Exercises — Kepler's second law — equal areas in equal times, from angular momentum conservation

(a) The areal velocity $\frac{dA}{dt}$ — the area swept by the radius vector per unit time — is constant. (b) Gravity is a central force ($\vec{F}\parallel \vec{r}$), so it exerts zero torque about the Sun; zero torque means angular momentum $L$ is conserved, and $\frac{dA}{dt}=\frac{L}{2m}$ is constant because $L$ and $m$ are both constant.

Faster at perihelion. Because $\frac{1}{2}{r}^2\dot{\theta }$ is fixed, a small $r$ forces a large angular rate $\dot{\theta }$, hence a large speed. The swept-area rate is the SAME at both points — that is the whole content of the law. Do not confuse "faster" (speed, which differs) with "area per second" (which is equal everywhere).

Both points are apsides, so $\vec{v}\perp \vec{r}$ and $L=mvr$ at each. Conservation of $L$ gives $v_p{r}_p=v_a{r}_a\Rightarrow v_a=v_p\frac{r_p}{r_a}.$ Why this step? Only at apsides can we drop the $\sin \varphi$ factor and write $L=mvr$ directly. $v_a=58.98\times \frac{4.60\times 10^{10}}{6.98\times 10^{10}}=58.98\times 0.6591\approx 38.87\text{km/s}.$

Areal velocity is $\frac{dA}{dt}=\frac{L}{2m}=\frac{8.4\times 10^{14}}{2}=4.2\times 10^{14}{\text{m}}^2/\text{s}$. One hour $=3600\text{s}$, and because $dA/dt$ is constant, area $=$ rate $\times$ time: $A=4.2\times 10^{14}\times 3600=1.512\times 10^{18}{\text{m}}^2.$ Why no integral? Constant rate means the total is just rate times elapsed time — the same reason distance $=$ speed $\times$ time when speed is constant.

The magnitude of the cross product is $|\vec{r}\times \vec{v}|=rv\sin \varphi$. So $\frac{dA}{dt}=\frac{1}{2}rv\sin \varphi =\frac{1}{2}(2.0\times 10^{11})(2.5\times 10^4)\sin 50°.$ With $\sin 50°=0.766$: $\frac{dA}{dt}=\frac{1}{2}(2.0\times 10^{11})(2.5\times 10^4)(0.766)\approx 1.915\times 10^{15}{\text{m}}^2/\text{s}.$ Why $\sin \varphi$ and not $\cos \varphi$? The cross product measures the perpendicular piece of the two vectors — the part of $\vec{v}$ that actually swings the radius vector sideways and paints area. The parallel part just lengthens/shortens $\vec{r}$ and paints nothing.

Speeds: at apsides $L=m{v}_p{r}_p=m{v}_a{r}_a$, so $v_p{r}_p=v_a{r}_a$, giving $\frac{v_p}{v_a}=\frac{r_a}{r_p}$. Kinetic energies: $KE=\frac{1}{2}m{v}^2$, so $\frac{K{E}_p}{K{E}_a}=\frac{\frac{1}{2}m{v}_p^2}{\frac{1}{2}m{v}_a^2}={\left(\frac{v_p}{v_a}\right)}^2={\left(\frac{r_a}{r_p}\right)}^2.$ Why square it? Kinetic energy scales with the square of speed, so any speed ratio becomes a squared ratio in energy.

(a) First $b=a\sqrt{1-e^2}=1.496\times 10^{11}\sqrt{1-0.0167^2}=1.496\times 10^{11}\times 0.99986=1.49579\times 10^{11}\text{m}$. $A=\pi ab=\pi (1.496\times 10^{11})(1.49579\times 10^{11})\approx 7.030\times 10^{22}{\text{m}}^2.$ (b) Over one full period the radius vector sweeps the entire ellipse once, so $A=\frac{L}{2m}T$. Solve for $L/m$: $\frac{L}{m}=\frac{2A}{T}=\frac{2(7.030\times 10^{22})}{3.156\times 10^7}\approx 4.455\times 10^{15}{\text{m}}^2/\text{s}.$ Why this works: $dA/dt$ is constant, so integrating it over one orbit just multiplies by $T$ — the swept area equals the whole ellipse because the planet returns to start exactly once per period.

Angular momentum is conserved everywhere, so $L$ at perihelion $=L$ at the new point: $m{v}_p{r}_p=mrv\sin \varphi \Rightarrow v=\frac{v_p{r}_p}{r\sin \varphi }.$ Why: at perihelion $\varphi =90°$ so $L=m{v}_p{r}_p$; at the general point $L=mrv\sin \varphi$. Set equal (mass cancels). $v=\frac{(9.0\times 10^3)(7.0\times 10^6)}{(1.0\times 10^7)(\sin 60°)}=\frac{6.30\times 10^{10}}{(1.0\times 10^7)(0.8660)}=\frac{6.30\times 10^{10}}{8.660\times 10^6}\approx 7.27\times 10^3\text{m/s}.$ So $v\approx 7.27\text{km/s}$.

(a) Position 2 is aphelion, an apsis, so ${\vec{v}}_2\perp {\vec{r}}_2$ and $\frac{dA}{dt}=\frac{1}{2}{r}_2{v}_2=\frac{1}{2}(5.0\times 10^{11})(4.0\times 10^3)=1.0\times 10^{15}{\text{m}}^2/\text{s}.$ (b) Areal velocity is constant, and in general $\frac{dA}{dt}=\frac{1}{2}r{v}_{\perp }$ (only the perpendicular part paints area). At position 1: $v_{\perp ,1}=\frac{2(dA/dt)}{r_1}=\frac{2(1.0\times 10^{15})}{8.0\times 10^{10}}=2.5\times 10^4\text{m/s}=25\text{km/s}.$ Why $v_{\perp }$ and not $v$? Area is swept only by the sideways swing of the radius vector; that is exactly the perpendicular velocity component. (c) The perpendicular component is $v_{\perp ,1}=v_1\sin {\varphi }_1$, so $\sin {\varphi }_1=\frac{v_{\perp ,1}}{v_1}=\frac{25}{30}=0.8333\Rightarrow {\varphi }_1=\arcsin (0.8333)\approx 56.4°.$ Sanity check: ${\varphi }_1<90°$ means the comet still has a radial velocity component — it is moving inward or outward, not at an apsis, which is consistent with $r_1$ being between perihelion and aphelion.

A central force means $\vec{F}=f(r)\hat{r}$ for some function $f$ — it always points along $\pm \hat{r}$. Then the torque about the centre is $\vec{\tau }=\vec{r}\times \vec{F}=\vec{r}\times f(r)\hat{r}=f(r)(\vec{r}\times \hat{r})=\vec{0},$ because $\vec{r}$ and $\hat{r}$ are parallel ($\vec{r}=r\hat{r}$, and any vector crossed with itself is zero). Zero torque gives $\frac{d\vec{L}}{dt}=\vec{\tau }=\vec{0}$, so $L=\text{const}$, hence $\frac{dA}{dt}=\frac{L}{2m}=\text{const}$. $■$

Why the shape isn't fixed: The second law used only that $f(r)$ points radially — it never used the form of $f(r)$. Different radial laws ($1/r^2$ gravity, $-kr$ spring, etc.) give different orbit shapes; the ellipse-with-Sun-at-focus specifically requires the inverse-square law (that is Kepler's First Law (Ellipse)). Equal areas is universal; the ellipse is special.