Satellites: Orbits and Energy — Question 3

Question 3

A tangential speed increase creates an ellipse rather than an immediate larger circle

A satellite initially has circular speed $v_c=\sqrt{\mu/r}$ at radius $r$. A short prograde burn increases its speed to $1.2v_c$ without changing position. Use specific energy $\epsilon=v^2/2-\mu/r=-\mu/(2a)$ for a bound ellipse.

1. Find the new semimajor axis and apoapsis radius.

2. Identify the burn point as periapsis and determine the speed-factor threshold for escape.

Question 3 - Solution

A tangential speed increase creates an ellipse rather than an immediate larger circle

(a) Derivation and calculation

Immediately after the burn, $$v'=1.2\sqrt{\mu/r},\quad \epsilon=\frac{v'^2}{2}-\frac\mu r=(.72-1)\frac\mu r=-.28\frac\mu r.$$ Hence $$\boxed{a=-\frac\mu{2\epsilon}=\frac{25r}{14}}.$$ The tangential burn leaves $\dot r=0$ and increases speed above circular, making this periapsis: $$\boxed{r_a=2a-r=\frac{18r}{7}}.$$

(b) Result and interpretation

The radial-coordinate acceleration verifies a local radius minimum: $$\ddot r=\frac{v'^2}{r}-\frac\mu{r^2} =(1.44-1)\frac\mu{r^2}>0.$$ For a tangential speed $kv_c$, $$\epsilon=\left(\frac{k^2}{2}-1\right)\frac\mu r.$$ $$\boxed{k_{\mathrm{escape}}=\sqrt2};\qquad k<\sqrt2\text{ is bound},\quad k>\sqrt2\text{ is unbound}.$$

Physical check

The new apoapsis exceeds the burn radius and the energy remains negative for $k=1.2$.