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Orbit transfers

Hohmann transfer

The Hohmann transfer orbit offers a solution to transferring between two circular orbits, via an elliptic transfer orbit, using the least possible velocity change (\(\Delta v\)). It was first recognized/demonstrated by Hohmann in 1925, before the rocket technology to achieve any orbit existed!

A typical use of the Hohmann transfer is to start at a lower-altitude circular “parking orbit”, then use the elliptical transfer orbit to reach a higher-altitude circular orbit. This involves two instantaneous changes in velocity (\(\Delta v\)), one at the perigee of the elliptical orbit and one at the apogee (the “apogee kick”).

Illustration of a Hohmann transfer orbit

Fig. 10 Illustration of a Hohmann transfer orbit, from a smaller circular orbit to a larger circular orbit. Source: Leafnode, CC BY-SA 2.5 via Wikimedia Commons.

The problem: Given the original orbit’s perigee and apogee distances (\(r_{p1}\) and \(r_{a1}\)), and the apogee distance of the desired elliptical orbit (which also may be the radius of the final circular orbit) \(r_{a2}\), find the required instantaneous change in velocity \(\Delta v\) to enter the elliptical orbit.

The change in velocity needed is the difference in perigee velocities of the two orbits. For orbit 1:

\[\begin{split} \begin{align} e_1 &= \frac{r_{a1} - r_{p1}}{r_{a1} + r_{p1}} \\ r_{p1} &= \frac{h_1^2 / \mu}{1 + e_1} \\ \rightarrow h_1 &= \sqrt{2 \mu} \sqrt{\frac{r_{p1} r_{a1}}{r_{p1} + r_{a1}}} \\ v_{p1} &= \frac{h_1}{r_{p1}} \;, \end{align} \end{split}\]

where \(v_{p1}\) is the velocity of orbit 1 at its perigee.

Similarly, for orbit 2:

\[\begin{split} \begin{align} h_2 &= \sqrt{2 \mu} \sqrt{\frac{r_{p2} r_{a2}}{r_{p1} + r_{a2}}} \\ v_{p2} &= \frac{h_2}{r_{p2}} \;, \end{align} \end{split}\]

where \(r_{p2} = r_{p1}\) is the perigee distance, shared between the orbits. Then, the instantaneous velocity change needed is

\[ \Delta v = v_{p2} - v_{p1} \;. \]

Orbit 2 may be an elliptical transfer orbit, used to reach a higher circular orbit via an “apogee kick”, which is a second velocity change at the apogee of the elliptical orbit. We can determine the velocity change needed to enter a circular orbit with a radius matching \(r_{a2}\):

\[\begin{split} \begin{align} h_3 &= \sqrt{2 \mu} \sqrt{\frac{r_{a2} r_{a2}}{r_{a2} + r_{a2}}} = \sqrt{\mu r_{a2}} \\ v_{3} &= \frac{h_3}{r_{a2}} \;, \\ \Delta v &= v_3 - v_{a2} \end{align} \end{split}\]

Apse line rotation

A more general orbital transfer is an apse line rotation, which involves an instantaneous velocity change applied at some location in one orbit and leads to anew orbit with a different eccentricity and perigee/apogee distances.

Problem: Given initial perigee distance \(r_{p1}\), initial apogee distance \(r_{a1}\), desired perigee distance \(r_{p2}\), desired apogee distance \(r_{a2}\), and apse rotation angle \(\eta\).

Find: true anomaly of the burn \(\theta_1\), required instantaneous velocity change \(\Delta v\), and the required direction/angle of thrust from the local horizontal direction.

Solution steps:

  1. Calculate \(e_1\), \(e_2\), \(h_1\), \(h_2\):

    \[\begin{split} \begin{align} e_1 &= \frac{r_{a1} - r_{p1}}{r_{a1} + r_{p1}} \\ r_p &= \frac{h_1^2 / \mu}{1 + e_1} \\ \rightarrow h_1 &= \sqrt{ r_p (1 + e_1) \mu} \;, \end{align} \end{split}\]

    then repeat for \(e_2\) and \(h_2\).

  2. Solve for the true anomalies in each orbit of the intercept, \(\theta_1\) and \(\theta_2\):

\[\begin{split} \begin{align} \eta &= \theta_1 - \theta_2 \rightarrow \theta_2 = \theta_1 - \eta \\ r_{1I} &= \frac{h_1^2 / \mu}{1 + e_1 \cos \theta_1} = r_{2I} = \frac{h_2^2 / \mu}{1 + e_2 \cos \theta_2} \end{align} \end{split}\]

Following some math, we can solve for the angles:

\[\begin{split} \begin{align} \theta_1 &= \phi + \cos^{-1} \left( \frac{c}{a} \cos \phi \right) \\ \theta_2 &= \theta_1 - \eta \\ \text{where} \quad \phi &= \tan^{-1} \left( \frac{b}{a} \right) \\ a &= e_1 h_2^2 - e_2 h_1^2 \cos \eta \\ b &= -e_2 h_1^2 \sin \eta \\ c &= h_1^2 - h_2^2 \;, \end{align} \end{split}\]

where \(\phi\) gives the direction of the local horizontal.

  1. Calculate the position and velocity in the initial orbit:

\[\begin{split} \begin{align} v_{theta 1} &= \frac{h_1}{r_1} \\ v_{r 1} &= \frac{\mu}{h_1} e_1 \sin \theta_1 \\ v_1 &= \sqrt{ v_{r1}^2 + v_{\theta 1}^2 } \\ \gamma_1 &= \tan^{-1} \left( \frac{v_{r1}}{v_{\theta 1}} \right) \end{align} \end{split}\]

  1. Calculate the position and velocity in the desired orbit:

\[\begin{split} \begin{align} v_{theta 1} &= \frac{h_2}{r_2} \\ v_{r 2} &= \frac{\mu}{h_2} e_2 \sin \theta_2 = \frac{\mu}{h_2} e_2 \sin (\theta_1 - \eta) \\ v_2 &= \sqrt{ v_{r2}^2 + v_{\theta 2}^2 } \\ \gamma_2 &= \tan^{-1} \left( \frac{v_{r2}}{v_{\theta 2}} \right) \end{align} \end{split}\]

  1. Calculate the instantaneous change in velocity needed:

\[ \Delta v = \sqrt{ v_1^2 + v_2^2 - 2 v_1 v_2 \cos \left( \theta_2 - \theta_1 \right)} \;, \]

at the intercept location.

  1. Calculate the direction/angle of thrust:

\[ I = \tan^{-1} \left( \frac{v_{r2} - v_{r1}}{v_{\theta 2} - v_{\theta 1}} \right) + 180^{\circ} \]

Cost of transfers

The cost of an orbit transfer, in terms of the mass of propellant needed, depends on a few factors:

  • propellant used

  • thrust developed

  • rate of fuel consumption (\(\dot{m}\))

The specific impulse, given in units of seconds, is a performance parameter that quantifies the amount of thrust developed to the mass flow rate of propellant required, where higher values indicate (theoretically) better performance:

\[ I_{sp} = \frac{T}{\dot{m} g_0} \;, \]

where \(T\) is the thrust force (N), \(\dot{m}\) is the mass flow rate (kg/s), and \(g_0\) is the acceleration due to gravity on the ground (9.81 m/s\(^2\)). The mass flow rate is directly related to the time rate of change of vehicle mass: \(\dot{m} = -\frac{dm}{dt}\).

Based on a free-body diagram, we can write the equation of motion in the tangential (axial) direction:

\[ m a_t = T - D - m g \sin \gamma = m \frac{dV}{dt} \;, \]

where in space we can neglect the effects of drag, or \(D \approx 0\), due to the extremely rarified atmosphere (for highly accurate calculations, we will need to account for the perturbuations due to atmospheric drag); we can also neglect the weight due to reduced gravity, so \(m g \approx 0\). We can also assume thrust is constant. Then:

\[\begin{split} \begin{align} T = m \frac{dV}{dt} &= I_{sp} g_0 \left( - \frac{dm}{dt} \right) \\ - I_{sp} g_0 \frac{dm}{dt} &= m \frac{dV}{dt} \\ \int_{m_0}^{m_f} - I_{sp} g_0 \frac{dm}{m} &= \int_0^{\Delta V} dV \\ \Delta v &= - I_{sp} g_0 \ln \left( \frac{m_f}{m_0} \right) \\ \rightarrow \Delta m &= m_0 \left( 1 - e^{-\Delta V / \left( I_{sp} g_0 \right)} \right) \;, \end{align} \end{split}\]

where \(\Delta m = m_0 - m_f\), which is the ideal rocket equation, originally developed by Tsiolkovsky. Given the specific impulse and initial mass of a rocket, this tells us the change in mass needed (via exhaust of propellant) to achieve a particular change in velocity.

Finite burn times

While some burns can reasonably be approximated as instantaneous, other orbit transfers will involve or require finite burn times. These involve a burn of the engine starting at some time (and some true anomaly in the initial orbit), and leads to a different orbit, similar to an apse line rotation.

To determine how a finite burn time affects an object’s orbit, we need to solve the orbital equation of motion, but with a thrust force:

\[ \vec{\ddot{r}} + \frac{\mu}{r^3} \vec{r} = \frac{\vec{F}_T}{m} = \frac{T \, \vec{v}}{m v} \;, \]

assuming that the thrust force \(\vec{F}_T\), with magnitude \(T\), is in the same direction as the velocity \(\vec{v}\), along with an equation for the changing mass of the vehicle, as it expels propellant:

\[ \frac{dm}{dt} = -\frac{T}{I_{sp} g_0} \;. \]

This can be solved by decomposing into a system of first-order equations, and integrating numerically:

\[\begin{split} \begin{align} z_1 &= x \quad \dot{z}_1 = z_4 \\ z_2 &= y \quad \dot{z}_2 = z_5 \\ z_3 &= z \quad \dot{z}_3 = z_6 \\ z_4 &= \dot{x} \quad \dot{z}_4 = \frac{-\mu}{r^3} z_1 + \frac{T z_4}{z_7 v} \\ z_5 &= \dot{y} \quad \dot{z}_5 = \frac{-\mu}{r^3} z_2 + \frac{T z_5}{z_7 v} \\ z_6 &= \dot{z} \quad \dot{z}_6 = \frac{-\mu}{r^3} z_3 + \frac{T z_6}{z_7 v} \\ z_7 &= m \quad \dot{z}_7 = -\frac{T}{I_{sp} g_0} \;, \end{align} \end{split}\]

where

\[\begin{split} \begin{align} r &= \sqrt{z_1^2 + z_2^2 + z_3^2} \\ v &= \sqrt{z_4^2 + z_5^2 + z_6^2} \;. \end{align} \end{split}\]

To solve a problem where the burn time \(\Delta t\) to reach a particular orbit is unknown, an iterative approach can be used.

Orbits in space Rendezvous