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Problem : What is the angular momentum of Mercury when it is located at $\vec{r} = (45 \times 10^6 \rm{km}, 57 \times 10^6 \rm{km}, 0)$ relative to the sun and has velocity $\vec{v} = (140 \rm{m/s}, 125 \rm{m/s}, 0)$, and a mass $m = 3.30 \times 10^{23}$ kg?

$\vec{L} = \vec{r} \times \vec{p}$ and as such it will be completely in the $\hat{z}$ direction. The magnitude is given by the mass of mercury multiplied by the determinant of the matrix: \begin{equation} \begin{array}{cc} 45 \times 10^9 & 57 \times 10^8 \\ 140 & 125 \end{array} \end{equation} And the angular momentum is $-2.36 \times 10^{13} \times 3.30 \times 10^{23} = 7.77 \times 10^{36}$ kgm$^2$/s.

Problem : Use Kepler's Second Law to explain why planets travel faster near the sun.

This is a direct consequence of the law of equal areas. When planets are furthest from the sun, a small motion will cause the radius to sweep out a large area, since the radius is long. Near the sun, the radius is shorter, and so planets must cover a larger distance to sweep out that same area. Since equal areas are swept out in equal times, we must conclude that planets travel fastest when their radius is short–-that is, when they area near the sun.

Problem : If an Inter-Continental Ballistic Missile (ICBM) is launched into an elliptical path, where in its trajectory will it be traveling the slowest?

Since Kepler's Second Law tells us that projectiles travel slowest when they are furthest from the object they are orbiting around, we can conclude that the ICBM must be traveling slowest when it is farthest from the earth--that is, at the very top of its trajectory.

Problem : Mercury has an aphelion distance of $69.8 \times 10^6$ kilometers and perihelion distance of $45.9 \times 10^6$ kilometers. What is the ratio $\frac{v_{a}}{v_p}$ where $v_a$ and $v_p$ are the speeds at the apogee and perigee respectively?

At the aphelion and perihelion the velocity is completely perpendicular to the radius. Since angular momentum is conserved we can write that $mv_ar_a\sin\theta_a = mv_pr_p\sin\theta_p$. But in this case $\theta_a = \theta_p = \pi /2$. Thus we have $r_av_a = r_pv_p$ and finally that: \begin{equation} \frac{v_a}{v_p} = \frac{r_p}{r_a} \approx 0.66 \end{equation}

Problem : Beginning with $\frac{dA}{dt} = \frac{L}{2m}$, which is just an expression of Kepler's Second Law, prove Kepler's Third Law. Use the facts that $A$, the area of an ellipse, is equal to $\pi ab$ and that the semimmajor axis length is given by $a = \frac{L^2}{GMm^2(1-\epsilon^2)}$.

Integrating $\frac{dA}{dt} = \frac{L}{2m}$ over the whole ellipse, we get $A = \frac{LT}{2m}$ (the integration is trivial). We can then square this and set it equal to the area $A^2 = \pi^2 a^2b^2$ and rearrange: \begin{equation} T^2 = \frac{4m^2\pi^2a^4(1 - \epsilon^2)}{L^2} \end{equation} Now using the given expression for $a$: \begin{equation} T^2 = \frac{4\pi^2 m^2 a^3 (1 - \epsilon^2)L^2}{(1 - \epsilon^2)GMm^2} = \frac{4\pi^2a^3}{GM} \end{equation} Which is exactly Kepler's Third Law.
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