Kepler定律

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这份力学讲义的52页 Figure 17: A particle moves from a point $P$ at time $t$ to a point $Q$ at time $t+\delta t$, with the angle subtended at the origin changing by a small amount $\delta \theta$, sweeping out a region $O P Q O$, of area $\delta A$ as shown in green. Let $r$ denote the distance $|O P|$, with $r+\delta r$ the distance $|O Q|, P^{\prime}$ the point on $O Q$ such that $\left|O P^{\prime}\right|=r$ and $Q^{\prime}$ the point on OP, such that $\left|O Q^{\prime}\right|=r+\delta r$. Proof of K2: Kepler's second law is a simple consequence of conservation of angular momentum. Recall from conservation of angular momentum we have $r^{2} \dot{\theta}=h=$ constant. A straight line from the Sun to a planet is simply the position vector $\mathbf{r}(t)$. In a small time interval $\delta t$, as shown in Figure 17) the planet sweeps out a region $O P Q O$ that is approximately triangular. In the first instance, with the points $O, P, P^{\prime}, Q, Q^{\prime}$ as given in Figure 17 , we consider $\delta r=$ $|O Q|-|O P| \geq 0$, with $r=|O P|$. Noting the area of the circular sector, of radius $s$, subtending an angle $\psi$ between its radii, is given by $\psi s^{2} / 2$, we have $$ \frac{1}{2} r^{2} \delta \theta \leq \delta A \leq \frac{1}{2}(r+\delta r)^{2} \delta \theta . $$ Dividing by $\delta t$ and taking $\delta t \rightarrow 0$, so that $\delta r \rightarrow 0, \delta \theta \rightarrow 0$, we have $$ \dot{A}=\frac{1}{2} r^{2} \dot{\theta}=\frac{1}{2} h=\text { constant }\tag{6.47} $$ which also holds by an analogous argument when $\delta r<0$.$\blacksquare$ Aside Being a consequence only of conservation of angular momentum, Kepler's second law holds for any central force (even non-conservative ones). Proof of K3: Recall that the area of an ellipse with semi-major axis $a$ and semi-minor axis $b$ is $$ A=\pi a b .\tag{6.48} $$ We know from K2 that this area is swept out at a constant rate $\dot{A}=\frac{1}{2} h$. Integrating this over one period we obtain $$ A=\int \mathrm{d} A=\frac{1}{2} h \int \mathrm{d} t=\frac{1}{2} h T .\tag{6.49} $$ Thus the square of the period $T$ is $$ T^{2}=\frac{4 A^{2}}{h^{2}}=\frac{4 \pi^{2} a^{2} b^{2}}{h^{2}}=\frac{4 \pi^{2}}{G_{N} M_{S}} \cdot \frac{b^{2}}{a r_{0}} \cdot a^{3},\tag{6.50} $$ where in the last step we have substituted $h^{2}=\kappa r_{0} / m=G_{N} M_{S} r_{0}$ using (6.20), where $\kappa=G_{N} m M_{S}$, $m$ is the mass of the planet and $M_{S}$ is the mass of the Sun. Recall from Eqn (6.24) for the equation of a Kepler problem ellipse, we have $$ a=\frac{r_{0}}{\left(1-e^{2}\right)}, \quad b=\frac{r_{0}}{\left(1-e^{2}\right)^{1 / 2}},\tag{6.51} $$and thus $$ \frac{b^{2}}{a r_{0}}=1 $$ Hence $$ T^{2}=\frac{4 \pi^{2}}{G_{N} M_{S}} a^{3},\tag{6.52} $$ which is precisely Kepler's third law.$\blacksquare$