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Assume delta-v isn't a problem and circular orbits. EDIT: Assume that you're already in orbit so you don't have to shift a massive load of fuel up, and the absolute ideal is something that has a calculable delta-V like a Hohmann transfer.

If there was some kind of industrial accident on Mars and we needed to leave right away, is there always a burn we can perform to get to where we want to be?

I can imagine that we could move into an orbit closer to the sun to 'catch up' Mars or further away than it to slow down, but would this be significantly quicker than just waiting for our ideal launch window?

Is there a way to optimise the arrival, similar to a Porkchop Plot?

Background - I'm thinking about a management game with orbital mechanics where there's a clear 'optimal' time to travel - this is the equivalent of the launch window - but there's an ability to travel at any time if you're willing to pay the extra cost for the rocket fuel.

So ideally the 'optimal' case is something like a Hohmann or Bi-elliptic transfer, and then the further you move away from your launch window, the more dV (and such the more cost) it will be.

Clearly it's a game so I can throw realism out the window, but I was wondering if there's a specific realistic trajectory that a rocket would take if an emergency mission needed to be mounted - plus, waiting two yeard for the optimal trajectory to restock your Mars base would be a bit of a drag.

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Hi Matt, you might want to reword your question, if delta-v isn't a problem then you can pretty much go anywhere anytime you want. You might want to give a little more background info upfront and restate it so that the assumption is that your interested in keeping normal physics, but you can assume idealized conditions. – user11547 Feb 5 '13 at 11:01

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I think you are describing the bi-elliptic transfer, where you raise the apoapsis above the target and then do a retrograde burn there to lower the periapsis. Technically this can save a bit of $\Delta v$ compared to a Hohmann transfer because the $\Delta v$ is spent more efficiently (the Oberth effect), but it also takes longer. If dv really isn't a problem you will always make the best time with a brachistochrone trajectory: accelerate continuously towards the target until you reach the halfway point, then flip around and deccelerate till you reach the destination at a standstill. This is really a generalised brachistochrone since you need to take into account the $1/r^2$ variation of gravity in the solar system. You can also take advantage of aerobraking at the target (if there is an acceptable atmosphere) to push the turnaround point even closer to the target and cut some time off the trip. Tweak as necessary - the idea still works. Porkchop plots are used to minimize the energy/$\Delta v$ requirement since in the real world lifting un-needed rocket fuel out of the Earth's gravity well is an expensive waste.

EDIT: I read an article from NASA describing work using genetic algorithms to optimize these sorts of continuously accelerating transfers. Sadly I don't have the reference any more, but its probably possible to find online.

EDIT 2: You can find a lot of information relating to real and fictional technologies from a hard sci-fi writing perspective at Atomic Rockets. If you are prepared to throw realistic $\Delta v$ requirements out the window then the only limitation on mission time is the acceleration your squishy human cargo is able to endure. For instance, if you pack your people in acceleration couches for the duration you can accelerate at, lets say, 3g long term (I'm not volunteering for this). An accelerate halfway/flip over/deccelerate mission takes (in the first approximation neglecting gravity) takes a time

$$ t = 2\sqrt{\frac{d}{a}} $$

to go a distance $d$ at an acceleration of $a$. To get a rough worst case figure take Earth and Mars at their furthest opposition $ d \approx 2.7 \mathrm{AU} $ so

$$ t_{max} \approx 65 \mathrm{hr}\ !!!$$

or if you only subject your passengers to 1g

$$ t_{max} \approx 113 \mathrm{hr}\ !!!$$

Clearly this is nothing like a real world mission profile. The reason is that the $\Delta v$ requirement for the 1g mission is

$$ \Delta v = a t = 4\times10^6 \frac{\mathrm{m}}{\mathrm{s}} $$

For the 3g mission it goes up to $6\times10^6 \frac{\mathrm{m}}{\mathrm{s}}$. You need a rocket engine with an exhaust velocity of the same order of magnitude if you want to achieve anything like a reasonable cargo/passenger capacity. This is, ahem, not currently feasible. But if you're willing to posit it then you hardly need to worry about complications like the sun's gravity - you can basically go in a straight line! (If you want to check my numbers, please do. I'm not entirely sure I believe them!)

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Hey - I think this is great, it's almost there. I've added a bit of background above as well. If I understand correctly the brachistochrone trajectory could be calculated for an interplanetary burn and it would essentially be the shortest route, using the gravity of the sun to 'pull' the ship in after its burn? – Matt Kemp Feb 5 '13 at 14:01
@MattKemp See my edit. Hope it helps - if not let me know! – Michael Brown Feb 5 '13 at 14:37
A brachistochrone is a shortest time trajectory. It is more complicated here than the case in the wikipedia article since the acceleration is changing in both magnitude and direction, but the principle is the same. – Michael Brown Feb 5 '13 at 14:39
I should also mention the Kzinti lesson: "a reaction drive's efficiency as a weapon is in direct proportion to its efficiency as a drive." Or equivalently Jon's law for sci-fi writers: "Any interesting space drive is a weapon of mass destruction." :) – Michael Brown Feb 5 '13 at 14:49
Great, thanks - it also includes one specific tidbit: "A Hohmann orbit is the maximum transit time / minimum deltaV mission. A "Brachistochrone" is a minimum transit time / maximum deltaV mission" which neatly explains why they don't degrade nicely. What would be the 'midpoint' to accelerate to? Would you constantly accelerate in the direction of the target, or towards the position of the target that you expect to intercept it at? – Matt Kemp Feb 5 '13 at 15:29

Can we transfer burn to another planet at any time?

Yes... if we have a big enough rocket. The problem of figuring an orbit that gets you from point A to point B in a certain time is called Lambert's problem, and it turns out that there is a solution that is relatively easy to calculate—by which I mean that it does require solving equations numerically, but doesn't require numerical integration.1

First off, let's define some variables. I'll refer to a transfer between Earth and Mars, but this method works for any transfer. The distance of the starting point from the central body (in this case the Sun) is $r_1$, and the distance from the central body to the destination is $r_2$. The angle between the two points, measured from the central body, is $\theta$. Note carefully that $\theta$ is the angle between Earth at the time of launch and Mars at the time of arrival. Finally, $t_f$ is the time spent in transfer orbit (a.k.a. time-of-flight). Note that in real life there is a relation between $\theta$, $t_f$, and the time of launch, due to the motion of the planets.

We can calculate a couple of quantities that will make the problem easier. Let $c$ be the linear distance between the launch and destination locations. By the law of cosines:

$$ c^2 = r_1^2 + r_2^2 - 2r_1 r_2 \cos\theta $$

Next, let $s=(r_1+r_2+c)/2$.

Now we can calculate an important limiting case for the problem: the time it would take to transfer from $r_1$ to $r_2$ on a parabolic orbit, $t_p$. If we go any faster than this a solution still exists, but it will be a hyperbolic orbit (and the steps that follow will no longer be valid). In this case a hyperbolic trajectory would be "too fast" for practicality, as it would require enough energy to escape the solar system.

$$ t_p = \frac{\sqrt{2}}{3}\frac{1}{\sqrt{\mu}}\left[s^{3/2}-\mathrm{sgn}(\sin\theta)(s-c)^{3/2}\right] $$

Two things to note: $\mu$ is the gravitational parameter of the central body (the Sun), equal to its mass times the gravitational constant.2 Second, $\mathrm{sgn}$ is the signum function, equal to $+1$ when its argument is positive, and $-1$ when its argument is negative. Therefore we have:

$$ \mathrm{sgn}(\sin\theta) = \left\{\begin{array}{rl}+1 & 0\leq\theta<\pi \\ -1 & \pi <\theta\leq 2\pi\end{array}\right. $$

This reveals a distinction between orbits with $\theta<\pi$ (the "short way") and $\theta>\pi$ (the "long way") that we'll see again later.

Now I'll introduce Lambert's equation:

$$ t_f = \frac{a^{3/2}}{\sqrt{\mu}}\left[\alpha - \beta - (\sin \alpha - \sin \beta)\right] $$

You may notice a complication: this is an equation for $t_f$, when we really need a solution for $a$ (the semimajor axis of the transfer orbit) in terms of $t_f$! Unfortunately we have to use an iterative solution method: we pick a guess for $a$, calculate $t_f$ and compare it to the desired value, and then update our guess. We repeat this process until we find the value of $a$ that solves the equation. Something like Newton's method is usually used.

To compute $t_f$ using Lambert's equation, we need two additional helper variables $\alpha$ and $\beta$:

$$ \begin{align} \sin\frac{\alpha}{2} &= \sqrt{\frac{s}{2a}} \\ \sin\frac{\beta}{2} &= \sqrt{\frac{s-c}{2a}} \\ \end{align} $$

The first of these equations constrains $a$ to be greater than $a_m=s/2$. Therefore $a_m$ is the smallest possible semimajor axis, and therefore the transfer orbit with the minimum energy (sort of the general equivalent of a Hohmann transfer). We can calculate the transfer time for this orbit, $t_m$, using Lambert's equation. Note that, in the minimum energy case, $\alpha$ and $\beta$ simplify:

$$ \begin{align} \sin\frac{\alpha_m}{2} &= 1 & \alpha_m = \pi \\ \sin\frac{\beta_m}{2} &= \sqrt{1-\frac{c}{s}} \\ \end{align} $$

Thus, we get:

$$ t_m = \frac{(s/2)^{3/2}}{\sqrt{\mu}}\left[\pi - 2\sin^{-1}\sqrt{1-\frac{c}{s}} - \left(1 - \sqrt{1-\frac{c}{s}}\right)\right] $$

We have one final complication to Lambert's equation that depends on the choice of sign for the square root in the equations for $\alpha$ and $\beta$. Calculate the initial values as above:

$$ \begin{align} \sin\frac{\alpha}{2} &= \sqrt{\frac{s}{2a}} \\ \sin\frac{\beta}{2} &= \sqrt{\frac{s-c}{2a}} \\ \end{align} $$

taking the positive square root, and principal branch of $\sin^{-1}$. Call these values $\alpha_0$ and $\beta_0$. Then handle the following special cases:

$$ \begin{align} \alpha &= \left\{\begin{array}{rl}\alpha_0 & t_f < t_m \\ 2\pi-\alpha_0 & t_f > t_m \end{array}\right. \\ \beta &= \left\{\begin{array}{rl}\beta_0 & \theta < \pi \\ -\beta_0 & \theta > \pi \end{array}\right. \end{align} $$

Now you can iterate Lambert's equation to find $a$ for a given $t_f$. Remember to recalculate $\alpha$ and $\beta$ on each iteration!

After calculating $a$, we can perform the final step: calculating the required initial velocity (as well as the resulting velocity at the destination). To do that we calculate a couple of helper variables, $\gamma_1$, $\gamma_2$, $A$, and $B$:

$$ \begin{align} r_2^2 &= r_1^2 + c^2 + 2r_1c\cos\gamma_1 \\ r_1^2 &= r_2^2 + c^2 + 2r_2c\cos\gamma_2 \\ A &= \sqrt{\frac{\mu}{4a}}\cot\frac{\alpha}{2} \\ B &= \sqrt{\frac{\mu}{4a}}\cot\frac{\beta}{2} \end{align} $$

I'll give the velocity in the tangential-normal basis; $\hat e_n$ points away from the central body; and $\hat e_t$ points perpendicular to $\hat e_n$ in the orbital plane.

$$ \begin{align} \vec v_1 &= \left[(\sin\gamma_1)A+(\sin\gamma_1)B\right]\hat e_t + \left[(-1-\cos\gamma_1)A+(1-\cos\gamma_1)B\right]\hat e_n \\ \vec v_2 &= \left[(\sin\gamma_2)A+(\sin\gamma_2)B\right]\hat e_t - \left[(-1-\cos\gamma_2)A+(1-\cos\gamma_2)B\right]\hat e_n \end{align} $$

To find out how much fuel you need, compute the difference between Earth's velocity and the required initial velocity. If you're assuming circular orbits, you can just subtract the average velocity from the $t$-component of the initial velocity. You can then find the magnitude of the required velocity, $v_1$.

A final thing to remember is that $v_1$ is the velocity that you need to have after leaving Earth's gravity well. To find the velocity you'd need from the Earth's surface, you need to factor in escape velocity:

$$ v_0^2 = v_\textit{esc}^2 + v_1^2 $$

Escape velocity $v_\textit{esc}$ at a distance $R$ from the center of a planet is just:

$$ v_\textit{esc} = \sqrt{\frac{2\mu}{R}} $$

Or $\sqrt{2}$ times the speed of a circular orbit at that altitude.

1 A different solution to Lambert's problem is presented in Fundamentals of Astrodynamics by Bate, Mueller, and White (pdf). They call it the Gauss problem, and it is the subject of chapter 5. If you are interested in orbital mechanics, I highly recommend this book.

2 Don't actually calculate $\mu$ by multiplying the mass and gravitational constant. The mass of astronomical bodies is hard to measure, and $G$ is also very hard to measure, so your value of $\mu$ will not be very precise. However, we can measure $\mu$ directly by looking at orbits around a body, and we can measure it with much more precision than either mass or $G$.

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