Practise the mathematics behind this
The practice portal carries the algebra, logarithms, calculus and vector work this article uses, with worked solutions that explain the reasoning rather than just the answer.
Open the practice portal →Everyone quotes the seven minutes of terror. Almost nobody explains why those seven minutes are hard, and the reason is mathematical rather than mechanical: the equation governing a Mars entry cannot be rearranged and solved. Here is the equation, the reason it resists, and the graph I got by solving it the way engineers actually do.
Landing on Mars takes about seven minutes, and for most of them nobody on Earth knows what is happening. The radio delay is longer than the descent. By the time a signal arrives saying the vehicle has entered the atmosphere, it has already landed or already failed. That fact gets quoted a great deal. What gets quoted much less often is the reason those seven minutes are hard, which is not distance, or radiation, or engineering nerve. It is that the problem cannot be solved with algebra.
This article works the problem the way it is actually worked. It uses ideas a student meets in a first calculus course, and it ends with a graph that I generated by solving the equations numerically rather than by looking one up. I want to show what a differential equation is for, because that is the thing school treatments most often leave out.
A vehicle falling towards Mars has gravity pulling it down, air resistance pushing it back along its path, and eventually engine thrust. Newton's second law says the acceleration is the total force divided by the mass, which for motion in the vertical direction gives:
Nothing here is exotic. A student who has met can read every symbol. But notice what has happened: the acceleration on the left is the second derivative of height, and the drag term on the right depends on both the height and the speed. The unknown appears on both sides, wrapped inside derivatives of itself. That is what makes it a differential equation rather than a formula, and it is why you cannot simply rearrange it and read off the answer.
Drag is the term that decides everything, and it is worth writing out:
Here is the air density, is the speed squared, is a drag coefficient that depends on the shape, and A is the frontal area. Two features of this expression drive the entire descent.
The first is that drag goes as the square of speed. Halving your speed cuts drag to a quarter. Braking is therefore violently effective at first and then rapidly stops helping, which is why a Mars entry vehicle sheds most of its speed in the first minute and then spends the rest of the descent in a much gentler regime.
The second is the behaviour of the density. Mars has an atmosphere about one percent as dense as Earth's at the surface, and like every atmosphere it thins out roughly exponentially with height:
with a scale height . That means the density falls by a factor of e for every 11 kilometres you climb, so it is a hundred times thinner 50 kilometres up than it is at the surface.
Put the two together and you have the whole difficulty of Mars in one sentence. At the top of the descent the vehicle is going very fast, so the speed squared term is enormous, but the density is almost nothing. Near the bottom the density is at its highest, but the speed has already fallen so the speed squared term has collapsed. Somewhere between the two there is a moment where the product is at its largest. Finding that moment is the engineering problem.
Substituting the drag expression into Newton's second law and dividing through by the mass gives the equation that governs the entry:
This has no neat closed form solution. There is a classical approximation, published by Allen and Eggers in 1958, which is genuinely elegant and still taught, but it needs assumptions that a real vehicle does not honour. So engineers do what engineers do: they step the equation forward in time in very small increments.
Compute the acceleration from where you are now, use it to update the speed and the height a hundredth of a second later, then do it again. Repeat about ten thousand times. That is the whole method, and it is called Euler stepping. It is not sophisticated. It is simply what you do when the mathematics refuses to give you a formula.
Here is the result of doing exactly that, for a vehicle entering the Martian atmosphere at 5.5 kilometres per second on a fifteen degree path, with a ballistic coefficient of 115 kilograms per square metre.
Read the solid curve from the top right. For the first 50 kilometres of descent almost nothing happens: the vehicle is moving at over five kilometres per second through air so thin it barely registers. Then, over roughly twenty kilometres of altitude, the speed collapses. The dashed curve shows why: the deceleration rises to a sharp peak near 22 kilometres up, reaching about 13 times Earth gravity in this model, then falls away almost as fast as it rose.
That peak is not a design choice. Nobody put it there. It emerged from solving the equation, and it exists because two competing exponentials cross: the density climbing as the vehicle falls, and the speed squared collapsing as the vehicle brakes. This is the single most important thing a student can take from the problem. The interesting features of a differential equation are usually not written anywhere in it. You have to solve it to find out they are there.
Two caveats, because the model above is deliberately simple. A real entry vehicle flies at an angle that generates lift, and steers with it, which spreads the deceleration out and lowers the peak considerably. And a real trajectory is two dimensional, so the vehicle travels several hundred kilometres downrange while it descends. Both refinements make the equations messier without changing their character. They are still differential equations, and they are still solved by stepping.
Look again at the drag term and you will notice that the mass, the drag coefficient and the area only ever appear in one combination:
This is the ballistic coefficient, and it is the number Mars engineers argue about. A low means a light vehicle with a large blunt face: it slows down high in the atmosphere, where there is still plenty of altitude left to deploy a parachute. A high means a heavy compact vehicle: it punches deep before slowing, and may run out of sky.
This is the reason Mars is harder to land on than either the Moon or the Earth, and the reason is worth stating precisely. On the Moon there is no atmosphere at all, so you cannot brake aerodynamically and must do the whole job with engines, but at least the physics is simple and the answer is just propellant. On Earth the atmosphere is thick enough to do nearly all the braking for you, and a parachute finishes the job. Mars is the cruel middle case: there is enough atmosphere to cause severe heating and to demand a heat shield, but not enough to slow a heavy vehicle to a safe landing speed. Every Mars landing has to combine a heat shield, a supersonic parachute and rocket engines, because no one of them is sufficient.
And the difficulty scales the wrong way. Make the vehicle heavier without making it wider, and goes up, and the vehicle arrives lower and faster. This is the reason landing large payloads on Mars remains an open engineering problem rather than a solved one.
I teach this problem to students who are comfortable with differentiation and have met the idea of a rate of change. They do not need to solve the equation analytically, because nobody solves it analytically. They need to do three things, and all three are within reach at sixteen.
First, write the equation down from the physics. Identify the forces, apply Newton's second law, and be careful about signs. This is harder than it sounds and it is where most of the learning is.
Second, step it forward by hand for the first few intervals, on paper, with a calculator. Ten steps is enough to feel what the method is doing and to see the answer start to bend.
Third, write ten lines of code that does the same thing ten thousand times, and plot the result. The figure above came from about fifteen lines. A student who does this has produced a graph that nobody handed them, from an equation they wrote themselves, describing a real spacecraft. In my experience that changes what a student thinks mathematics is for, in a way that no amount of exam practice does.
It is also honest about what the subject is like. Practising engineers spend very little time finding exact solutions and a great deal of time solving equations approximately and then arguing about whether the approximation is good enough. A student who meets that at sixteen is not being given a simplified version of the discipline. They are being given the real one.
Because Mars sits in a cruel middle ground. The Moon has no atmosphere, so you brake entirely with engines and the problem is just propellant. The Earth has a thick atmosphere that does nearly all the braking, with a parachute to finish. Mars has enough atmosphere to demand a heat shield and cause severe heating, but not enough to slow a heavy vehicle to a safe landing speed. Every Mars landing therefore needs a heat shield, a supersonic parachute and rocket engines together.
The acceleration of the vehicle depends on the drag, the drag depends on the air density and the speed, the density depends on the altitude, and the altitude and speed are exactly the quantities you are trying to find. The unknown appears inside its own derivatives on both sides of the equation, which is the defining feature of a differential equation.
Two effects compete. Air density rises exponentially as the vehicle descends, which increases drag. Speed falls as the vehicle brakes, and drag depends on speed squared, which decreases drag. High up there is speed but no air; low down there is air but no speed. The product is largest somewhere in between, and for a typical entry that is around twenty kilometres up.
It is the mass of the vehicle divided by its drag coefficient times its frontal area. It is the only combination of those three quantities that appears in the entry equations, so it alone decides how deep into the atmosphere a vehicle penetrates before it slows. A low value means light and blunt, slowing high up. A high value means heavy and compact, arriving low and fast.
Differentiation, the idea of a rate of change, and a willingness to use a calculator or a few lines of code. The equation is written down from Newton's second law, which is physics rather than advanced mathematics, and it is solved by repeated arithmetic rather than by any clever technique.
Yes, numerically, which is how professionals solve it too. Stepping the equation forward by hand for ten intervals shows what the method does, and about fifteen lines of code will reproduce the graph in this article. No analytic solution is required, because in practice no analytic solution is used.
The practice portal carries the algebra, logarithms, calculus and vector work this article uses, with worked solutions that explain the reasoning rather than just the answer.
Open the practice portal →