Planets and Satellites: Kepler's Laws — Question 7

Question 7

The area law gives a period from geometry and transverse motion

An elliptic orbit has semiaxes $a=5\,\mathrm m$ and $b=4\,\mathrm m$. Its specific angular momentum is $h=rv_t=6\,\mathrm{m^2/s}$. Use ellipse area $\pi ab$.

1. Find its period from areal velocity.

2. Find its eccentricity and explain why using $2\pi a/v$ would need additional justification.

Question 7 - Solution

The area law gives a period from geometry and transverse motion

(a) Derivation and calculation

The constant areal speed and ellipse area are $$\dot A=\frac h2=3\ \mathrm{m^2/s},\qquad A=\pi ab=20\pi\ \mathrm{m^2}.$$ Hence $$\boxed{T=\frac A{\dot A}=\frac{2\pi ab}h =\frac{20\pi}{3}\ \mathrm s\simeq20.944\ \mathrm s}.$$

(b) Result and interpretation

The geometry gives $$\boxed{e=\sqrt{1-b^2/a^2}=\sqrt{1-16/25}=.6}.$$ A general time calculation would be $$T=\oint\frac{ds}{v(s)}.$$ Neither the path length $2\pi a$ nor one constant speed is appropriate for this eccentric ellipse; the area law avoids both assumptions.

Physical check

Area divided by areal speed has units of time; the formula reduces to the circular period when $a=b$ and $h=av$.