Point Break: Mars!

My mind works in rather silly ways sometimes. Posting about skydiving soon after posting about avalanches on Mars, I got to thinking about how cool it would be to skydive off of a Martian cliff. Then I started to wonder: could you? What would be a skydiver’s terminal velocity in freefall on Mars? Well, with a little prior knowledge and some physics, we can actually make a rough estimate of this!

The maximum, or ‘terminal’, velocity a skydiver experiences in freefall is determined by two opposing forces: the force of gravity pulling the skydiver towards the ground and the wind resistance experienced by the skydiver moving through the air pushing in the opposite direction:

Terminal velocity is achieved when the strength of the two forces cancel out, i.e. when $F_g=F_w$ . A common misconception people have about skydiving is the idea that in freefall you’ll feel ‘weightless’, or feel like your stomach is dropping out. In fact, weightlessness only occurs when one is freely accelerating towards the ground at the acceleration of gravity. At terminal velocity, you are moving at constant speed and experience your full weight.

The force of gravity is given by $F_g=mg$ , where m is the mass of the skydiver and g is the acceleration of gravity. On Earth, $g=9.8 ,\mbox{m}/\mbox{s}^2$ , while on Mars $g=3.8 \,\mbox{m}/\mbox{s}^2$ . The acceleration of gravity is smaller on Mars, and consequently the gravitational force and terminal velocity must be smaller, right? Well, no; the atmospheric density on Mars is also much smaller than on Earth, meaning that there is less wind resistance on Mars. Numbers culled from the internet (I’m hoping they’re accurate) are $\rho = 0.020 \,\mbox{kg}/\mbox{m}^3$ for Mars, $\rho = 1.23\, \mbox{kg}/\mbox{m}^3$ on Earth, a significant difference.

So how do we determine the terminal velocity on Mars? We need a more-or-less universal mathematical formula for the force of air resistance, which is in fact provided by the so-called Prandtl expression:

$F_w = C_w \frac{\rho}{2}v^2 A$ . (1)

In this formula, ρ is the density of the air, v is the velocity of the skydiver, A is the surface area of the skydiver, and $C_w$ is a dimensionless drag coefficient, which is a measure of the ‘aerodynamic drag’ produced by the object, and is determined by the specific shape and roughness. The drag coefficient is essentially a constant independent of all other parameters, except that it changes drastically when the object speed passes the speed of sound.

The terminal velocity $v_t$ of our skydiver is the velocity at which the air resistance balances the force of gravity, or:

$mg= C_w \frac{\rho}{2}v_t^2 A$ . (2)

We can solve this equation algebraically for the terminal velocity, and find

$v_t=\sqrt{\frac{2mg}{C_wA \rho}}$ . (3)

Now we’ve got a problem! We don’t know the drag coefficient or the area of the skydiver, so it seems like this formula won’t do us any good. However, we do know the terminal velocity of a skydiver on Earth, so we can solve our formula for $C_wA$ using these known properties:

$C_wA = \frac{2mg}{\rho v_t^2}$ . (4)

Let’s use me as an example. With all my skydiving gear, I weigh about 170 lbs (77 kg). An average terminal velocity for me is about 120 mph (53.6 m/s). Using $\rho = 1.23\, \mbox{kg}/\mbox{m}^3$ and $g=9.8 \,\mbox{m}/\mbox{s}^2$ , we find that

$C_wA = .427 \,\mbox{m}^2$ . (5)

This number in principle should be nearly the same for Earth and Mars, since it is independent of the density of the atmosphere and the acceleration of gravity. Using all of our Mars numbers in Eq. (3) above, we find the result:

$v_t= 261\, \mbox{m}/\mbox{s} = 585\, \mbox{mph}$ !

This is pretty damn fast. It is still below the speed of sound (which is roughly 344 m/s in an ideal gas, density independent), so our assumption about the constancy of the drag coefficient seems a good one. Skydiving would seem to be a very bad idea on Mars. Even opening a 150 square foot parachute, which would increase body surface area roughly by a factor of 20, would only decrease the velocity to 131 mph.

Now that I’ve finished this post, a quick Google search shows I’m treading ground that has been tread before! But this gives us a good set of answers to compare our results to. A page explicitly looking at Martian skydiving gives a terminal velocity of 597 mph, in rough agreement with our own. A nice NASA page allows one to estimate terminal velocities on Earth and on Mars, and their default parameters give a value of 285 m/s, also in rough agreement with our own estimate.

Incidentally, such freefall speeds are possible on Earth – as long as you skydive from high enough! In 1960, Joseph Kittinger, an Air Force pilot, made a jump from a balloon at 102,800 feet! This was done as part of the military’s research into high-altitude aircraft bailouts. In the thin upper atmosphere, he reached a maximum velocity of 614 mph over the course of a 4 minute and 36 second freefall. Kittinger is a crazy mo-fo, and I mean that in a good way!

There are two lessons to be learned in this exercise: 1. We can use some relatively basic physics concepts and some real-world numbers to estimate cool effects on other planets, and 2. Skydiving on Mars is a really bad idea!

6 Responses to Point Break: Mars!

Personal Demon says:

March 7, 2008 at 5:46 pm

You make a terrific point about not feeling “weightless” once you reach terminal velocity. That had never occurred to me before, although it makes perfect sense. It’s a classic example of how “common sense” frequently leads to erroneous conclusions. Specifically, it is common knowledge that when you fall you feel weightless. It’s natural to think of skydiving as a really long fall, and therefore assume that you feel weightless the whole time.

You should teach physics or something 😉

skullsinthestars says:

March 11, 2008 at 11:21 am

PD: I suspect that fear of that “weightless” feeling scares more people away from skydiving than the fear of death itself! The only time I’ve felt truly weightless in freefall was jumping out of a hot air balloon. There, you genuinely start accelerating from rest, and the feeling is quite unsettling (and cool). My friends mocked me after the jump for shouting “Holy crap!” as I left the balloon.

How annoying… my post lost all of the backslashes in the LaTeX commands for some reason over the past few days. I’ve put them back in now, but my apologies to anyone who found the equations completely incomprehensible. WordPress does weird things sometimes…

Magnulus says:

May 1, 2011 at 3:07 pm

Holy crap, sometimes I just absolutely LOVE the Internet and the way it connects people with silly questions to people who’ve already asked and answered said silly questions.

I didn’t understand most of the math, but you certainly confirmed my suspicions about the effect of the thin atmosphere on the speed. But nearly 600?! Man, that’s quite some speed…

- skullsinthestars says:
  
  May 2, 2011 at 4:03 pm
  
  It kinda blew my mind when I first did the calculation myself! It explains why a number of elaborate means such as retrothrusters and big balloon cushions need to be used to help slow down Mars landers.
  
Bud Bundy says:

June 9, 2015 at 8:46 am

Nice article. I’m not sure though why you would use a tiny parachute such as your 150 square foot example, I see tandem chutes over 500 sq ft. in use. In thinner Martian air, you should be able to use one much bigger with the same materials, no?

Bud Bundy says:

June 9, 2015 at 8:50 am

It also occurred to me that the terminal velocity concept is that of balanced forces, gravity vs. air pressure. Does this not mean that much smaller Martian gravity is balanced by much smaller air pressure? In other words, opening your huge parachute at 600 mph on Mars should not put any more stress on it than at 120 mph on Earth, as you are counter-acting the much lighter force of the thinner air…