Satellites: Orbits and Energy — Question 6

Question 6

A two-burn transfer must match the outer circular speed at arrival

A coplanar transfer from a circular orbit of radius $r$ to one of radius $4r$ uses an ellipse tangent to both, with two prograde burns. Let $v_0=\sqrt{\mu/r}$. Use $v^2=\mu(2/r_{local}-1/a)$ and $T=2\pi\sqrt{a^3/\mu}$.

1. Find both burn magnitudes in units of $v_0$.

2. Find the transfer time in units of the initial circular period $T_0$.

Question 6 - Solution

A two-burn transfer must match the outer circular speed at arrival

(a) Derivation and calculation

The transfer ellipse has $$a_t=\frac{r+4r}{2}=\frac{5r}{2}.$$ Vis-viva at its two apsides gives $$\frac{v_{t,p}^2}{v_0^2}=2-\frac25=\frac85,\qquad \frac{v_{t,a}^2}{v_0^2}=\frac24-\frac25=\frac1{10}.$$ Since $v_{c,\mathrm{out}}=v_0/2$, $$\boxed{\frac{\Delta v_1}{v_0}=\sqrt{8/5}-1\simeq.264911},$$ $$\boxed{\frac{\Delta v_2}{v_0}=\frac12-\frac1{\sqrt{10}}\simeq.183772}.$$

(b) Result and interpretation

The transfer takes half the elliptic period: $$t=\pi\sqrt{a_t^3/\mu},\qquad T_0=2\pi\sqrt{r^3/\mu}.$$ $$\boxed{t/T_0=\tfrac12(a_t/r)^{3/2} =\tfrac12(5/2)^{3/2}=1.97642}.$$ At arrival $v_{t,a}<v_{c,\mathrm{out}}$, so a second prograde burn is required.

Physical check

Transfer angular momentum gives $rv_{t,p}=4rv_{t,a}$, confirming both apsis speeds.