# Hohmann trajectory

## Contents

## Travelling to Mars via Hohmann

The Hohmann trajectory is used as the flightroute between Earth and Mars for most Mars mission plans (human and robotic). The rocket goes from Earth to Mars in two burns. The first burn is called the injection burn. This burn puts the rocket on an elliptical transfer orbit between Earth and Mars. This burn causes the rocket to depart from Earth in the same direction that the Earth is traveling. It gives the rocket just enough energy to reach the orbit of Mars.

After 8.6 months the rocket arrives at Mars on the opposite side of the Sun from where it left Earth. Upon arrival at Mars the rocket does a second burn called the circularization burn to match the orbit of Mars.

The transfer orbit is such that the point where it departs from Earth is the closest point on the transfer ellipse to the Sun and the point where the transfer orbit intersects Mars is the furthest point in the orbit from the Sun.

## Minimum energy Transfer

The Hohmann orbit uses as little fuel as is possible for all transfer orbits based on the two body problem. The reason it is minimum energy is because all thrust is in the direction the body is traveling. Upon departure the rocket leaves in the direction that the Earth is traveling and upon arrival the object is traveling in the same direction as Mars prior to the circularization burn.

Intuitively one would say all work is done to accelerate the object and no work is done by the rocket to change the direction it travels. Mathematically we recognize that if the rocket fires in the direction it is traveling, the rocket speeds up as a consequence the rocket burn occurs over a larger distance. Since work is the product of force and distance a greater amount of work is done to increase the velocity of the rocket.

Consequently the Hohmann transfer is not necessarily the only minimum energy transfer for the two body problem. However, if the rocket takes longer to accelerate it will spend more time at slower velocities. Since the acceleration is approximately infinite for a Hohmann transfer it is the fastest of all the minimum energy transfers for the two body problem.

**see also:**

Three body problem

Interplanetary Super Highway

“Zero Energy Transfer”

## Mathematics

If the periapsis and period of the starting orbit and the apoapsis and period of the destination orbit are known, then Kepler's Laws of Planetary Motion can be used to estimate the time required for a Hohmann trajectory transfer orbit.

The semi-major axis of the transfer orbit will be roughly:

<math>a_{t} = {r_{1} + r_{2} \over 2}</math>

where:

<math>a_{t}</math> is the semi-major axis of the transfer orbit

<math>r_{1}</math> is the periapsis (distance furthest from the sun) of the starting orbit, measured in AU

<math>r_{2}</math> is the apoapsis (distance closest to the sun) of the destination orbit, measured in AU

The period of the complete Hohmann orbit (of which the Hohmann transfer tranjectory is a part) is then given by Kepler's Third Law of Planetary Motion, using Earth's orbit as a reference:

<math>P_{Hohmann} = P_{Earth} \cdot {a_{t} \over a_{Earth}}^{3/2}</math>

<math>P_{Hohmann} = 2 P_{Transfer}</math>

where:

P_{Hohmann} is the period of the complete Hohmann orbit

P_{Transfer} is the time required for the Hohmann Transfer

P_{Earth} is the orbital period of Earth's orbit, equal to one Earth year

a_{Earth} is the semimajor axis of the Earth's Orbit, which is approximately 1 AU

The period and velocity of an orbit can also be calculated using the standard gravitational parameter:

<math>\mu = G M</math>

where:

M the mass of the object being orbited (The Mass of sun equals <math>2 \cdot 10^{30} kg</math>)

G is the gravitational constant (<math>6.6742 \cdot 10^{-11} {m^{3} \over s^{2} kg} </math>)

<math>P = \sqrt{\left( {\mu \over r_{1}} \right)} \cdot \sqrt{\left( {2 \cdot r_{2} \over (r_{1} + r_{2}) - 1} \right)} </math>

The total delta v required for the Hohmann transfer will be the sum of the the velocity magnitude change required for orbital insertion and the velocity magnitude change required for the circularization burn. If the initial and final orbits are assumed circular, then these velocities are:

<math>\Delta v_{1} = \sqrt{2 \mu ( {1 \over r_{1}} - {1 \over r_{1}+r_{2}} )}- \sqrt{\mu \over r_{1}} </math>

<math>\Delta v_{2} = \sqrt{\mu \over r_{2}} - \sqrt{2 \mu ( {1 \over r_{2}} - {1 \over r_{1}+r_{2}} )} </math>

<math>\Delta v_{Total} = \Delta v_{1} + \Delta v_{2}</math>

where:

<math>\Delta v_{1}</math> is the delta v required for insertion into the transfer orbit, starting in orbit 1

<math>\Delta v_{2}</math> is the delta v required for circularization from the transfer orbit, finishing in orbit 2

<math>\Delta v_{1}</math> is the total delta v required for the Hohmann Transfer, from beginning to end.

## Example of Hohmann Transfer

Assume travel is from Earth orbit to a circular orbit at the mean orbital distance of Mars. (Mars orbit is not exactly circular, but this provides a good approximation.)

Let:

<math>r_{1} = 1 AU = 149.6 \cdot 10^{6} km</math>

<math>r_{2} = 1.52 AU = 227.4 \cdot 10^{6} km</math>

This implies that r_{t} = 188.5 * 10^{6} km.

From this, we can employ Kepler's Laws to estimate the period of the entire transfer orbit, P = 1.414 Earth Years or 516.6 days. The time required for the transfer is half that, or 258.3 days. If a circularization burn were not used at the end of the transfer, a spacecraft in a Hohmann orbit would continue to cycle back and forth between the Earth and Mars every 516.6 days. Neither Earth nor mars would be nearby every time the transfer orbit coincided, however - a return to the original orientation would take years. Thus minimum energy Hohmann trajectories have limited utility for an Earth-Mars Cycler vehicle.

The velocity required for the orbital injection is approximately 2.9 km/s. The value required for the orbital circularization is approximately -2.4 km/s. This implies a total delta v of approximately 5.3 km/s for the entire orbital transfer. The actual figure will be slightly more due to the fact that Mars and Earth orbits are not exactly circular, and due to necessary orbital correction that will occur during flight.

## External Links

Rocket and Space Technology Orbital Mechanics Page