\subsection{Circular Motion and Orbital Dynamics}

This subsection resolves motion on a circle into radial and tangential parts. In AP mechanics, the key idea is that even when the speed stays constant, the velocity direction changes, so there is still an inward acceleration. In a circular orbit, gravity supplies that inward acceleration.

\dfn{Radial and tangential directions for circular motion}{Let a particle move on a circle of radius $r$ centered at a fixed point. Let $\vec{r}$ denote the particle's position vector from the center, let $\hat{r}=\vec{r}/r$ denote the outward radial unit vector, let $\hat{t}$ denote the unit vector tangent to the path in the direction of motion, let $v$ denote the speed, and let $\vec{a}$ denote the acceleration.

The acceleration can be decomposed into radial and tangential parts:
\[
\vec{a}=a_r\hat{r}+a_t\hat{t}.
\]
The tangential component $a_t$ changes the speed, while the radial component changes the direction of the velocity. For circular motion, the radial acceleration points toward the center, so its direction is $-\hat{r}$.}

\thm{Centripetal acceleration and circular-orbit speed}{Let a particle of mass $m$ move on a circle of radius $r$ with speed $v$. Let $\hat{r}$ denote the outward radial unit vector and let $\hat{t}$ denote the tangential unit vector in the direction of motion. Then the acceleration is
\[
\vec{a}=-\frac{v^2}{r}\hat{r}+\frac{dv}{dt}\hat{t}.
\]
In particular, for uniform circular motion, $dv/dt=0$, so
\[
\vec{a}=-\frac{v^2}{r}\hat{r},
\qquad
a_r=\frac{v^2}{r}\text{ inward}.
\]

Now let the same particle be in a circular orbit around a spherically symmetric body of mass $M$, with orbital radius $r$. If gravity is the only significant radial force, then
\[
\frac{GMm}{r^2}=\frac{mv^2}{r},
\]
so the orbital speed is
\[
v=\sqrt{\frac{GM}{r}}.
\]}

\pf{Short derivation from radial force balance}{Let $F_r$ denote the net inward radial force on the particle. For circular motion of radius $r$ and speed $v$, the required inward acceleration has magnitude $v^2/r$, so Newton's second law in the radial direction gives
\[
F_r=m\frac{v^2}{r}.
\]
Because inward is the $-\hat{r}$ direction,
\[
\vec{a}_r=-\frac{v^2}{r}\hat{r}.
\]
If the speed changes, the tangential component is
\[
\vec{a}_t=\frac{dv}{dt}\hat{t}.
\]
Therefore,
\[
\vec{a}=\vec{a}_r+\vec{a}_t=-\frac{v^2}{r}\hat{r}+\frac{dv}{dt}\hat{t}.
\]
For a circular orbit, gravity provides the entire inward force, so
\[
F_r=\frac{GMm}{r^2}=m\frac{v^2}{r}.
\]
Canceling $m$ and solving for $v$ gives
\[
v=\sqrt{\frac{GM}{r}}.
\]}

\cor{Circular-orbit period}{Let $T$ denote the period of a circular orbit of radius $r$ around a spherically symmetric body of mass $M$. Since one orbit has circumference $2\pi r$,
\[
v=\frac{2\pi r}{T}.
\]
Combine this with
\[
v=\sqrt{\frac{GM}{r}}
\]
to obtain
\[
T=2\pi\sqrt{\frac{r^3}{GM}}.
\]
Near Earth's surface, if $r\approx R_E$ and $g=GM/R_E^2$, this becomes
\[
T\approx 2\pi\sqrt{\frac{R_E}{g}}.
\]}

\qs{Worked example}{A satellite moves in a circular orbit at altitude $h=4.00\times 10^5\,\mathrm{m}$ above Earth's surface. Let Earth have mass $M_E=5.97\times 10^{24}\,\mathrm{kg}$, radius $R_E=6.37\times 10^6\,\mathrm{m}$, and let the gravitational constant be $G=6.67\times 10^{-11}\,\mathrm{N\,m^2/kg^2}$.

Find the satellite's orbital radius $r$, orbital speed $v$, orbital period $T$, and centripetal acceleration magnitude $a_r$.}

\sol First compute the orbital radius from Earth's center:
\[
r=R_E+h=6.37\times 10^6\,\mathrm{m}+4.00\times 10^5\,\mathrm{m}=6.77\times 10^6\,\mathrm{m}.
\]

For a circular orbit, gravity supplies the centripetal force, so
\[
\frac{GM_E m}{r^2}=\frac{mv^2}{r}.
\]
Cancel $m$ and solve for $v$:
\[
v=\sqrt{\frac{GM_E}{r}}.
\]
Substitute the given values:
\[
v=\sqrt{\frac{\left(6.67\times 10^{-11}\,\mathrm{N\,m^2/kg^2}\right)\left(5.97\times 10^{24}\,\mathrm{kg}\right)}{6.77\times 10^6\,\mathrm{m}}}.
\]
This gives
\[
v\approx 7.67\times 10^3\,\mathrm{m/s}.
\]

Now find the period from
\[
T=\frac{2\pi r}{v}.
\]
Thus,
\[
T=\frac{2\pi\left(6.77\times 10^6\,\mathrm{m}\right)}{7.67\times 10^3\,\mathrm{m/s}}
\approx 5.55\times 10^3\,\mathrm{s}.
\]
In minutes,
\[
T\approx \frac{5.55\times 10^3\,\mathrm{s}}{60}\approx 92.4\,\mathrm{min}.
\]

Finally, the centripetal acceleration magnitude is
\[
a_r=\frac{v^2}{r}.
\]
Using the speed just found,
\[
a_r=\frac{\left(7.67\times 10^3\,\mathrm{m/s}\right)^2}{6.77\times 10^6\,\mathrm{m}}
\approx 8.69\,\mathrm{m/s^2}.
\]

Therefore the satellite's orbital radius is
\[
r=6.77\times 10^6\,\mathrm{m},
\]
its orbital speed is
\[
v\approx 7.67\times 10^3\,\mathrm{m/s},
\]
its orbital period is
\[
T\approx 5.55\times 10^3\,\mathrm{s}\approx 92.4\,\mathrm{min},
\]
and its centripetal acceleration magnitude is
\[
a_r\approx 8.69\,\mathrm{m/s^2}.
\]
