Planets and Satellites: Kepler's Laws — Question 4

Question 4

Orbital period depends on semimajor axis rather than eccentricity

Two small bodies orbit the same central mass. Body A has period $T_A$ and semimajor axis $a_A$; body B has period $8T_A$. Use $T^2=4\pi^2a^3/(GM)$.

1. Find $a_B/a_A$.

2. If B has eccentricity $.75$, compare its periapsis radius with $a_A$ and explain why its long period does not imply it is always farther away.

Question 4 - Solution

Orbital period depends on semimajor axis rather than eccentricity

(a) Derivation and calculation

For the same central mass, $$\frac{T_B^2}{T_A^2}=\frac{a_B^3}{a_A^3}.$$ Since $T_B/T_A=8$, $$\boxed{\frac{a_B}{a_A}=8^{2/3}=4}.$$

(b) Result and interpretation

With $e_B=.75$, $$\boxed{r_{p,B}=a_B(1-e_B)=4a_A(.25)=a_A}.$$ The far apsis is $$r_{a,B}=a_B(1+e_B)=4a_A(1.75)=7a_A.$$ Thus the long period is compatible with an approach to $a_A$; the body spends much of its orbit at larger radii.

Physical check

The apsis average is $(a_A+7a_A)/2=4a_A$, recovering the inferred semimajor axis.