Satellites: Orbits and Energy — Question 4

Question 4

A radial impulse changes energy while preserving angular momentum

A satellite in a circular orbit of radius $r$ and speed $v_c=\sqrt{\mu/r}$ receives an outward radial velocity increment $u=.5v_c$. Its transverse speed stays $v_c$. Use $\epsilon=-\mu/(2a)$ and $e^2=1+2\epsilon h^2/\mu^2$, where $h=rv_t$.

1. Find its new $a$ and $e$.

2. Find both apsis radii and explain why the burn point is not an apsis.

Question 4 - Solution

A radial impulse changes energy while preserving angular momentum

(a) Derivation and calculation

The radial impulse is perpendicular to the original velocity: $$v'^2=v_c^2+u^2=1.25\frac\mu r,\qquad \epsilon=\frac{v'^2}{2}-\frac\mu r=-\frac{3\mu}{8r}.$$ Thus $$\boxed{a=\frac{4r}{3}},\qquad h^2=(rv_c)^2=\mu r,$$ $$e^2=1+\frac{2\epsilon h^2}{\mu^2}=1-\frac34, \qquad\boxed{e=\frac12}.$$

(b) Result and interpretation

The new apsides are $$\boxed{r_p=a(1-e)=\frac{2r}{3}},\qquad \boxed{r_a=a(1+e)=2r}.$$ The burn radius lies strictly between them because $\dot r=u\ne0$. The angular-momentum change is $$\Delta\vec h=\vec r\times(u\hat r)=\vec0,$$ while the specific energy increases by $u^2/2$.

Physical check

The radial impulse preserves angular momentum but cannot leave a circular orbit when the radial speed is nonzero.