# Hohmann Transfer¶

If the initial orbit and final, or target, orbit do not intersect, we need at least two impulses to transfer between the orbits. In 1925, Walter Hohmann [Hoh60] showed that the most efficient way to do this with two impulses, when the initial and final orbits are circular, is to connect opposite sides of the initial and target orbits with an ellipse. This transfer is called a Hohmann transfer.

Fig. 61 The Hohmann transfer orbit. Leafnode CC BY-SA 2.5, via Wikimedia Commons.

In Fig. 61, the spacecraft is initially in orbit 1 with radius $$\mathsf{R}$$ and is targeting orbit 3 with radius $$\mathsf{R}'$$. To achieve the transfer, the spacecraft boosts its velocity into an elliptical orbit, 2. When the spacecraft arrives at the far side of the ellipse, it boosts its velocity again to move into orbit 3.

The periapsis distance of this transfer orbit is equal to $$\mathsf{R}$$ and the apoapsis distance is $$\mathsf{R}'$$. This means that the transfer orbit intersects both orbits 1 and 3. By providing impulses at both intersection points, it is possible to move from orbit 1 to orbit 3.

## Energy Considerations¶

The specific energy of the initial and final orbits are given by Eq. (123), repeated here for reference:

$E = -\frac{\mu}{2r}$

where $$r$$ is the radius of the circular orbit. Let subscripts $$i$$ and $$f$$ represent the initial and final orbits, respectively. Then, if $$r_i < r_f$$, the specific energy increases (gets closer to zero) as a result of the transfer. On the other hand, if $$r_i > r_f$$, the specific energy decreases (gets more negative).

The specific energy of the elliptical transfer orbit is given by Eq. (134), repeated here for reference:

$E = -\frac{\mu}{2a}$

For the Hohmann transfer, the semi-major axis of the transfer orbit is between the radii of the initial and final orbits. Therefore, the specific energy of the elliptical transfer orbit is also between the values for the initial and final orbits. We will use the subscript $$t$$ for quantities associated with the transfer orbit.

## Calculating $$\Delta v$$¶

At the locations where orbit 2 intersects with orbits 1 and 3, the velocity vectors are parallel. The reason the Hohmann transfer is the most efficient two-impulse maneuver is because only the magnitude of the velocity needs to change, not its direction as well. This means that the minimum propellant is used to achieve the necessary $$\Delta \vector{v}$$.

For multiple-impulse maneuvers, we are interested in determining the total $$\Delta v$$. Since a maneuver to decrease the spacecraft velocity consumes the same amount of propellant as one that increases the velocity we are concerned with the magnitude of all of the necessary velocity changes.

For each impulse, we need to calculate the difference in velocity between the circular orbits and the transfer orbit. Since the velocity vectors are parallel at both impulse locations we can work entirely with velocity magnitudes.

The velocity on the circular orbits is given by Eq. (121):

$v_{\text{circular}} = \sqrt{\frac{\mu}{r}}$

Since the impulses occur at periapsis and apoapsis of the transfer orbit, the transfer orbit velocity is given by:

(284)\begin{aligned} v_{t,p} &= \frac{h_t}{r_p} & v_{t,a} &= \frac{h_t}{r_a} \end{aligned}

From the equations for elliptical orbits, we can solve for the specific angular momentum in terms of $$r_p$$ and $$r_a$$:

(285)$h_t = \sqrt{2\mu \frac{r_a r_p}{r_a + r_p}}$

Alternatively, we can use the vis viva equation, Eq. (93), to find the velocity on the transfer orbit

(286)\begin{aligned} v_{t,p} &= \sqrt{\mu\left(\frac{2}{r_{t,p}} - \frac{1}{a_t}\right)} & v_{t,a} &= \sqrt{\mu\left(\frac{2}{r_{t,a}} - \frac{1}{a_t}\right)} \end{aligned}

where the semi-major axis of the elliptical transfer orbit is:

(287)$a_t = \frac{r_{t,a} + r_{t,p}}{2}$

Finally, the total $$\Delta v$$ required for the Hohmann transfer is:

(288)\begin{split}\begin{aligned} \Delta v &= \lvert v_f - v_{t,a}\rvert + \lvert v_i - v_{t,p}\rvert & \text{If r_i < r_f} \\ \Delta v &= \lvert v_f - v_{t,p}\rvert + \lvert v_i - v_{t,a}\rvert & \text{If r_f < r_i} \end{aligned}\end{split}

### Transit Time¶

Since the Hohmann transfer traverses half of the ellipse, the transfer time is given as half the period of the elliptical orbit from Eq. (138):

(289)$t_{12} = \frac{T}{2} = \pi\sqrt{\frac{a_t^3}{\mu}}$

where $$t_{12}$$ is the transfer time and $$a_t$$ is the semi-major axis of the transfer orbit.

## Elliptical Initial or Target Orbits¶

The idea of a Hohmann transfer can be extended to the case where one or both of the initial and final orbits are ellipses. Recalling that the Hohmann transfer traverses half of the elliptical transfer orbit, the departure and arrival points on the initial and final orbits will be on opposite sides of the attracting body. In addition, the definition of the Hohmann transfer is that the transfer orbit at the departure and arrival points should be tangent to the initial and final orbits, respectively.

To start, let’s assume that both the initial and final orbits are ellipses. The orbits have a common focus and their eccentricity vectors point in the same direction. For the transfer orbit to be tangent to the initial and final orbits, we have two choices:

1. Depart at periapsis of the initial orbit, arriving at apoapsis of the final orbit

2. Depart at apoapsis of the initial orbit, arriving at periapsis of the final orbit

It is not clear which transfer orbit is more efficient in terms of the $$\Delta v$$ requirement to achieve the transfer. We can analyze each transfer orbit in terms of the energy equation to determine which transfer is more efficient.

First, we will assume that the initial orbit has a smaller semi-major axis than the final orbit. Therefore, the final orbit has a higher energy (closer to zero) than the initial orbit. Likewise, the transfer orbit will have a higher energy than the initial orbit and a smaller energy than the final orbit.

To change from the initial orbit to the transfer orbit, the kinetic energy of the spacecraft must increase. The point on the initial orbit where the kinetic energy is closest to the transfer orbit is when the velocity of the initial orbit is highest: at periapsis.

To change from the transfer orbit to the final orbit, the kinetic energy must increase again. The point on the final orbit where the kinetic energy is closest to the transfer orbit is when the velocity of the final orbit is lowest: at apoapsis.

Therefore, we see that option 1 is the case that provides the lowest energy transfer. Although we conducted this analysis for the case of $$a_i < a_f$$, the same type of reasoning will apply. See if you can decide which option is best for the case of $$a_i > a_f$$ before you open the answer below.