Planets and Satellites: Kepler's Laws — Question 9

Question 9

A projected ellipse need not be an eccentric physical orbit

A circular orbit of radius $a$ is viewed at inclination $i$ from face-on. Under orthographic projection its apparent semiaxes are $a$ and $a\cos i$. The observed minor-to-major ratio is $.60$.

1. Find $i$ and the eccentricity of the apparent ellipse.

2. Can that apparent eccentricity be inserted into the physical periapsis formula $a(1-e)$?

Question 9 - Solution

A projected ellipse need not be an eccentric physical orbit

(a) Derivation and calculation

The projection ratio is $$\frac{b_{\mathrm{app}}}{a_{\mathrm{app}}}=\cos i=.60,\qquad \boxed{i=\cos^{-1}(.60)=53.1301^\circ}.$$ Its apparent eccentricity is $$\boxed{e_{\mathrm{app}}=\sqrt{1-(b_{\mathrm{app}}/a_{\mathrm{app}})^2} =\sqrt{1-.36}=.80}.$$

(b) Result and interpretation

The physical orbit is circular: $$e_{\mathrm{physical}}=0,\qquad \boxed{r_p=r_a=a}.$$ Incorrectly using the projected eccentricity would give $$a(1-e_{\mathrm{app}})=.2a,$$ which contradicts the actual constant radius. Inclination changes the image, not the orbit’s source distance.

Physical check

Viewed face-on, $i=0$ and the apparent eccentricity returns to zero; viewing geometry must be removed before orbital interpretation.