_0=\sqrt{\frac{G M}{r}}$
$v_0 \propto \frac{1}{\sqrt{r}}$


Hence, when $r$ increases of circular orbit then orbital velocity, $v$ decreases.
(ii) Angular velocity of the body $\omega=\frac{2 \pi}{T}$
By Kepler's third law of period,
$$
T^2 \propto r^3 \Rightarrow T=k r^{3 / 2}
$$
where $k$ is a constant.
$$
\begin{aligned}
&\therefore \omega=\frac{2 \pi}{k r^{3 / 2}} \Rightarrow \omega \propto \frac{1}{r^{3 / 2}} \Rightarrow \omega \propto \frac{1}{\sqrt{r^3}} \\
&\left(\because \omega=\frac{2 \pi}{T}\right)
\end{aligned}
$$
Hence, when $r$ increases then angular velocity $\omega$ decreases.
(iii) Kinetic energy of the body (KE)
$$
=\frac{1}{2} m v^2=\frac{1}{2} m \times \frac{G M}{r}=\frac{G M m}{2 r} \Rightarrow(\mathrm{KE}) \propto \frac{1}{r}
$$
Hence, when $r$ increases, the K.E. decreases.
(iv) Gravitational potential energy of the body.
$$
\text { Hence P.E. }=-\frac{G M m}{r} \Rightarrow P E \propto\left(-\frac{1}{r}\right)
$$
Hence, when $r$ increases PE becomes less negative but (P.E.) will also be increases.
(v) Total energy of the body
$$
E=K E+P E=\frac{G M m}{2 r}+\left(-\frac{G M m}{r}\right)=-\frac{G M m}{2 r}
$$
Hence, when $r$ increases then total energy becomes less negative. But energy will also be increases.
(vi) Angular momentum of the body,
$$
\begin{aligned}
&L=m v r=m r \sqrt{\frac{G M}{r}}=m \sqrt{G M r} \\
&\therefore L \propto \sqrt{r}
\end{aligned}
$$
Hence, when $r$ increases, angular momentum $L$ is also increases.