Hohmann Orbit Transfer: Minimize Delta V for Bigger Orbit

[greg_rack]

Hi guys,

just a short question of Hohmann transfer.
I got the derivation for the required $\Delta v$, composed by the sum of two impulses, for establishing on a larger orbit... but how do we demonstrate it's actually the transfer which requires the smallest $\Delta v$?

 

[etotheipi]

Let the initial and final circular orbits have radii $a$ and $b$ respectively. Right after the initial burn at point $P_1$ let the spaceship have velocity $\mathbf{u}$, and right before the final burn at point $P_2$ let the spaceship have velocity $\mathbf{v}$. Angular momentum is conserved during the transfer orbit from $P_1$ to $P_2$, so$$a u_{\theta} = b v_{\theta} \implies v_{\theta} = \frac{a}{b} u_{\theta}$$Energy is also conserved during the transfer orbit, so \begin{align*}
u^2 - \frac{2GM}{a} &= v^2 - \frac{2GM}{b} \\

u_r^2 + u_{\theta}^2 - \frac{2GM}{a} &= v_r^2 + v_{\theta}^2 - \frac{2GM}{b}
\end{align*}now eliminate $v_{\theta}$ using the previous relation, i.e.\begin{align*}
v_r^2 = \left(1- \frac{a^2}{b^2} \right)u_{\theta}^2 + u_r^2 - 2GM \left( \frac{1}{a} - \frac{1}{b} \right)
\end{align*}Now consider the two burns. In the initial and final circular orbits the velocities are $\mathbf{e}_{\theta} \sqrt{\frac{2GM}{a}}$ and $\mathbf{e}_{\theta}\sqrt{\frac{2GM}{b}}$ respectively. The changes in the velocity due to the first burn is\begin{align*}

\mathbf{b}_1 &= \left( u_{\theta} - \sqrt{\frac{2GM}{a}} \right) \mathbf{e}_{\theta} + u_r \mathbf{e}_r \\

|\mathbf{b}_1|^2 &= \left( u_{\theta} - \sqrt{\frac{2GM}{a}} \right)^2 + u_r^2

\end{align*}Similarly for the second burn you can write \begin{align*}
|\mathbf{b}_2|^2 &= \left( \sqrt{\frac{2GM}{b}} - v_{\theta} \right)^2 + v_r^2 \\

&= \left( u_{\theta} - \frac{a}{b} \sqrt{\frac{2GM}{b}} \right)^2 + u_r^2 + 2GM \left( \frac{3}{b} - \frac{2}{a} - \frac{a^2}{b^3}\right)
\end{align*}having used the previous relation to eliminate $v_r$ and $v_{\theta}$. The most efficient orbit means minimising $\mathcal{Q} = |\mathbf{b}_1| + |\mathbf{b}_2|$. Holding $u_{\theta}$ constant, both $|\mathbf{b}_1|$ and $|\mathbf{b}_2|$ increase monotonically with $u_r$. Consider decreasing $u_r$ until either (i) $u_r = 0$ (in which case $P_1$ is the periapsis of the transfer orbit) or otherwise (ii) until some critical non-zero value $u_r = k$ below which the rocket won't reach the final circular orbit (in which case $P_2$ is the apoapsis of the connecting orbit). Recall that the total energy of an orbit of semi-major axis $\alpha$ is $-GM/2\alpha$. In either case, to reach the final circular orbit we must have $\alpha \geq \dfrac{1}{2} \left(a + b \right)$.

For case (i), if $P_1$ is the periapsis of the transfer orbit ($u_{r} = 0$ at $P_1$) then\begin{align*}

u_{\theta}^2 - \frac{2GM}{a} = - \frac{GM}{\alpha} \implies u_{\theta}^2 \geq GM\left( \frac{b}{a(a+b)}\right)

\end{align*}You can prove yourself that this constraint implies both $|\mathbf{b}_1|$ and $|\mathbf{b}_2|$ are monotonically increasing functions of $u_{\theta}$, and hence that $\mathcal{Q}$ is minimised for this $u_{\theta} = \sqrt{GM\left( \dfrac{b}{a(a+b)}\right)}$.

Similarly for case (ii), if $P_2$ is the apoapsis of the transfer orbit ($v_{r} = 0$ at $P_2$) then\begin{align*}
v_{\theta}^2 - \frac{2GM}{b} &= -\frac{GM}{\alpha} \\

\frac{a^2}{b^2} u_{\theta}^2 &= GM\left( \frac{2}{b} - \frac{1}{\alpha} \right) \\

u_{\theta}^2 &\geq \frac{b^2}{a^2} GM \left( \frac{a}{b(a+b)}\right) = GM \left( \frac{b}{a(a+b)}\right)
\end{align*}as before.

Notice then, that the transfer orbit corresponding to the minimum $u_{\theta}$ has both $P_1$ as a periapsis and $P_2$ as an apoapsis; i.e. the Hohmann transfer!

 

[Filip Larsen]

Note that for some situations the three-burn bi-elliptic transfer orbit requires less delta-V that the corresponding two-burn Hohmann transfer orbit.