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concepts/mechanics/u6/m6-5-orbits.tex

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+\subsection{Circular Orbits, Satellite Speed, and Orbital Energy}
+
+This subsection focuses on Newtonian circular orbits around a much more massive central body. The main AP results are the orbital-speed formula and the linked kinetic, potential, and total-energy relations for a satellite in a circular orbit.
+
+\dfn{Circular-orbit setup and gravitational potential-energy reference}{Let a satellite of mass $m$ move in a circular orbit around a spherically symmetric body of mass $M$. Let $O$ denote the center of the central body. Let $\vec{r}$ denote the satellite's position vector from $O$, let $r=|\vec{r}|$ denote the constant orbital radius, let $\hat{r}=\vec{r}/r$ denote the outward radial unit vector, let $\vec{v}$ denote the satellite's velocity, and let $v=|\vec{v}|$ denote its speed. Let $\vec{L}_O=\vec{r}\times m\vec{v}$ denote the satellite's angular momentum about $O$. Let $K$ denote kinetic energy, let $U$ denote gravitational potential energy, and let $E=K+U$ denote total mechanical energy.
+
+Choose the gravitational potential-energy reference so that $U=0$ when the separation is infinite. Then for separation $r$,
+\[
+U(r)=-\frac{GMm}{r}.
+\]
+The gravitational force on the satellite is
+\[
+\vec{F}_g=-\frac{GMm}{r^2}\hat{r}.
+\]
+}
+
+\thm{Circular-orbit speed and energy relations}{Let a satellite of mass $m$ move in a circular orbit of radius $r$ around a spherically symmetric body of mass $M$. Because $\vec{F}_g$ is parallel to $\vec{r}$,
+\[
+\vec{\tau}_O=\vec{r}\times \vec{F}_g=\vec{0},
+\]
+so the orbital angular momentum about the center is conserved.
+
+For a circular orbit, gravity supplies the centripetal force:
+\[
+\frac{GMm}{r^2}=\frac{mv^2}{r}.
+\]
+Therefore,
+\[
+v=\sqrt{\frac{GM}{r}}.
+\]
+The kinetic energy is then
+\[
+K=\frac12 mv^2=\frac{GMm}{2r}.
+\]
+Using $U=-GMm/r$, the total mechanical energy is
+\[
+E=K+U=\frac{GMm}{2r}-\frac{GMm}{r}=-\frac{GMm}{2r}.
+\]
+Thus, for a circular Newtonian orbit,
+\[
+v=\sqrt{\frac{GM}{r}},
+\qquad
+K=\frac{GMm}{2r},
+\qquad
+U=-\frac{GMm}{r},
+\qquad
+E=-\frac{GMm}{2r}.
+\]
+}
+
+\ex{Illustrative example}{Two satellites of the same mass orbit the same planet in circular orbits. Satellite A has orbital radius $r$, and satellite B has orbital radius $4r$.
+
+Since $v=\sqrt{GM/r}$,
+\[
+v_B=\sqrt{\frac{GM}{4r}}=\frac12\sqrt{\frac{GM}{r}}=\frac12 v_A.
+\]
+Since $K=GMm/(2r)$,
+\[
+K_B=\frac{GMm}{2(4r)}=\frac14 K_A.
+\]
+Also,
+\[
+U_B=-\frac{GMm}{4r}=\frac14 U_A,
+\qquad
+E_B=-\frac{GMm}{8r}=\frac14 E_A.
+\]
+So a larger circular orbit has a lower speed and a less negative total energy.}
+
+\nt{With the reference choice $U(\infty)=0$, any bound gravitational orbit has negative total mechanical energy. For a circular orbit specifically, $E=-K=U/2<0$. Also, the formulas $v=\sqrt{GM/r}$, $K=GMm/(2r)$, and $E=-GMm/(2r)$ are for \emph{circular} Newtonian orbits only. In a noncircular orbit, the speed is not constant, so these same expressions do not apply at every point of the motion.}
+
+\qs{Worked example}{An Earth satellite of mass $m=850\,\mathrm{kg}$ moves in a circular orbit at altitude $h=4.00\times 10^5\,\mathrm{m}$ above Earth's surface. Let Earth's mass be $M_E=5.97\times 10^{24}\,\mathrm{kg}$, Earth's radius be $R_E=6.37\times 10^6\,\mathrm{m}$, and the gravitational constant be $G=6.67\times 10^{-11}\,\mathrm{N\cdot m^2/kg^2}$.
+
+Find:
+\begin{enumerate}[label=(\alph*)]
+\item why the satellite's angular momentum about Earth's center is conserved,
+\item the orbital speed $v$,
+\item the kinetic energy $K$,
+\item the gravitational potential energy $U$, and
+\item the total mechanical energy $E$.
+\end{enumerate}}
+
+\sol Let $r$ denote the orbital radius measured from Earth's center. First compute it from the given altitude:
+\[
+r=R_E+h=6.37\times 10^6\,\mathrm{m}+4.00\times 10^5\,\mathrm{m}=6.77\times 10^6\,\mathrm{m}.
+\]
+
+For part (a), the only significant force on the satellite is Earth's gravitational force, which points along the line from Earth to the satellite. Therefore $\vec{F}_g$ is parallel to $\vec{r}$, so the torque about Earth's center is
+\[
+\vec{\tau}_O=\vec{r}\times \vec{F}_g=\vec{0}.
+\]
+Since
+\[
+\frac{d\vec{L}_O}{dt}=\vec{\tau}_O,
+\]
+it follows that
+\[
+\frac{d\vec{L}_O}{dt}=\vec{0},
+\]
+so the satellite's angular momentum about Earth's center is conserved.
+
+For part (b), use the circular-orbit speed formula:
+\[
+v=\sqrt{\frac{GM_E}{r}}.
+\]
+Substitute the values:
+\[
+v=\sqrt{\frac{\left(6.67\times 10^{-11}\right)\left(5.97\times 10^{24}\right)}{6.77\times 10^6}}.
+\]
+This gives
+\[
+v\approx 7.67\times 10^3\,\mathrm{m/s}.
+\]
+
+For part (c), the kinetic energy is
+\[
+K=\frac12 mv^2.
+\]
+Using $m=850\,\mathrm{kg}$ and the value of $v^2=GM_E/r$,
+\[
+K=\frac12(850)\left(7.67\times 10^3\right)^2\,\mathrm{J}
+\approx 2.50\times 10^{10}\,\mathrm{J}.
+\]
+
+For part (d), the gravitational potential energy is
+\[
+U=-\frac{GM_E m}{r}.
+\]
+Substitute the numbers:
+\[
+U=-\frac{\left(6.67\times 10^{-11}\right)\left(5.97\times 10^{24}\right)(850)}{6.77\times 10^6}\,\mathrm{J}
+\approx -5.00\times 10^{10}\,\mathrm{J}.
+\]
+
+For part (e), the total mechanical energy is
+\[
+E=K+U.
+\]
+So,
+\[
+E=(2.50\times 10^{10})+(-5.00\times 10^{10})\,\mathrm{J}
+\approx -2.50\times 10^{10}\,\mathrm{J}.
+\]
+This agrees with the circular-orbit relation
+\[
+E=-\frac{GM_E m}{2r}.
+\]
+
+Therefore,
+\[
+v\approx 7.67\times 10^3\,\mathrm{m/s},
+\qquad
+K\approx 2.50\times 10^{10}\,\mathrm{J},
+\]
+\[
+U\approx -5.00\times 10^{10}\,\mathrm{J},
+\qquad
+E\approx -2.50\times 10^{10}\,\mathrm{J}.
+\]
+The negative total energy shows that the satellite is in a bound orbit.