Path to Mars

 While the scope of these discussions is not really intending to delve too deeply into all the technical issues, there may be some value in outlining some of the detail related to the flight path to Mars and back, as it might provide an insight to the distances involved and the time duration based on certain assumptions. While there are many possible permutations, this discussion will consider the basic idea encapsulated in terms of a Hohmann transfer path from Earth to Mars. However, the simplified derivations to be followed assumes that the orbits of Earth and Mars are both circular around the Sun, where we might estimate the radius of the Earth's orbit [r1] to be 150 million kilometres or one astronomical unit (AU).

Note: for the purposes of this discussion, we can simplify the calculations by using relative ratios of orbital distance and time with respect to Earth. As such, the Earth has an orbital radius [r1=1] and an orbital period [T1=1]. In comparison, Mars has an orbital radius [r2=1.5237] and an orbital period [T2=1.8822].

Based on the initial diagram left, the distance between Earth and Mars would be 0.5237 AU’s, which approximates to 78 million kilometres. However, most practical trajectories between Earth and Mars have to take into account the orbital velocity of the Earth [30km/s] and Mars [24km/s], such that the destination end-point in either direction is always a moving target. Therefore, these practical trajectories never align to the minimum of 78 million kilometres and are further complicated by many other issues, which will not be detailed. Again, for simplicity, we might assume that the spaceship is already in an Earth orbit, where its gravitational effects can be ignored, although we might realise that a spaceship in Earth orbit will still align to Earth’s orbital velocity [30km/s] around the Sun. As such, we might realise that the path defined by [V0] in the first diagram is not a practical solution, which leads us to consider a ‘Hohmann Transfer Orbit’ between Earth and Mars as illustrated in the second diagram right. In addition to the spaceship’s arrival being synchronised to the actual orbital position of Mars, we need the approach velocity of the spaceship to be compatible with the orbital velocity of Mars. In the second diagram, the position [P] aligns to the perihelion of Earth’s orbit, i.e. closest to the Sun, while [A] aligns to the aphelion of Mars, furthest from the Sun. However, we might recognise that the initial position of Mars, at the time of the spaceship’s launch has to be calculated such that it arrives at point [A] at the same time as the spaceship. To determine that position, we need to know the duration of the flight from [P] to [A], which can be calculated using Kepler's 3rd law. This law can be expressed in the concise mathematical form of [1]:

[1]   

Where [T] is the orbital period and [a] the semi-major axis, half the length of the orbital ellipse, while the constant applies to all objects orbiting the Sun, including the Earth. However, the value of the constant depends on the units in which [T] and [a] are defined and can be simplified, if we adopt he ratio units with respect to Earth as follows:

[2]   

As a consequence, the unity value of the constant can be used for any planet, such that [2] reduces to:

[3]   

However, [3] holds true for any orbit around the Sun, including the spaceship in a transfer ellipse, where all distances are defined in AU units.

[4]   

 We can now calculate [T] from [3] by substituting for [a]:

[5]   

However, the result in [5] represents the total time between [PAP], not [PA], such that the trajectory time is [PA=0.7087] or approximately 258 days. Knowing the trajectory time from [P] to [A], now allows us to calculate the position of Mars when the spaceship leaves [P], which in angular terms is the ratio of the trajectory time [0.7087] to the orbit time of Mars in the same units, as shown in [6] and the diagram right:

[6]   

However, the trajectory chosen for any round-trip mission to Mars, and back, can be complicated by all manner of trade-offs between travel time and stay duration. For example, a lower-energy outbound trajectory invariably leads to a relatively unfavourable alignment for an energy-efficient return. This trade-off can be described in terms of two distinct classes of round-trip Mars missions, i.e. opposition-class or conjunction-class, where the former leads to a short-stay mission, while the latter leads to a longer-stay mission.

The previous diagram is simply reflective of the trade-off between these two different classes of trajectory, where the duration times can be subject to considerable variance based on when the mission is scheduled. Again, the logistical differences between the opposition-class and conjunction-class trajectories is driven by the fact that the Earth has an orbital velocity of 30m/s, while Mars has a lower orbital velocity of 24m/s, such that the shortest distance between these planets occurs every 2 years. At one level, the planning of manned missions to Mars has often favoured the conjunction-class, which while suggesting a longer mission duration, leads to a minimum energy trajectory, i.e. a lower propellant requirement. However, in terms of reduced mission risk, the shorter mission duration associated with the opposition-class trajectory might be better, especially on the initial manned missions. However, while opposition-class trajectories invariably require more energy, i.e. propellant, the SpaceX Mars mission plan might mitigate this issue if propellant can be produced on Mars.

But what velocities are involved?

While this next section becomes a little more involved in mathematical manipulation, the final results may be generally useful for later reference. We can readily calculate the orbital velocity of Earth and Mars around the Sun with mass [MS] as follows, where [r1, r2] are the orbital radius of Earth and Mars respectively.

[7]   

However, we might realise that without an injection of the delta-velocities [∆v1] and [∆v2], any spaceship will not move onto the Hohmann transfer path starting at [E] and arrive at [M] with the correct velocity, such that we might express this requirement as follows:

[8]   

We can proceed from [8] by using the conservation of momentum, where [m] relates to the mass of the spaceship, but which can be immediately cancelled out.

[9]   

The next step to determine [vP] is a little more protracted based on the conservation of energy:

[10] 

Again, we can immediately cancel the mass of the spaceship throughout and substitute for [vA]

[11] 

At this point, we want to try to isolate the terms associated with [vP] on one side:

[12] 

Again, we can manipulate the form in [12] using standard algebra:

[13] 

Isolating in terms of [vP] we get:

[14] 

Having formed the equations for both [vA] and [VP] in [9] and [14], we can now rearrange [8] in terms of the delta-v [∆v] values and substitute in the expressions derived.

[15] 

We will start with [∆v1]:

[16] 

And now [∆v2]:

[17] 

At this point, we have all the equations to quantify a Hohmann transfer path between Earth and Mars, although it needs to be remembered that these equations are based on several simplifying approximations. However, we might use the results to provide a general insight.

[18] 

In [18], we see the starting velocity [vP] is the sum of the Earth’s orbital velocity around the Sun [v1=30km/s] and the delta-velocity [∆v1=2.94km/s] needed to escape into the required Hohmann transfer path with velocity [vP=32.94km/s]. However, [9] tells us that this velocity is not maintain all the way to Mars due to the gravitational effects of the Sun while in transit:

[19] 

We can complement [18] with a corresponding delta-v [∆v] process at Mars.

[20] 

 While there is some inaccuracy in these figures, they are a reasonable approximation. However, as indicated, the actual process involves considerably more complexity as neither Earth or Mars have perfectly circular orbits, where Mars can be affected by the relative position of Jupiter. There is also the complexity of the amount of fuel used in the delta-v manoeuvres, which we might simply outline in terms of the Tsiolkovsky rocket equation.

[21] 

Where [Δv] is the maximum change of velocity, [ve] is the exhaust velocity, [m0] is the initial mass of the spaceship, including propellant and [mf] is the final mass of the spaceship without propellant. However, we might rearrange [21] in order to show the fuel dependency on [Δv] and [ve].

[22] 

In [22], we see the exponential sensitivity of the mass ratio, i.e. fuel usage, to changes in the ratio of [Δv/ve] associated with manoeuvring the spaceship along some given trajectory, which is also illustrated in more detail in the pe graph right.